MATH 334 - DIFFERENTIAL
EQUATIONS - Matrix methods and
eigenvalue problems
Question Bank - Set 5
Liberty University
Question 1
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. If the
eigenvectors corresponding to λ1and λ2are v1=
1
1
0
and v2=
1
0
1
, find the
eigenvector corresponding to λ3.
Solution
Step 1: The eigenvector corresponding to λ3can be found by solving the equa-
tion (A−λI)v=0, where λ=λ3=−3 and vis the eigenvector we are looking
for.
Step 2: Substitute λ=−3 into the equation:
(A−(−3)I)v=0
(A+ 3I)v=0
Step 3: Given that Ais a 3 ×3 matrix, we can write Aand Iexplicitly:
A=
a b c
d e f
g h i
I=
100
010
001
Step 4: Substitute Aand Iinto the equation:
a+ 3 b c
d e + 3 f
g h i + 3
x
y
z
=
0
0
0
Step 5: Solve the system of equations to find the eigenvector corresponding
to λ3=−3:
(a+ 3)x+by +cz = 0
dx + (e+ 3)y+fz = 0
gx +hy + (i+ 3)z= 0
Step 6: The solution to the system of equations will give us the eigenvector
corresponding to λ3.
Question 2
Question
Let Abe a 3×3 matrix with eigenvalues −2,1,and 4. Given that the eigenvector
corresponding to the eigenvalue −2 is
1
−1
2
, find the eigenvectors correspond-
ing to the eigenvalues 1 and 4.
Solution
Step 1: Recall that for an eigenvalue λand its corresponding eigenvector v, the
relationship Av=λvholds.
Step 2: Let’s find the eigenvector corresponding to the eigenvalue 1. We
have:
Av=λv
Av= 1v
Av=
−200
0 1 0
0 0 4
x
y
z
=
x
y
z
Step 3: Solving the system of equations, we get:
−2x=x, y =y, 4z=z
Step 4: Simplifying the equations, we see that x= 0, ycan be any value,
and zcan be any value. Thus, the eigenvector corresponding to the eigenvalue
1 is any vector of the form
0
1
0
.
2
Step 5: Now, let’s find the eigenvector corresponding to the eigenvalue 4.
We have:
Av= 4v
Av=
−200
0 1 0
0 0 4
x
y
z
= 4
x
y
z
Step 6: Solving the system of equations, we get:
−2x= 4x, y = 4y, 4z= 4z
Step 7: Simplifying the equations, we see that x= 0, y= 0, and zcan be
any value. Therefore, the eigenvector corresponding to the eigenvalue 4 is any
vector of the form
0
0
1
.
Question 3
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
2
1
and v2=
1
1
0
,
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that each eigenvalue corresponds to a linearly independent set
of eigenvectors forming an eigenspace. To find the basis for the eigenspace
corresponding to λ3= 4, we need to find the eigenvectors corresponding to λ3.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to λ3= 4. Since
Av3=λ3v3, we have:
Av3= 4v3
Step 3: Substituting v3into the above equation, we get:
A
x
y
z
= 4
x
y
z
Step 4: This can be written as a system of equations:
Ax = 4x
Ay = 4y
Az = 4z
3
Step 5: Solving the system of equations, we obtain the equation Ax = 4x
where x,y, and zare variables. After solving, we will have the eigenvector
corresponding to λ3.
Step 6: Now, let’s solve the system of equations Ax = 4x:
A
x
y
z
=
4x
4y
4z
a b c
d e f
g h i
x
y
z
=
4x
4y
4z
Step 7: Solve the system of equations to find the eigenvector v3=
x
y
z
.
Question 4
Question
Let Abe a 3 ×3 matrix such that the characteristic equation of Ais given by
λ3−6λ2+ 11λ−6 = 0. Find the eigenvalues of Aand for each eigenvalue, find
a basis for the corresponding eigenspace.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation λ3−6λ2+ 11λ−6 = 0. Step 2: Factorize the characteristic equation
to obtain: (λ−1)(λ−2)(λ−3) = 0. Step 3: From the factorization, we see that
the eigenvalues are λ1= 1, λ2= 2, and λ3= 3. Step 4: To find the eigenvectors
corresponding to each eigenvalue, we solve the system (A−λiI)v= 0, where v
is the eigenvector corresponding to eigenvalue λi. Step 5: For λ1= 1, we solve
the system (A−I)v1= 0. This gives the equation:
000
011
002
x
y
z
=
0
0
0
Step 6: Solving this system gives the eigenvector v1=
0
0
1
. Step 7: For λ2= 2,
we solve the system (A−2I)v2= 0. This gives the equation:
−100
0 0 1
0 0 1
x
y
z
=
0
0
0
4
Step 8: Solving this system gives the eigenvector v2=
0
1
0
. Step 9: For λ3= 3,
we solve the system (A−3I)v3= 0. This gives the equation:
−200
0−1 1
0 0 0
x
y
z
=
0
0
0
Step 10: Solving this system gives the eigenvector v3=
0
1
1
. Step 11: There-
fore, the eigenvalues of matrix Aare 1, 2, and 3, and the corresponding eigen-
vectors are
0
0
1
,
0
1
0
, and
0
1
1
respectively.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues 6,2,and −3. Find the eigenvalues of
the matrix A2−4A+ 3I, where Iis the 3 ×3 identity matrix.
Solution
Step 1: First, find the eigenvalues of A2−4A+ 3Iby substituting Awith its
eigenvalues.
Step 2: The eigenvalues of A2−4A+ 3Iare given by the roots of the
characteristic equation det(A2−4A+ 3I−λI) = 0.
Step 3: Substitute Awith its eigenvalues 6,2,and −3:
det
620 0
0 220
0 0 (−3)2
−4
6 0 0
0 2 0
0 0 −3
+ 3
100
010
001
−λ
100
010
001
= 0
Step 4: Simplify and solve for λto find the eigenvalues of A2−4A+ 3I.
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Given
that the eigenvector corresponding to λ1is (1,0,1), find the eigenvectors corre-
sponding to λ2and λ3.
5
Solution
Step 1: To find the eigenvectors corresponding to λ2= 2 and λ3= 3, we will
solve the system of equations (A−λI)v=0for each eigenvalue.
For λ2= 2:
A−2I=
1 0 1
0 0 1
−101
We solve the system (A−2I)v=0:
1 0 1
0 0 1
−101
x
y
z
=
0
0
0
This leads to the system of equations:
x+z= 0
z= 0
Solving this system yields x= 0, y= 1, z= 0. Therefore, the eigenvector
corresponding to λ2= 2 is v2= (0,1,0).
Step 2: For λ3= 3:
A−3I=
−201
0−1 1
−100
We solve the system (A−3I)v=0:
−201
0−1 1
−100
x
y
z
=
0
0
0
This leads to the system of equations:
−2x+z= 0
−y+z= 0
−x= 0
Solving this system yields x= 0, y= 1, z= 0. Therefore, the eigenvector
corresponding to λ3= 3 is v3= (0,1,0).
Question 7
Question
Let Abe a 3 ×3 matrix given by
A=
310
020
012
.
6
Determine the eigenvalues and corresponding eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
given by det(A−λI) = 0, where λis the eigenvalue and Iis the identity matrix.
A−λI =
3−λ1 0
0 2 −λ0
0 1 2 −λ
.
Setting the determinant of this matrix to zero, we get
det
3−λ1 0
0 2 −λ0
0 1 2 −λ
= (3 −λ)((2 −λ)2)−0−0=0.
Step 2: Simplifying the determinant equation, we get
(3 −λ)((2 −λ)2) = (3 −λ)(4 −4λ+λ2)=0.
This equation simplifies to λ3−7λ2+ 14λ−8 = 0.
Step 3: Solving the cubic equation using synthetic division or a numerical
method, we find the eigenvalues to be λ= 2,2,3. Notice that 2 appears twice
as a repeated eigenvalue.
Step 4: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute each eigenvalue back into the equation (A−λI)v=0and solve for
v.
For λ= 2, we have
A−2I=
110
000
010
which leads to the eigenvector v1=
−1
0
1
.
For the repeated eigenvalue λ= 2, we have another eigenvector v2=
−1
0
1
.
For λ= 3, we have
A−3I=
0 1 0
0−1 0
0 1 −1
which leads to the eigenvector v3=
1
0
−1
.
Therefore, the eigenvalues of Aare λ= 2,2,3 with corresponding eigenvec-
tors v1=
−1
0
1
,v2=
−1
0
1
, and v3=
1
0
−1
.
7
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
B=A2−2A+I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
matrix B.
Solution
To find the eigenvalues of matrix B, we need to first find the matrix Band then
determine its eigenvalues.
Step 1: Find matrix B=A2−2A+I:Given that Ais a 3×3 matrix with
eigenvalues λ1= 1, λ2= 2, and λ3= 3, wecanexpressAas :A=P DP −1,where
Pis the matrix of eigenvectors and Dis the diagonal matrix of eigenvalues.
Thus, Acan be written as:
A=P DP −1=v1v2v3
100
020
003
v1v2v3−1,
where
Next, we can calculate A2−2A+Ito find matrix B:
B=A2−2A+I.
Step 2: Calculate B:Since we have A=P DP −1, we have:
A2= (P DP −1)(P DP −1) = P D(P−1P)DP −1=P D2P−1.
Substitute this into B=A2−2A+I:
B=P D2P−1−2P DP −1+I.
Step 3: Evaluate B:Substitute the expressions for A=P DP −1and D
into the equation for Bto obtain the matrix B.
Step 4: Find the eigenvalues of B:Once you have the matrix B, find
its eigenvalues µ1,µ2, and µ3.
The resulting eigenvalues µ1,µ2, and µ3are the eigenvalues of matrix B.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If the
eigenvectors corresponding to λ1and λ2are
v1=
1
2
3
and v2=
4
5
6
,
find an eigenvector v3corresponding to the eigenvalue λ3= 3.
8
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3, we
can represent Ain terms of its eigenvectors and eigenvalues as
A=P DP −1,
where
P= [v1v2v3] and D=
λ10 0
0λ20
0 0 λ3
.
Since v1and v2are the eigenvectors corresponding to λ1= 1 and λ2= 2,
respectively, we have
P=
1 4 a
2 5 b
3 6 c
.
Now, to find v3, we will equate Dto P−1AP and solve for v3.
P−1AP =D
1 4 a
2 5 b
3 6 c
−1
p11 p12 p13
p21 p22 p23
p31 p32 p33
1 4 a
2 5 b
3 6 c
=
100
020
003
.
Now, solve for v3using P−1AP =D. Once v3is found, normalize v3to
obtain an eigenvector corresponding to λ3= 3.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. If the
eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
1
, find
the eigenvectors corresponding to λ3.
Solution
Step 1: The eigenvectors v3corresponding to λ3can be found by solving the
equation (A−λI)v=0, where λ= 4. This gives us:
(A−4I)v=
−200
0 1 0
0 0 0
x
y
z
=
0
0
0
Step 2: Solving the system of equations, we get:
−2x= 0, y = 0, z = 0
9
Step 3: So, the eigenvector v3corresponding to λ3= 4 is
0
0
1
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 5, and λ3= 8. If
the eigenvectors corresponding to λ1and λ3are v1=
1
0
1
and v3=
0
1
2
respectively, find the eigenvector corresponding to λ2.
Solution
Step 1: Begin by setting up the equation to find the eigenvector v2corresponding
to λ2:
Av2=λ2v2
Step 2: Knowing the eigenvalues and eigenvectors, we can write Ain terms
of its diagonalized form:
A=P DP −1
where Dis a diagonal matrix with the eigenvalues on the diagonal, and the
columns of Pare the corresponding eigenvectors.
Step 3: From the given information, we have:
D=
200
050
008
and
P=
100
011
120
Step 4: Now we can write the equation Av2=λ2v2as:
P DP −1v2=λ2v2
Step 5: Premultiplying by P−1, we get:
DP −1v2=λ2P−1v2
Step 6: Substituting the values of Dand Pinto the equation, we find:
200
050
008
100
011
120
v2= 5
100
011
120
v2
10
Step 7: Multiplying out the matrix product on the left side gives:
200
050
008
1
1
2
= 5
1
1
2
Step 8: Solving the equation above will give us the eigenvector v2corre-
sponding to the eigenvalue λ2.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 4. Given
that det(A) = 10, find the eigenvalues of 3A−2A−1.
Solution
Step 1: First, let’s find the eigenvalues of A−1. Since Ahas eigenvalues λ1=
2, λ2=−1,and λ3= 4, the eigenvalues of A−1are 1
λ1,1
λ2,and 1
λ3. Therefore,
the eigenvalues of A−1are 1
2,−1,and 1
4.
Step 2: Next, let’s find the eigenvalues of 3A−2A−1. The eigenvalues of
3A−2A−1are given by 3λi−2µi, where λiare the eigenvalues of Aand µiare
the eigenvalues of A−1. Substituting the given values, we have:
3λ1−2µ1= 3(2) −21
2= 6 −1=5,
3λ2−2µ2= 3(−1) −2(−1) = −3 + 2 = −1,
3λ3−2µ3= 3(4) −21
4= 12 −1
2=23
2.
Therefore, the eigenvalues of 3A−2A−1are 5, -1, and 23
2.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. If
the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
1
,
respectively, find a matrix Psuch that P−1AP is a diagonal matrix.
11
Solution
Step 1: Recall that if λis an eigenvalue of matrix Awith corresponding eigen-
vector v, then the matrix Pformed by stacking the eigenvectors as columns
diagonalizes A. That is, P−1AP =D, where Dis a diagonal matrix whose
diagonal entries are the eigenvalues.
Let P=v1v2v3=
1 0 a
0 1 b
1 1 c
Step 2: We know that Av1=λ1v1and Av2=λ2v2. So let’s calculate Av1
and Av2.
Av1=A
1
0
1
=
2−1 0
0 1 −1
2−1 1
1
0
1
=
2
0
2
= 2
1
0
1
= 2v1
Step 3: Now, let’s calculate Av2.
Av2=A
0
1
1
=
2−1 0
0 1 −1
2−1 1
0
1
1
=
−1
1
0
= 1
0
1
1
=v2
Step 4: Since Av1= 2v1and Av2=v2, we can see that the eigenvectors v1
and v2are already corresponding to the correct eigenvalues.
Step 5: We can choose v3arbitrarily for the remaining eigenvalue λ3=−3.
Let’s choose v3=
1
−2
1
.
Step 6: Now, form the matrix P=v1v2v3.
P=
1 0 1
0 1 −2
1 1 1
Step 7: Calculate P−1using the inverse formula P−1=1
det(P)adj(P).
P−1=1
−23−1 1Question14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. If v1,
v2, and v3are the corresponding eigenvectors, find the matrix A.
12
Solution
Step 1: Recall the definition of an eigenvector and eigenvalue. For a matrix A
and a scalar λ, the eigenvector vsatisfies the equation Av=λv.
Step 2: Since v1,v2, and v3are eigenvectors of Acorresponding to the
eigenvalues λ1= 2, λ2=−1, and λ3= 3, respectively, we have the following
three equations: Av1= 2v1,Av2=−1v2,Av3= 3v3.
Step 3: Therefore, we can write the matrix Ain terms of its eigenvectors
and eigenvalues as: A=v1v2v3
λ10 0
0λ20
0 0 λ3
vT
1
vT
2
vT
3
.
Step 4: Substituting the given values for the eigenvalues and eigenvectors
into the equation above, we get: A=v1v2v3
200
0−1 0
003
vT
1
vT
2
vT
3
.
Step 5: Therefore, the matrix Acan be written as: A=2v1−v23v3.
Question 15
Question
Let Abe a 3×3 matrix with eigenvalues −2, 1, and 4. Given that the eigenvector
corresponding to the eigenvalue −2 is
1
2
1
, find:
A2
1
2
1
Solution
To find A2
1
2
1
, we first need to find the matrix Ausing its eigenvalues and
eigenvectors. Let v1=
1
2
1
be the eigenvector corresponding to the eigenvalue
−2. Since Ais a 3 ×3 matrix, the set {v1}forms a basis for R3. Additionally,
we are given that the eigenvalues are −2, 1, and 4, so the characteristic equation
for matrix Awould be:
det(A−λI) = 0
det
a b c
d e f
g h i
−λ
100
010
001
= 0
13
det
a−λ b c
d e −λ f
g h i −λ
= 0
Plugging in the given eigenvalues, we get:
det
a+ 2 b c
d e −1f
g h i −4
= 0
Expanding this determinant, we can find the matrix A. Next, we calculate A2:
A2=A·A
A2= Matrix multiplication of Aitself
Finally, we find A2
1
2
1
by multiplying A2with the given eigenvector
1
2
1
.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2=−2, and λ3= 5.
Given that the eigenvectors corresponding to λ1and λ2are
1
2
−1
and
−1
1
2
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Since Ais a 3 ×3 matrix, it will have 3 eigenvalues and 3 corresponding
eigenvectors.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to the eigenvalue
λ3= 5.
Step 3: The eigenvectors of a matrix corresponding to distinct eigenvalues
are linearly independent. Since
1
2
−1
and
−1
1
2
are independent,
1
2
−1
and
−1
1
2
are independent to
x
y
z
.
Step 4: Since v3is an eigenvector corresponding to λ3= 5, we have Av3=
λ3v3.
Step 5: We can set up the following equation based on Step 4:
14
A
x
y
z
= 5
x
y
z
Step 6: Multiplying the matrices on the left side of the equation, we have:
3−1∗
∗ −2 1
∗ ∗ 5
x
y
z
=
5x
5y
5z
Step 7: Solving the resulting system of equations, we find that v3=
−2
1
1
is the eigenvector corresponding to the eigenvalue λ3= 5.
Question 17
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Given that one eigenvec-
tor corresponding to the eigenvalue −2 is
1
0
1
, find a basis for the eigenspace
corresponding to the eigenvalue 1.
Solution
Step 1: To find a basis for the eigenspace corresponding to the eigenvalue 1, we
need to solve the system (A−λI)v=0, where λis the eigenvalue (1 in this
case) and vis the eigenvector we want to find.
Step 2: Start by writing the matrix Aand the identity matrix I:
A=
a b c
d e f
g h i
and I=
100
010
001
Step 3: Substitute the eigenvalue λ= 1 into the equation (A−λI)v=0:
(A−I)v=
a−1b c
d e −1f
g h i −1
x
y
z
=
0
0
0
Step 4: Writing the system of equations, we get:
(a−1)x+by +cz = 0
dx + (e−1)y+fz = 0
gx +hy + (i−1)z= 0
Step 5: Since we are looking for a basis, we will not be able to solve for the
specific values of x, y, and z. Instead, we can set z= 1 and solve for xand y.
15
Step 6: Substituting z= 1 into the system of equations gives:
(a−1)x+by +c= 0
dx + (e−1)y+f= 0
gx +hy + (i−1) = 0
Step 7: To simplify, we can write the system of equations in matrix form:
a−1b
d e −1
g h
x
y=
−c
−f
−(i−1)
Step 8: This system of equations may have multiple solutions, thus to find a
basis, we can solve the system using row reduction to find the parametric form
of the solutions.
Step 9: The solutions to the system will give us a basis for the eigenspace
corresponding to the eigenvalue 1.
Question 18
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 0, and λ3=−1.
Given that the eigenvector corresponding to λ1= 2 is
1
1
2
and the eigenvector
corresponding to λ2= 0 is
1
1
1
, find the eigenvector corresponding to λ3=−1.
Solution
Given a matrix Awith eigenvalues λ1, λ2, λ3and corresponding eigenvectors
v1,v2,v3, we can find the eigenvector corresponding to λ3by solving the system
of equations Ax=λ3x.
Step 1: We have the following information:
A=
|||
v1v2v3
|||
, λ3=−1
Step 2: We can set up our system of equations as follows:
Ax=λ3x⇒Ax=−x
Step 3: Expressing Aand xin terms of column vectors:
|||
v1v2v3
|||
x1
x2
x3
=−
x1
x2
x3
16
This gives us a system of equations:
v1x1+v2x2+v3x3=−x1
v1x1+v2x2+v3x3=−x2
v1x1+v2x2+v3x3=−x3
Step 4: We can substitute the values of the eigenvectors v1and v2:
111
111
210
x1
x2
x3
=
−x1
−x2
−x3
Step 5: Solving the system of equations, we get the eigenvector correspond-
ing to λ3=−1:
x=
1
1
−2
Therefore, the eigenvector corresponding to λ3=−1 is
1
1
−2
.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,1,−1 and corresponding eigen-
vectors v1=
1
2
0
,v2=
0
1
−1
, and v3=
1
0
1
. Define the matrix Pwhose
columns are these eigenvectors.
(a) Calculate P−1.
(b) Determine A5.
Solution
(a) To find P−1, we first construct the matrix Pusing the given eigenvectors:
P= [v1v2v3] =
101
210
0−1 1
Next, we find the inverse of Pusing the formula P−1=1
det(P)adj(P), where
adj(P) is the adjugate of P. To do this, we calculate the determinant of P:
det(P) =
101
210
0−1 1
= 1(1 ·1−(−1) ·0) −0−1·1=2
17
Then, we find the cofactor matrix of P:
Cof(P) =
1−1 1
0 1 0
−221
Next, we calculate the adjugate of P:
adj(P) = Cof(P)T=
1 0 −2
−1 1 2
101
Finally, we compute P−1:
P−1=1
det(P)adj(P) = 1
2
1 0 −2
−1 1 2
101
=
1
20−1
−1
2
1
21
1
201
2
(b) To determine A5, we use the formula A=PΛP−1, where Λ is the
diagonal matrix of eigenvalues. We form Λ using the given eigenvalues:
Λ =
2 0 0
0 1 0
0 0 −1
Substitute the values of Pand P−1into the formula A=PΛP−1:
A=
101
210
0−1 1
2 0 0
0 1 0
0 0 −1
1
20−1
−1
2
1
21
1
201
2
Multiplying these matrices will give us A, and to find A5, we raise Ato the
power of 5.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2= 2, and λ3=−1. If the
eigenvectors corresponding to λ2= 2 are v2=
1
0
1
and w2=
−1
1
0
, find a
basis for the eigenspace corresponding to λ1= 3.
Solution
Step 1: Recall that the eigenspace Eλassociated with an eigenvalue λis the set
of all eigenvectors corresponding to λtogether with the zero vector.
18
Step 2: Since v2and w2are eigenvectors corresponding to λ2= 2, the
eigenspace E2is given by:
E2= span
1
0
1
,
−1
1
0
Step 3: To find the eigenvectors corresponding to λ1= 3, we can use the
fact that the sum of each diagonal element in a matrix is equal to the sum of
its eigenvalues.
Step 4: Let v1=
x
y
z
be an eigenvector corresponding to λ1= 3. Then
Av1= 3v1.
Step 5: This gives us the system of equations:
3x= 3x
3y= 3y
3z= 3z
Step 6: Since x,y, and zcan be any real number, the eigenvectors corre-
sponding to λ1= 3 are vectors of the form
1
0
0
,
0
1
0
, and
0
0
1
.
Step 7: Therefore, a basis for the eigenspace corresponding to λ1= 3 is
given by:
B3=
1
0
0
,
0
1
0
,
0
0
1
Question 21
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3= 3. Suppose
the corresponding eigenvectors are
v1=
1
1
1
,v2=
1
0
1
,and v3=
0
1
1
.
Determine the matrix A.
Solution
Step 1: The matrix Ais similar to a diagonal matrix Dwhose diagonal en-
tries are the eigenvalues of A. Let Pbe the matrix formed by stacking the
eigenvectors v1,v2, and v3as columns.
19
P=
110
101
111
Step 2: The matrix Dis given by
D=P−1AP
where Ais the matrix we are trying to find.
Step 3: To find P−1, we first compute the determinant of P.
det(P) = 1(0 −1) −1(1 −1) + 0(1 −1) = −1
Step 4: Next, we find the adjugate of P, denoted as adj(P), by taking the
transpose of the matrix of cofactors of P.
adj(P) =
0−1 1
1−1 0
−1 2 −1
Step 5: The inverse of Pis given by
P−1=1
det(P)·adj(P) =
0 1 −1
−1 1 0
1−2 1
Step 6: We can now find the matrix Aby using the formula D=P−1AP ,
which gives
A=P DP −1
A=
110
101
111
200
010
003
0 1 −1
−1 1 0
1−2 1
A=
312
221
323
Therefore, the matrix Ais
312
221
323
.
20
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. Given
that B=A2−5A+6I, where Iis the 3 ×3 identity matrix, find the eigenvalues
of B.
Solution
Step 1: Find the eigenvalues of B. Given B=A2−5A+ 6I, we can rewrite B
as:
B=A2−5A+ 6I=A2−5A+ 6
100
010
001
Step 2: First, find the eigenvalues of A2. Since Ahas eigenvalues λ1= 2,
λ2= 3, and λ3= 4, the eigenvalues of A2are λ2
1= 4, λ2
2= 9, and λ2
3= 16.
Step 3: Find Bin terms of A. Now, we can express Bin terms of the
eigenvalues of A:
B=A2−5A+ 6I=
4 0 0
0 9 0
0 0 16
−5
200
030
004
+ 6
100
010
001
=
−600
0−3 0
0 0 6
+
12 0 0
0 15 0
0 0 20
=
6 0 0
0 12 0
0 0 26
Step 4: Find the eigenvalues of B. The eigenvalues of Bare the diagonal
elements of the matrix we found in Step 3. Therefore, the eigenvalues of Bare
λ′
1= 6, λ′
2= 12, and λ′
3= 26.
Question 23
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 0. If
B=A2−3A+ 2I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
21
Solution
Step 1: First, let’s find the eigenvalues of matrix B.
B=A2−3A+ 2I
=A2−3A+ 2I
=A2−3A+ 2
100
010
001
=A2−3A+ 2
100
010
001
=A2−3A+
200
020
002
Step 2: Now, let’s find the eigenvalues of B. The eigenvalues of Bare the
solutions to the characteristic equation |B−λI|= 0, where λrepresents the
eigenvalue.
|B−λI|=
A2−3A+
200
020
002
−λ
100
010
001
=|A2−3A+
2−λ0 0
0 2 −λ0
0 0 2 −λ
|
=|A−2I||A+I||A|
= (2 −λ)(−1−λ)λ
Therefore, the eigenvalues of matrix Bare λ= 2,−1,0.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 5.
Find a matrix Psuch that A=PDP−1, where Dis the diagonal matrix with
eigenvalues of Aon the diagonal arranged in the same order as in P.
Solution
Step 1: Start by constructing the matrix Dusing the eigenvalues of A.
D=
200
0−1 0
005
22
Step 2: Next, construct the matrix Pusing the eigenvectors corresponding
to each eigenvalue. Let
P=v1v2v3
where v1,v2,and v3are the eigenvectors corresponding to the eigenvalues 2,−1,
and 5 respectively.
Step 3: Since λ1= 2, solve (A−2I)v1=0to find v1. Let
v1=
a
b
c
Solving the equation (A−2I)v1=0, we get
0−3 1
−1 0 0
0 1 −2
a
b
c
=
0
0
0
Solving this system of equations gives the eigenvector v1=
1
1
1
after normal-
ization.
Step 4: Repeat the same process for the remaining eigenvalues to find the
eigenvectors v2and v3. After finding all the eigenvectors, construct the matrix
Pas:
P=
1??
1??
1??
Step 5: Finally, calculate P−1and verify that A=PDP−1.
Question 25
Question
Let A=2 1
1 2be a matrix. Find the eigenvectors corresponding to the
eigenvalues of A.
Solution
Step 1: Find the eigenvalues by solving the characteristic equation det(A−λI) =
0.
det 2−λ1
1 2 −λ= (2 −λ)(2 −λ)−1·1
=λ2−4λ+ 3
= (λ−3)(λ−1) = 0.
23
Therefore, the eigenvalues are λ1= 1 and λ2= 3.
Step 2: Find the eigenvector corresponding to λ1= 1 by solving the system
(A−λI)v= 0. For λ1= 1:
2−1 1
1 2 −1x
y=0
0.
This gives the system of equations:
(x+y= 0
y= 0 .
Solving this system, we find that x= 0 and y= 0. Therefore, the eigenvector
corresponding to λ1= 1 is v1=0
0.
Step 3: Find the eigenvector corresponding to λ2= 3 by solving the system
(A−λI)v= 0. For λ2= 3:
2−3 1
1 2 −3x
y=0
0.
This gives the system of equations:
(−x+y= 0
y= 0 .
Solving this system, we find that x= 0 and y= 0. Therefore, the eigenvector
corresponding to λ2= 3 is v2=0
0.
Thus, the eigenvectors corresponding to the eigenvalues of matrix Aare 0
0
for both eigenvalues 1 and 3.
Question 26
Question
Let Abe a 3 ×3 matrix given by
A=
2 1 −1
0 2 0
1 0 2
.
Determine the eigenvalues and eigenvectors of A.
24
Solution
Step 1: Find the characteristic equation.
The characteristic equation for a matrix Ais given by det(A−λI) = 0,
where Iis the identity matrix of the same size as Aand λis the eigenvalue we
are trying to solve for. For our 3 ×3 matrix A, we have
A−λI =
2−λ1−1
0 2 −λ0
1 0 2 −λ
.
Step 2: Find the determinant and solve for λ.
Calculating the determinant, we get
det(A−λI) = (2 −λ)2(2 −λ)+1·1(2 −λ) = (2 −λ)((2 −λ)2+ 1).
Setting this equal to 0, we have
(2 −λ)((2 −λ)2+ 1) = 0.
Solving for λ, we find that the eigenvalues are λ= 2,2 + i, and 2 −i.
Step 3: Find the eigenvectors for each eigenvalue.
For λ= 2: Setting λ= 2 in (A−λI)v= 0, we get
0 1 −1
0 0 0
1 0 0
x
y
z
=
0
0
0
,
which gives us the equation x=zand y= 0. Therefore, the eigenvector
corresponding to the eigenvalue λ= 2 is
1
0
1
.
For λ= 2 + i: Setting λ= 2 + iin (A−λI)v= 0, we get
−i1−1
0−i0
1 0 −i
x
y
z
=
0
0
0
,
which gives us the equation −ix +y−z= 0, −iy = 0, and x=iz. From this,
we can find the eigenvector corresponding to the eigenvalue λ= 2 + i.
For λ= 2 −i: Similarly, we can find the eigenvector corresponding to the
eigenvalue λ= 2 −i.
Question 27
Question
Let A=4−1
2 1 . Find a matrix Psuch that P−1AP is diagonal, and calculate
P−1.
25
Solution
Step 1: Find the eigenvalues of matrix Aby solving the characteristic equation
|A−λI|= 0.
|A−λI|=
4−λ−1
2 1 −λ
= (4 −λ)(1 −λ)−(−1)(2)
=λ2−5λ+ 6
= (λ−2)(λ−3) = 0
So, the eigenvalues are λ1= 2 and λ2= 3.
Step 2: Find the eigenvectors corresponding to each eigenvalue by solving
(A−λI)x= 0. For λ= 2:
A−2I=2−1
2−1→2−1
0 0 →1−1
2
0 0
Letting x2=t, we have x1=1
2t. Thus, the eigenvector corresponding to λ= 2
is v1=1
2. Similarly, for λ= 3:
A−3I=1−1
2−2→1−1
0 0 →1−1
0 0
Again, letting x2=t, we have x1=t. Thus, the eigenvector corresponding to
λ= 3 is v2=1
1.
Step 3: Assemble the eigenvectors into the matrix P=v1v2. So,
P=1 1
2 1.
Step 4: Calculate P−1. To find the inverse of a 2 ×2 matrix, we use the
formula
P−1=1
ad −bc d−b
−c a .
Thus, P−1=1
11−1
−2 1 =1−1
−2 1 .
Therefore, the matrix Psuch that P−1AP is diagonal is
P=1 1
2 1
and
P−1=1−1
−2 1 .
26
Question 28
Question
Let Abe a 3 ×3 matrix with eigenvalues −1,2,and 3. Find the eigenvalues of
the matrix A2+ 3A−2I, where Iis the 3 ×3 identity matrix.
Solution
Given that Ahas eigenvalues −1,2, and 3, we can find the eigenvalues of A2+
3A−2Iby first finding the matrix A2+ 3A−2Iin terms of A.
Step 1: Compute A2Since Ahas eigenvalues −1,2, and 3, the characteristic
equation of Ais given by (A+ 1)(A−2)(A−3) = 0. Expanding this equation
gives A3−4A2−5A+ 6I= 0. Rearranging terms, we have
A2= 4A+ 5I−6I= 4A−I
Step 2: Find A2+ 3A−2INow we can substitute A2= 4A−Iinto the
expression A2+ 3A−2Ito get
A2+ 3A−2I= (4A−I)+3A−2I= 7A−3I
Step 3: Find the eigenvalues of A2+ 3A−2IThe eigenvalues of A2+ 3A−
2I= 7A−3Iare simply 7 times the eigenvalues of Aminus 3. Therefore,
the eigenvalues of A2+ 3A−2Iare 7 × −1−3 = −10, 7 ×2−3 = 11, and
7×3−3 = 18. Thus, the eigenvalues of the matrix A2+ 3A−2Iare −10,11,
and 18.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. Find the eigenvalues of
the matrix B=A3−2A2+A.
Solution
Given that Ahas eigenvalues 1, 2, and 3, we have:
det(A−λI) = 0
This gives us the characteristic equation for Aas:
det(A−λI) = (A−λI)(A−2I)(A−3I)=0
From this, we can express A3,A2, and Ain terms of λusing the eigenvalues
1, 2, and 3. Let B=A3−2A2+A:
27
B=A3−2A2+A
= (13−2·12+ 1)I+ (23−2·22+ 2)A+ (33−2·32+ 3)A2
= 0 ·I+ 2 ·A+ 6 ·A2
= 0 ·I+ 2 ·A+ 6 ·2A−4·I
= 12 ·A−4·I
= 4(3A−I)
Therefore, the eigenvalues of Bare 3 ×4−1 = 11, 2 ×4−1 = 7, and
1×4−1 = 3.
Question 30
Question
Let A=2 1
1 2be a 2 ×2 matrix. Determine all eigenvalues and eigenvectors
of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0 where Iis the identity matrix.
|A−λI|=
2−λ1
1 2 −λ
= (2 −λ)2−1=0
⇒(2 −λ)2−1=0
⇒λ2−4λ+ 3 = 0
⇒(λ−3)(λ−1) = 0
⇒λ1= 3, λ2= 1
Step 2: To find the eigenvectors corresponding to λ= 3, we solve (A−3I)v=
0.
A−3I=2−3 1
1 2 −3=−1 1
1−1
−1 1
1−1x
y=0
0
⇒ −x+y= 0 and x−y= 0
⇒No non-trivial solution exists
28
Step 3: To find the eigenvectors corresponding to λ= 1, we solve (A−I)v=
0.
A−I=2−1 1
1 2 −1=1 1
1 1
1 1
1 1x
y=0
0
⇒x+y= 0
⇒One eigenvector is 1
−1
Therefore, the eigenvalues of Aare λ= 3 and λ= 1, with corresponding
eigenvectors being 1
−1and no eigenvector for λ= 3.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
2−1 0
−1 2 −1
0−1 2
.
Find the eigenvectors and eigenvalues of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation
det(A−λI) = 0,
where Iis the identity matrix. Substitute Aand λinto the equation to get
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0.
Step 2: Expand the determinant to get
(2 −λ)((2 −λ)(2 −λ)−(−1)(−1)) + (−1)(−1) −0
= (2 −λ)((λ2−4λ+ 4) + 1) + 1
= (2 −λ)(λ2−4λ+ 5) + 1
= 2λ2−8λ+ 10 −λ3+ 4λ2−5λ+ 1.
=−λ3+ 6λ2−13λ+ 11.
29
Step 3: Set the characteristic equation equal to zero:
−λ3+ 6λ2−13λ+ 11 = 0.
This cubic equation is difficult to factorize so we can find the eigenvalues using
numerical methods such as Newton’s method. By computing the roots, we find
that the eigenvalues are λ1≈1.32, λ2≈1.0 and λ3≈2.68.
Step 4: To find the eigenvectors associated with each eigenvalue, we substi-
tute the eigenvalues back into the equation (A−λI)v=0and solve for v. We
find the eigenvectors v1,v2, and v3associated with the eigenvalues λ1,λ2, and
λ3, respectively.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, λ3= 4. If tr(A2) = 33
(where tr denotes the trace of a matrix), find the possible values of tr(A).
Solution
Step 1: Recall that the trace of a matrix is the sum of its eigenvalues. Therefore,
we have tr(A) = λ1+λ2+λ3.
Step 2: Substituting the given eigenvalues, we have tr(A) = 2 + 4 + 4 = 10.
Step 3: The trace of A2is equal to the sum of the squares of the eigenvalues
of A. Therefore, we have tr(A2) = λ2
1+λ2
2+λ2
3.
Step 4: Substituting the given eigenvalues, we have tr(A2) = 22+ 42+ 42=
4 + 16 + 16 = 36.
Step 5: Since we are given that tr(A2) = 33, this leads to a contradiction.
Therefore, there are no possible values of tr(A) that satisfy the given conditions.
Question 33
Question
Let Abe a 3 ×3 matrix with eigenvalues −1, 2, and 4. Find the eigenvalues of
the matrix B=A2−4A+ 4I, where Iis the 3 ×3 identity matrix.
Solution
Step 1: Let’s first find the eigenvalues of B. Given that Ahas eigenvalues
−1, 2, and 4, the eigenvalues of A2are simply the squares of these eigenvalues:
Eigenvalues of A2: (−1)2= 1, 22= 4, and 42= 16.
Step 2: Next, calculate the eigenvalues of −4Aby multiplying each eigen-
value of Aby −4: Eigenvalues of −4A:−4×(−1) = 4, −4×2 = −8, and
−4×4 = −16.
30
Step 3: Now, let’s find the eigenvalues of 4I: Since 4Iis a scalar multiple of
the identity matrix, all its eigenvalues are equal to 4.
Step 4: Substituting these results into the expression for B=A2−4A+ 4I,
we get: Eigenvalues of B: 1−4+4 = 1, 4−(−8)+4 = 16, and 16−(−16)+4 = 36.
Therefore, the eigenvalues of Bare 1, 16, and 36.
Question 34
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Find
the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we solve the equation
(A−2I)v1=0, where v1is the eigenvector corresponding to λ1and Iis the
3×3 identity matrix. Step 2: Subtract 2Ifrom Ato get:
−121
1 1 1
2 2 2
Step 3: Applying row operations, we get the row echelon form of the matrix as:
111
010
000
Step 4: The system of equations corresponding to the row echelon form is:
v11 +v12 +v13 = 0
v22 = 0
Step 5: Thus, the eigenvector corresponding to λ1= 2 is
−1
0
1
after fixing a
free variable.
Step 6: Repeat steps 1-5 for λ2=−1 and λ3= 4 to find the eigenvectors
corresponding to these eigenvalues.
Step 7: The eigenvectors corresponding to λ2=−1 and λ3= 4 are
1
−2
1
and
1
1
0
respectively.
31
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Suppose B= 3A2−4A.
Find the eigenvalues of B.
Solution
Step 1: First, let’s find the eigenvalues of Busing the given eigenvalues of A.
Let λbe an eigenvalue of Awith corresponding eigenvector v. Then, we have
Av=λv.
Step 2: Now, we can find Bvin terms of Av.
Bv= (3A2−4A)v= 3A2v−4Av.
Step 3: Since Av=λv, we can simplify Bv.
Bv= 3A2v−4Av= 3λ2v−4λv.
Step 4: Therefore, Bv= (3λ2−4λ)v. This shows that if λis an eigenvalue
of A, then 3λ2−4λis an eigenvalue of B.
Step 5: Using the eigenvalues of A(−2,1,and 3), let’s find the eigenvalues
of B. For λ=−2: 3λ2−4λ= 3(−2)2−4(−2) = 12 + 8 = 20. For λ= 1:
3λ2−4λ= 3(1)2−4(1) = 3 −4 = −1. For λ= 3: 3λ2−4λ= 3(3)2−4(3) =
27 −12 = 15.
Step 6: Therefore, the eigenvalues of Bare 20,−1,and 15.
32
Step 4: Substitute Aand Iinto the equation:
a+ 3 b c
d e + 3 f
g h i + 3
x
y
z
=
0
0
0
Step 5: Solve the system of equations to find the eigenvector corresponding
to λ3=−3:
(a+ 3)x+by +cz = 0
dx + (e+ 3)y+fz = 0
gx +hy + (i+ 3)z= 0
Step 6: The solution to the system of equations will give us the eigenvector
corresponding to λ3.
Question 2
Question
Let Abe a 3×3 matrix with eigenvalues −2,1,and 4. Given that the eigenvector
corresponding to the eigenvalue −2 is
1
−1
2
, find the eigenvectors correspond-
ing to the eigenvalues 1 and 4.
Solution
Step 1: Recall that for an eigenvalue λand its corresponding eigenvector v, the
relationship Av=λvholds.
Step 2: Let’s find the eigenvector corresponding to the eigenvalue 1. We
have:
Av=λv
Av= 1v
Av=
−200
0 1 0
0 0 4
x
y
z
=
x
y
z
Step 3: Solving the system of equations, we get:
−2x=x, y =y, 4z=z
Step 4: Simplifying the equations, we see that x= 0, ycan be any value,
and zcan be any value. Thus, the eigenvector corresponding to the eigenvalue
1 is any vector of the form
0
1
0
.
2
Step 5: Now, let’s find the eigenvector corresponding to the eigenvalue 4.
We have:
Av= 4v
Av=
−200
0 1 0
0 0 4
x
y
z
= 4
x
y
z
Step 6: Solving the system of equations, we get:
−2x= 4x, y = 4y, 4z= 4z
Step 7: Simplifying the equations, we see that x= 0, y= 0, and zcan be
any value. Therefore, the eigenvector corresponding to the eigenvalue 4 is any
vector of the form
0
0
1
.
Question 3
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. Given
that the eigenvectors corresponding to λ1and λ2are v1=
1
2
1
and v2=
1
1
0
,
respectively, find a basis for the eigenspace corresponding to λ3.
Solution
Step 1: Recall that each eigenvalue corresponds to a linearly independent set
of eigenvectors forming an eigenspace. To find the basis for the eigenspace
corresponding to λ3= 4, we need to find the eigenvectors corresponding to λ3.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to λ3= 4. Since
Av3=λ3v3, we have:
Av3= 4v3
Step 3: Substituting v3into the above equation, we get:
A
x
y
z
= 4
x
y
z
Step 4: This can be written as a system of equations:
Ax = 4x
Ay = 4y
Az = 4z
3
Step 5: Solving the system of equations, we obtain the equation Ax = 4x
where x,y, and zare variables. After solving, we will have the eigenvector
corresponding to λ3.
Step 6: Now, let’s solve the system of equations Ax = 4x:
A
x
y
z
=
4x
4y
4z
a b c
d e f
g h i
x
y
z
=
4x
4y
4z
Step 7: Solve the system of equations to find the eigenvector v3=
x
y
z
.
Question 4
Question
Let Abe a 3 ×3 matrix such that the characteristic equation of Ais given by
λ3−6λ2+ 11λ−6 = 0. Find the eigenvalues of Aand for each eigenvalue, find
a basis for the corresponding eigenspace.
Solution
Step 1: To find the eigenvalues of matrix A, we need to solve the characteristic
equation λ3−6λ2+ 11λ−6 = 0. Step 2: Factorize the characteristic equation
to obtain: (λ−1)(λ−2)(λ−3) = 0. Step 3: From the factorization, we see that
the eigenvalues are λ1= 1, λ2= 2, and λ3= 3. Step 4: To find the eigenvectors
corresponding to each eigenvalue, we solve the system (A−λiI)v= 0, where v
is the eigenvector corresponding to eigenvalue λi. Step 5: For λ1= 1, we solve
the system (A−I)v1= 0. This gives the equation:
000
011
002
x
y
z
=
0
0
0
Step 6: Solving this system gives the eigenvector v1=
0
0
1
. Step 7: For λ2= 2,
we solve the system (A−2I)v2= 0. This gives the equation:
−100
0 0 1
0 0 1
x
y
z
=
0
0
0
4
Step 8: Solving this system gives the eigenvector v2=
0
1
0
. Step 9: For λ3= 3,
we solve the system (A−3I)v3= 0. This gives the equation:
−200
0−1 1
0 0 0
x
y
z
=
0
0
0
Step 10: Solving this system gives the eigenvector v3=
0
1
1
. Step 11: There-
fore, the eigenvalues of matrix Aare 1, 2, and 3, and the corresponding eigen-
vectors are
0
0
1
,
0
1
0
, and
0
1
1
respectively.
Question 5
Question
Let Abe a 3 ×3 matrix with eigenvalues 6,2,and −3. Find the eigenvalues of
the matrix A2−4A+ 3I, where Iis the 3 ×3 identity matrix.
Solution
Step 1: First, find the eigenvalues of A2−4A+ 3Iby substituting Awith its
eigenvalues.
Step 2: The eigenvalues of A2−4A+ 3Iare given by the roots of the
characteristic equation det(A2−4A+ 3I−λI) = 0.
Step 3: Substitute Awith its eigenvalues 6,2,and −3:
det
620 0
0 220
0 0 (−3)2
−4
6 0 0
0 2 0
0 0 −3
+ 3
100
010
001
−λ
100
010
001
= 0
Step 4: Simplify and solve for λto find the eigenvalues of A2−4A+ 3I.
Question 6
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. Given
that the eigenvector corresponding to λ1is (1,0,1), find the eigenvectors corre-
sponding to λ2and λ3.
5
Solution
Step 1: To find the eigenvectors corresponding to λ2= 2 and λ3= 3, we will
solve the system of equations (A−λI)v=0for each eigenvalue.
For λ2= 2:
A−2I=
1 0 1
0 0 1
−101
We solve the system (A−2I)v=0:
1 0 1
0 0 1
−101
x
y
z
=
0
0
0
This leads to the system of equations:
x+z= 0
z= 0
Solving this system yields x= 0, y= 1, z= 0. Therefore, the eigenvector
corresponding to λ2= 2 is v2= (0,1,0).
Step 2: For λ3= 3:
A−3I=
−201
0−1 1
−100
We solve the system (A−3I)v=0:
−201
0−1 1
−100
x
y
z
=
0
0
0
This leads to the system of equations:
−2x+z= 0
−y+z= 0
−x= 0
Solving this system yields x= 0, y= 1, z= 0. Therefore, the eigenvector
corresponding to λ3= 3 is v3= (0,1,0).
Question 7
Question
Let Abe a 3 ×3 matrix given by
A=
310
020
012
.
6
Determine the eigenvalues and corresponding eigenvectors of A.
Solution
Step 1: To find the eigenvalues of A, we need to solve the characteristic equation
given by det(A−λI) = 0, where λis the eigenvalue and Iis the identity matrix.
A−λI =
3−λ1 0
0 2 −λ0
0 1 2 −λ
.
Setting the determinant of this matrix to zero, we get
det
3−λ1 0
0 2 −λ0
0 1 2 −λ
= (3 −λ)((2 −λ)2)−0−0=0.
Step 2: Simplifying the determinant equation, we get
(3 −λ)((2 −λ)2) = (3 −λ)(4 −4λ+λ2)=0.
This equation simplifies to λ3−7λ2+ 14λ−8 = 0.
Step 3: Solving the cubic equation using synthetic division or a numerical
method, we find the eigenvalues to be λ= 2,2,3. Notice that 2 appears twice
as a repeated eigenvalue.
Step 4: To find the eigenvectors corresponding to each eigenvalue, we sub-
stitute each eigenvalue back into the equation (A−λI)v=0and solve for
v.
For λ= 2, we have
A−2I=
110
000
010
which leads to the eigenvector v1=
−1
0
1
.
For the repeated eigenvalue λ= 2, we have another eigenvector v2=
−1
0
1
.
For λ= 3, we have
A−3I=
0 1 0
0−1 0
0 1 −1
which leads to the eigenvector v3=
1
0
−1
.
Therefore, the eigenvalues of Aare λ= 2,2,3 with corresponding eigenvec-
tors v1=
−1
0
1
,v2=
−1
0
1
, and v3=
1
0
−1
.
7
Question 8
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If
B=A2−2A+I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
matrix B.
Solution
To find the eigenvalues of matrix B, we need to first find the matrix Band then
determine its eigenvalues.
Step 1: Find matrix B=A2−2A+I:Given that Ais a 3×3 matrix with
eigenvalues λ1= 1, λ2= 2, and λ3= 3, wecanexpressAas :A=P DP −1,where
Pis the matrix of eigenvectors and Dis the diagonal matrix of eigenvalues.
Thus, Acan be written as:
A=P DP −1=v1v2v3
100
020
003
v1v2v3−1,
where
Next, we can calculate A2−2A+Ito find matrix B:
B=A2−2A+I.
Step 2: Calculate B:Since we have A=P DP −1, we have:
A2= (P DP −1)(P DP −1) = P D(P−1P)DP −1=P D2P−1.
Substitute this into B=A2−2A+I:
B=P D2P−1−2P DP −1+I.
Step 3: Evaluate B:Substitute the expressions for A=P DP −1and D
into the equation for Bto obtain the matrix B.
Step 4: Find the eigenvalues of B:Once you have the matrix B, find
its eigenvalues µ1,µ2, and µ3.
The resulting eigenvalues µ1,µ2, and µ3are the eigenvalues of matrix B.
Question 9
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3. If the
eigenvectors corresponding to λ1and λ2are
v1=
1
2
3
and v2=
4
5
6
,
find an eigenvector v3corresponding to the eigenvalue λ3= 3.
8
Solution
Given that Ais a 3 ×3 matrix with eigenvalues λ1= 1, λ2= 2, and λ3= 3, we
can represent Ain terms of its eigenvectors and eigenvalues as
A=P DP −1,
where
P= [v1v2v3] and D=
λ10 0
0λ20
0 0 λ3
.
Since v1and v2are the eigenvectors corresponding to λ1= 1 and λ2= 2,
respectively, we have
P=
1 4 a
2 5 b
3 6 c
.
Now, to find v3, we will equate Dto P−1AP and solve for v3.
P−1AP =D
1 4 a
2 5 b
3 6 c
−1
p11 p12 p13
p21 p22 p23
p31 p32 p33
1 4 a
2 5 b
3 6 c
=
100
020
003
.
Now, solve for v3using P−1AP =D. Once v3is found, normalize v3to
obtain an eigenvector corresponding to λ3= 3.
Question 10
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. If the
eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
1
, find
the eigenvectors corresponding to λ3.
Solution
Step 1: The eigenvectors v3corresponding to λ3can be found by solving the
equation (A−λI)v=0, where λ= 4. This gives us:
(A−4I)v=
−200
0 1 0
0 0 0
x
y
z
=
0
0
0
Step 2: Solving the system of equations, we get:
−2x= 0, y = 0, z = 0
9
Step 3: So, the eigenvector v3corresponding to λ3= 4 is
0
0
1
.
Question 11
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 5, and λ3= 8. If
the eigenvectors corresponding to λ1and λ3are v1=
1
0
1
and v3=
0
1
2
respectively, find the eigenvector corresponding to λ2.
Solution
Step 1: Begin by setting up the equation to find the eigenvector v2corresponding
to λ2:
Av2=λ2v2
Step 2: Knowing the eigenvalues and eigenvectors, we can write Ain terms
of its diagonalized form:
A=P DP −1
where Dis a diagonal matrix with the eigenvalues on the diagonal, and the
columns of Pare the corresponding eigenvectors.
Step 3: From the given information, we have:
D=
200
050
008
and
P=
100
011
120
Step 4: Now we can write the equation Av2=λ2v2as:
P DP −1v2=λ2v2
Step 5: Premultiplying by P−1, we get:
DP −1v2=λ2P−1v2
Step 6: Substituting the values of Dand Pinto the equation, we find:
200
050
008
100
011
120
v2= 5
100
011
120
v2
10
Step 7: Multiplying out the matrix product on the left side gives:
200
050
008
1
1
2
= 5
1
1
2
Step 8: Solving the equation above will give us the eigenvector v2corre-
sponding to the eigenvalue λ2.
Question 12
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 4. Given
that det(A) = 10, find the eigenvalues of 3A−2A−1.
Solution
Step 1: First, let’s find the eigenvalues of A−1. Since Ahas eigenvalues λ1=
2, λ2=−1,and λ3= 4, the eigenvalues of A−1are 1
λ1,1
λ2,and 1
λ3. Therefore,
the eigenvalues of A−1are 1
2,−1,and 1
4.
Step 2: Next, let’s find the eigenvalues of 3A−2A−1. The eigenvalues of
3A−2A−1are given by 3λi−2µi, where λiare the eigenvalues of Aand µiare
the eigenvalues of A−1. Substituting the given values, we have:
3λ1−2µ1= 3(2) −21
2= 6 −1=5,
3λ2−2µ2= 3(−1) −2(−1) = −3 + 2 = −1,
3λ3−2µ3= 3(4) −21
4= 12 −1
2=23
2.
Therefore, the eigenvalues of 3A−2A−1are 5, -1, and 23
2.
Question 13
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3=−3. If
the eigenvectors corresponding to λ1and λ2are v1=
1
0
1
and v2=
0
1
1
,
respectively, find a matrix Psuch that P−1AP is a diagonal matrix.
11
Solution
Step 1: Recall that if λis an eigenvalue of matrix Awith corresponding eigen-
vector v, then the matrix Pformed by stacking the eigenvectors as columns
diagonalizes A. That is, P−1AP =D, where Dis a diagonal matrix whose
diagonal entries are the eigenvalues.
Let P=v1v2v3=
1 0 a
0 1 b
1 1 c
Step 2: We know that Av1=λ1v1and Av2=λ2v2. So let’s calculate Av1
and Av2.
Av1=A
1
0
1
=
2−1 0
0 1 −1
2−1 1
1
0
1
=
2
0
2
= 2
1
0
1
= 2v1
Step 3: Now, let’s calculate Av2.
Av2=A
0
1
1
=
2−1 0
0 1 −1
2−1 1
0
1
1
=
−1
1
0
= 1
0
1
1
=v2
Step 4: Since Av1= 2v1and Av2=v2, we can see that the eigenvectors v1
and v2are already corresponding to the correct eigenvalues.
Step 5: We can choose v3arbitrarily for the remaining eigenvalue λ3=−3.
Let’s choose v3=
1
−2
1
.
Step 6: Now, form the matrix P=v1v2v3.
P=
1 0 1
0 1 −2
1 1 1
Step 7: Calculate P−1using the inverse formula P−1=1
det(P)adj(P).
P−1=1
−23−1 1Question14
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 3. If v1,
v2, and v3are the corresponding eigenvectors, find the matrix A.
12
Solution
Step 1: Recall the definition of an eigenvector and eigenvalue. For a matrix A
and a scalar λ, the eigenvector vsatisfies the equation Av=λv.
Step 2: Since v1,v2, and v3are eigenvectors of Acorresponding to the
eigenvalues λ1= 2, λ2=−1, and λ3= 3, respectively, we have the following
three equations: Av1= 2v1,Av2=−1v2,Av3= 3v3.
Step 3: Therefore, we can write the matrix Ain terms of its eigenvectors
and eigenvalues as: A=v1v2v3
λ10 0
0λ20
0 0 λ3
vT
1
vT
2
vT
3
.
Step 4: Substituting the given values for the eigenvalues and eigenvectors
into the equation above, we get: A=v1v2v3
200
0−1 0
003
vT
1
vT
2
vT
3
.
Step 5: Therefore, the matrix Acan be written as: A=2v1−v23v3.
Question 15
Question
Let Abe a 3×3 matrix with eigenvalues −2, 1, and 4. Given that the eigenvector
corresponding to the eigenvalue −2 is
1
2
1
, find:
A2
1
2
1
Solution
To find A2
1
2
1
, we first need to find the matrix Ausing its eigenvalues and
eigenvectors. Let v1=
1
2
1
be the eigenvector corresponding to the eigenvalue
−2. Since Ais a 3 ×3 matrix, the set {v1}forms a basis for R3. Additionally,
we are given that the eigenvalues are −2, 1, and 4, so the characteristic equation
for matrix Awould be:
det(A−λI) = 0
det
a b c
d e f
g h i
−λ
100
010
001
= 0
13
det
a−λ b c
d e −λ f
g h i −λ
= 0
Plugging in the given eigenvalues, we get:
det
a+ 2 b c
d e −1f
g h i −4
= 0
Expanding this determinant, we can find the matrix A. Next, we calculate A2:
A2=A·A
A2= Matrix multiplication of Aitself
Finally, we find A2
1
2
1
by multiplying A2with the given eigenvector
1
2
1
.
Question 16
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2=−2, and λ3= 5.
Given that the eigenvectors corresponding to λ1and λ2are
1
2
−1
and
−1
1
2
respectively, find the eigenvector corresponding to λ3.
Solution
Step 1: Since Ais a 3 ×3 matrix, it will have 3 eigenvalues and 3 corresponding
eigenvectors.
Step 2: Let v3=
x
y
z
be the eigenvector corresponding to the eigenvalue
λ3= 5.
Step 3: The eigenvectors of a matrix corresponding to distinct eigenvalues
are linearly independent. Since
1
2
−1
and
−1
1
2
are independent,
1
2
−1
and
−1
1
2
are independent to
x
y
z
.
Step 4: Since v3is an eigenvector corresponding to λ3= 5, we have Av3=
λ3v3.
Step 5: We can set up the following equation based on Step 4:
14
A
x
y
z
= 5
x
y
z
Step 6: Multiplying the matrices on the left side of the equation, we have:
3−1∗
∗ −2 1
∗ ∗ 5
x
y
z
=
5x
5y
5z
Step 7: Solving the resulting system of equations, we find that v3=
−2
1
1
is the eigenvector corresponding to the eigenvalue λ3= 5.
Question 17
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Given that one eigenvec-
tor corresponding to the eigenvalue −2 is
1
0
1
, find a basis for the eigenspace
corresponding to the eigenvalue 1.
Solution
Step 1: To find a basis for the eigenspace corresponding to the eigenvalue 1, we
need to solve the system (A−λI)v=0, where λis the eigenvalue (1 in this
case) and vis the eigenvector we want to find.
Step 2: Start by writing the matrix Aand the identity matrix I:
A=
a b c
d e f
g h i
and I=
100
010
001
Step 3: Substitute the eigenvalue λ= 1 into the equation (A−λI)v=0:
(A−I)v=
a−1b c
d e −1f
g h i −1
x
y
z
=
0
0
0
Step 4: Writing the system of equations, we get:
(a−1)x+by +cz = 0
dx + (e−1)y+fz = 0
gx +hy + (i−1)z= 0
Step 5: Since we are looking for a basis, we will not be able to solve for the
specific values of x, y, and z. Instead, we can set z= 1 and solve for xand y.
15
Step 6: Substituting z= 1 into the system of equations gives:
(a−1)x+by +c= 0
dx + (e−1)y+f= 0
gx +hy + (i−1) = 0
Step 7: To simplify, we can write the system of equations in matrix form:
a−1b
d e −1
g h
x
y=
−c
−f
−(i−1)
Step 8: This system of equations may have multiple solutions, thus to find a
basis, we can solve the system using row reduction to find the parametric form
of the solutions.
Step 9: The solutions to the system will give us a basis for the eigenspace
corresponding to the eigenvalue 1.
Question 18
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 0, and λ3=−1.
Given that the eigenvector corresponding to λ1= 2 is
1
1
2
and the eigenvector
corresponding to λ2= 0 is
1
1
1
, find the eigenvector corresponding to λ3=−1.
Solution
Given a matrix Awith eigenvalues λ1, λ2, λ3and corresponding eigenvectors
v1,v2,v3, we can find the eigenvector corresponding to λ3by solving the system
of equations Ax=λ3x.
Step 1: We have the following information:
A=
|||
v1v2v3
|||
, λ3=−1
Step 2: We can set up our system of equations as follows:
Ax=λ3x⇒Ax=−x
Step 3: Expressing Aand xin terms of column vectors:
|||
v1v2v3
|||
x1
x2
x3
=−
x1
x2
x3
16
This gives us a system of equations:
v1x1+v2x2+v3x3=−x1
v1x1+v2x2+v3x3=−x2
v1x1+v2x2+v3x3=−x3
Step 4: We can substitute the values of the eigenvectors v1and v2:
111
111
210
x1
x2
x3
=
−x1
−x2
−x3
Step 5: Solving the system of equations, we get the eigenvector correspond-
ing to λ3=−1:
x=
1
1
−2
Therefore, the eigenvector corresponding to λ3=−1 is
1
1
−2
.
Question 19
Question
Let Abe a 3 ×3 matrix with eigenvalues λ= 2,1,−1 and corresponding eigen-
vectors v1=
1
2
0
,v2=
0
1
−1
, and v3=
1
0
1
. Define the matrix Pwhose
columns are these eigenvectors.
(a) Calculate P−1.
(b) Determine A5.
Solution
(a) To find P−1, we first construct the matrix Pusing the given eigenvectors:
P= [v1v2v3] =
101
210
0−1 1
Next, we find the inverse of Pusing the formula P−1=1
det(P)adj(P), where
adj(P) is the adjugate of P. To do this, we calculate the determinant of P:
det(P) =
101
210
0−1 1
= 1(1 ·1−(−1) ·0) −0−1·1=2
17
Then, we find the cofactor matrix of P:
Cof(P) =
1−1 1
0 1 0
−221
Next, we calculate the adjugate of P:
adj(P) = Cof(P)T=
1 0 −2
−1 1 2
101
Finally, we compute P−1:
P−1=1
det(P)adj(P) = 1
2
1 0 −2
−1 1 2
101
=
1
20−1
−1
2
1
21
1
201
2
(b) To determine A5, we use the formula A=PΛP−1, where Λ is the
diagonal matrix of eigenvalues. We form Λ using the given eigenvalues:
Λ =
2 0 0
0 1 0
0 0 −1
Substitute the values of Pand P−1into the formula A=PΛP−1:
A=
101
210
0−1 1
2 0 0
0 1 0
0 0 −1
1
20−1
−1
2
1
21
1
201
2
Multiplying these matrices will give us A, and to find A5, we raise Ato the
power of 5.
Question 20
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 3, λ2= 2, and λ3=−1. If the
eigenvectors corresponding to λ2= 2 are v2=
1
0
1
and w2=
−1
1
0
, find a
basis for the eigenspace corresponding to λ1= 3.
Solution
Step 1: Recall that the eigenspace Eλassociated with an eigenvalue λis the set
of all eigenvectors corresponding to λtogether with the zero vector.
18
Step 2: Since v2and w2are eigenvectors corresponding to λ2= 2, the
eigenspace E2is given by:
E2= span
1
0
1
,
−1
1
0
Step 3: To find the eigenvectors corresponding to λ1= 3, we can use the
fact that the sum of each diagonal element in a matrix is equal to the sum of
its eigenvalues.
Step 4: Let v1=
x
y
z
be an eigenvector corresponding to λ1= 3. Then
Av1= 3v1.
Step 5: This gives us the system of equations:
3x= 3x
3y= 3y
3z= 3z
Step 6: Since x,y, and zcan be any real number, the eigenvectors corre-
sponding to λ1= 3 are vectors of the form
1
0
0
,
0
1
0
, and
0
0
1
.
Step 7: Therefore, a basis for the eigenspace corresponding to λ1= 3 is
given by:
B3=
1
0
0
,
0
1
0
,
0
0
1
Question 21
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 1, and λ3= 3. Suppose
the corresponding eigenvectors are
v1=
1
1
1
,v2=
1
0
1
,and v3=
0
1
1
.
Determine the matrix A.
Solution
Step 1: The matrix Ais similar to a diagonal matrix Dwhose diagonal en-
tries are the eigenvalues of A. Let Pbe the matrix formed by stacking the
eigenvectors v1,v2, and v3as columns.
19
P=
110
101
111
Step 2: The matrix Dis given by
D=P−1AP
where Ais the matrix we are trying to find.
Step 3: To find P−1, we first compute the determinant of P.
det(P) = 1(0 −1) −1(1 −1) + 0(1 −1) = −1
Step 4: Next, we find the adjugate of P, denoted as adj(P), by taking the
transpose of the matrix of cofactors of P.
adj(P) =
0−1 1
1−1 0
−1 2 −1
Step 5: The inverse of Pis given by
P−1=1
det(P)·adj(P) =
0 1 −1
−1 1 0
1−2 1
Step 6: We can now find the matrix Aby using the formula D=P−1AP ,
which gives
A=P DP −1
A=
110
101
111
200
010
003
0 1 −1
−1 1 0
1−2 1
A=
312
221
323
Therefore, the matrix Ais
312
221
323
.
20
Question 22
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 3, and λ3= 4. Given
that B=A2−5A+6I, where Iis the 3 ×3 identity matrix, find the eigenvalues
of B.
Solution
Step 1: Find the eigenvalues of B. Given B=A2−5A+ 6I, we can rewrite B
as:
B=A2−5A+ 6I=A2−5A+ 6
100
010
001
Step 2: First, find the eigenvalues of A2. Since Ahas eigenvalues λ1= 2,
λ2= 3, and λ3= 4, the eigenvalues of A2are λ2
1= 4, λ2
2= 9, and λ2
3= 16.
Step 3: Find Bin terms of A. Now, we can express Bin terms of the
eigenvalues of A:
B=A2−5A+ 6I=
4 0 0
0 9 0
0 0 16
−5
200
030
004
+ 6
100
010
001
=
−600
0−3 0
0 0 6
+
12 0 0
0 15 0
0 0 20
=
6 0 0
0 12 0
0 0 26
Step 4: Find the eigenvalues of B. The eigenvalues of Bare the diagonal
elements of the matrix we found in Step 3. Therefore, the eigenvalues of Bare
λ′
1= 6, λ′
2= 12, and λ′
3= 26.
Question 23
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 0. If
B=A2−3A+ 2I, where Iis the 3 ×3 identity matrix, find the eigenvalues of
B.
21
Solution
Step 1: First, let’s find the eigenvalues of matrix B.
B=A2−3A+ 2I
=A2−3A+ 2I
=A2−3A+ 2
100
010
001
=A2−3A+ 2
100
010
001
=A2−3A+
200
020
002
Step 2: Now, let’s find the eigenvalues of B. The eigenvalues of Bare the
solutions to the characteristic equation |B−λI|= 0, where λrepresents the
eigenvalue.
|B−λI|=
A2−3A+
200
020
002
−λ
100
010
001
=|A2−3A+
2−λ0 0
0 2 −λ0
0 0 2 −λ
|
=|A−2I||A+I||A|
= (2 −λ)(−1−λ)λ
Therefore, the eigenvalues of matrix Bare λ= 2,−1,0.
Question 24
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1,and λ3= 5.
Find a matrix Psuch that A=PDP−1, where Dis the diagonal matrix with
eigenvalues of Aon the diagonal arranged in the same order as in P.
Solution
Step 1: Start by constructing the matrix Dusing the eigenvalues of A.
D=
200
0−1 0
005
22
Step 2: Next, construct the matrix Pusing the eigenvectors corresponding
to each eigenvalue. Let
P=v1v2v3
where v1,v2,and v3are the eigenvectors corresponding to the eigenvalues 2,−1,
and 5 respectively.
Step 3: Since λ1= 2, solve (A−2I)v1=0to find v1. Let
v1=
a
b
c
Solving the equation (A−2I)v1=0, we get
0−3 1
−1 0 0
0 1 −2
a
b
c
=
0
0
0
Solving this system of equations gives the eigenvector v1=
1
1
1
after normal-
ization.
Step 4: Repeat the same process for the remaining eigenvalues to find the
eigenvectors v2and v3. After finding all the eigenvectors, construct the matrix
Pas:
P=
1??
1??
1??
Step 5: Finally, calculate P−1and verify that A=PDP−1.
Question 25
Question
Let A=2 1
1 2be a matrix. Find the eigenvectors corresponding to the
eigenvalues of A.
Solution
Step 1: Find the eigenvalues by solving the characteristic equation det(A−λI) =
0.
det 2−λ1
1 2 −λ= (2 −λ)(2 −λ)−1·1
=λ2−4λ+ 3
= (λ−3)(λ−1) = 0.
23
Therefore, the eigenvalues are λ1= 1 and λ2= 3.
Step 2: Find the eigenvector corresponding to λ1= 1 by solving the system
(A−λI)v= 0. For λ1= 1:
2−1 1
1 2 −1x
y=0
0.
This gives the system of equations:
(x+y= 0
y= 0 .
Solving this system, we find that x= 0 and y= 0. Therefore, the eigenvector
corresponding to λ1= 1 is v1=0
0.
Step 3: Find the eigenvector corresponding to λ2= 3 by solving the system
(A−λI)v= 0. For λ2= 3:
2−3 1
1 2 −3x
y=0
0.
This gives the system of equations:
(−x+y= 0
y= 0 .
Solving this system, we find that x= 0 and y= 0. Therefore, the eigenvector
corresponding to λ2= 3 is v2=0
0.
Thus, the eigenvectors corresponding to the eigenvalues of matrix Aare 0
0
for both eigenvalues 1 and 3.
Question 26
Question
Let Abe a 3 ×3 matrix given by
A=
2 1 −1
0 2 0
1 0 2
.
Determine the eigenvalues and eigenvectors of A.
24
Solution
Step 1: Find the characteristic equation.
The characteristic equation for a matrix Ais given by det(A−λI) = 0,
where Iis the identity matrix of the same size as Aand λis the eigenvalue we
are trying to solve for. For our 3 ×3 matrix A, we have
A−λI =
2−λ1−1
0 2 −λ0
1 0 2 −λ
.
Step 2: Find the determinant and solve for λ.
Calculating the determinant, we get
det(A−λI) = (2 −λ)2(2 −λ)+1·1(2 −λ) = (2 −λ)((2 −λ)2+ 1).
Setting this equal to 0, we have
(2 −λ)((2 −λ)2+ 1) = 0.
Solving for λ, we find that the eigenvalues are λ= 2,2 + i, and 2 −i.
Step 3: Find the eigenvectors for each eigenvalue.
For λ= 2: Setting λ= 2 in (A−λI)v= 0, we get
0 1 −1
0 0 0
1 0 0
x
y
z
=
0
0
0
,
which gives us the equation x=zand y= 0. Therefore, the eigenvector
corresponding to the eigenvalue λ= 2 is
1
0
1
.
For λ= 2 + i: Setting λ= 2 + iin (A−λI)v= 0, we get
−i1−1
0−i0
1 0 −i
x
y
z
=
0
0
0
,
which gives us the equation −ix +y−z= 0, −iy = 0, and x=iz. From this,
we can find the eigenvector corresponding to the eigenvalue λ= 2 + i.
For λ= 2 −i: Similarly, we can find the eigenvector corresponding to the
eigenvalue λ= 2 −i.
Question 27
Question
Let A=4−1
2 1 . Find a matrix Psuch that P−1AP is diagonal, and calculate
P−1.
25
Solution
Step 1: Find the eigenvalues of matrix Aby solving the characteristic equation
|A−λI|= 0.
|A−λI|=
4−λ−1
2 1 −λ
= (4 −λ)(1 −λ)−(−1)(2)
=λ2−5λ+ 6
= (λ−2)(λ−3) = 0
So, the eigenvalues are λ1= 2 and λ2= 3.
Step 2: Find the eigenvectors corresponding to each eigenvalue by solving
(A−λI)x= 0. For λ= 2:
A−2I=2−1
2−1→2−1
0 0 →1−1
2
0 0
Letting x2=t, we have x1=1
2t. Thus, the eigenvector corresponding to λ= 2
is v1=1
2. Similarly, for λ= 3:
A−3I=1−1
2−2→1−1
0 0 →1−1
0 0
Again, letting x2=t, we have x1=t. Thus, the eigenvector corresponding to
λ= 3 is v2=1
1.
Step 3: Assemble the eigenvectors into the matrix P=v1v2. So,
P=1 1
2 1.
Step 4: Calculate P−1. To find the inverse of a 2 ×2 matrix, we use the
formula
P−1=1
ad −bc d−b
−c a .
Thus, P−1=1
11−1
−2 1 =1−1
−2 1 .
Therefore, the matrix Psuch that P−1AP is diagonal is
P=1 1
2 1
and
P−1=1−1
−2 1 .
26
Question 28
Question
Let Abe a 3 ×3 matrix with eigenvalues −1,2,and 3. Find the eigenvalues of
the matrix A2+ 3A−2I, where Iis the 3 ×3 identity matrix.
Solution
Given that Ahas eigenvalues −1,2, and 3, we can find the eigenvalues of A2+
3A−2Iby first finding the matrix A2+ 3A−2Iin terms of A.
Step 1: Compute A2Since Ahas eigenvalues −1,2, and 3, the characteristic
equation of Ais given by (A+ 1)(A−2)(A−3) = 0. Expanding this equation
gives A3−4A2−5A+ 6I= 0. Rearranging terms, we have
A2= 4A+ 5I−6I= 4A−I
Step 2: Find A2+ 3A−2INow we can substitute A2= 4A−Iinto the
expression A2+ 3A−2Ito get
A2+ 3A−2I= (4A−I)+3A−2I= 7A−3I
Step 3: Find the eigenvalues of A2+ 3A−2IThe eigenvalues of A2+ 3A−
2I= 7A−3Iare simply 7 times the eigenvalues of Aminus 3. Therefore,
the eigenvalues of A2+ 3A−2Iare 7 × −1−3 = −10, 7 ×2−3 = 11, and
7×3−3 = 18. Thus, the eigenvalues of the matrix A2+ 3A−2Iare −10,11,
and 18.
Question 29
Question
Let Abe a 3 ×3 matrix with eigenvalues 1, 2, and 3. Find the eigenvalues of
the matrix B=A3−2A2+A.
Solution
Given that Ahas eigenvalues 1, 2, and 3, we have:
det(A−λI) = 0
This gives us the characteristic equation for Aas:
det(A−λI) = (A−λI)(A−2I)(A−3I)=0
From this, we can express A3,A2, and Ain terms of λusing the eigenvalues
1, 2, and 3. Let B=A3−2A2+A:
27
B=A3−2A2+A
= (13−2·12+ 1)I+ (23−2·22+ 2)A+ (33−2·32+ 3)A2
= 0 ·I+ 2 ·A+ 6 ·A2
= 0 ·I+ 2 ·A+ 6 ·2A−4·I
= 12 ·A−4·I
= 4(3A−I)
Therefore, the eigenvalues of Bare 3 ×4−1 = 11, 2 ×4−1 = 7, and
1×4−1 = 3.
Question 30
Question
Let A=2 1
1 2be a 2 ×2 matrix. Determine all eigenvalues and eigenvectors
of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation |A−
λI|= 0 where Iis the identity matrix.
|A−λI|=
2−λ1
1 2 −λ
= (2 −λ)2−1=0
⇒(2 −λ)2−1=0
⇒λ2−4λ+ 3 = 0
⇒(λ−3)(λ−1) = 0
⇒λ1= 3, λ2= 1
Step 2: To find the eigenvectors corresponding to λ= 3, we solve (A−3I)v=
0.
A−3I=2−3 1
1 2 −3=−1 1
1−1
−1 1
1−1x
y=0
0
⇒ −x+y= 0 and x−y= 0
⇒No non-trivial solution exists
28
Step 3: To find the eigenvectors corresponding to λ= 1, we solve (A−I)v=
0.
A−I=2−1 1
1 2 −1=1 1
1 1
1 1
1 1x
y=0
0
⇒x+y= 0
⇒One eigenvector is 1
−1
Therefore, the eigenvalues of Aare λ= 3 and λ= 1, with corresponding
eigenvectors being 1
−1and no eigenvector for λ= 3.
Question 31
Question
Let Abe a 3 ×3 matrix given by
A=
2−1 0
−1 2 −1
0−1 2
.
Find the eigenvectors and eigenvalues of A.
Solution
Step 1: To find the eigenvalues of A, we solve the characteristic equation
det(A−λI) = 0,
where Iis the identity matrix. Substitute Aand λinto the equation to get
2−λ−1 0
−1 2 −λ−1
0−1 2 −λ
= 0.
Step 2: Expand the determinant to get
(2 −λ)((2 −λ)(2 −λ)−(−1)(−1)) + (−1)(−1) −0
= (2 −λ)((λ2−4λ+ 4) + 1) + 1
= (2 −λ)(λ2−4λ+ 5) + 1
= 2λ2−8λ+ 10 −λ3+ 4λ2−5λ+ 1.
=−λ3+ 6λ2−13λ+ 11.
29
Step 3: Set the characteristic equation equal to zero:
−λ3+ 6λ2−13λ+ 11 = 0.
This cubic equation is difficult to factorize so we can find the eigenvalues using
numerical methods such as Newton’s method. By computing the roots, we find
that the eigenvalues are λ1≈1.32, λ2≈1.0 and λ3≈2.68.
Step 4: To find the eigenvectors associated with each eigenvalue, we substi-
tute the eigenvalues back into the equation (A−λI)v=0and solve for v. We
find the eigenvectors v1,v2, and v3associated with the eigenvalues λ1,λ2, and
λ3, respectively.
Question 32
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2= 4, λ3= 4. If tr(A2) = 33
(where tr denotes the trace of a matrix), find the possible values of tr(A).
Solution
Step 1: Recall that the trace of a matrix is the sum of its eigenvalues. Therefore,
we have tr(A) = λ1+λ2+λ3.
Step 2: Substituting the given eigenvalues, we have tr(A) = 2 + 4 + 4 = 10.
Step 3: The trace of A2is equal to the sum of the squares of the eigenvalues
of A. Therefore, we have tr(A2) = λ2
1+λ2
2+λ2
3.
Step 4: Substituting the given eigenvalues, we have tr(A2) = 22+ 42+ 42=
4 + 16 + 16 = 36.
Step 5: Since we are given that tr(A2) = 33, this leads to a contradiction.
Therefore, there are no possible values of tr(A) that satisfy the given conditions.
Question 33
Question
Let Abe a 3 ×3 matrix with eigenvalues −1, 2, and 4. Find the eigenvalues of
the matrix B=A2−4A+ 4I, where Iis the 3 ×3 identity matrix.
Solution
Step 1: Let’s first find the eigenvalues of B. Given that Ahas eigenvalues
−1, 2, and 4, the eigenvalues of A2are simply the squares of these eigenvalues:
Eigenvalues of A2: (−1)2= 1, 22= 4, and 42= 16.
Step 2: Next, calculate the eigenvalues of −4Aby multiplying each eigen-
value of Aby −4: Eigenvalues of −4A:−4×(−1) = 4, −4×2 = −8, and
−4×4 = −16.
30
Step 3: Now, let’s find the eigenvalues of 4I: Since 4Iis a scalar multiple of
the identity matrix, all its eigenvalues are equal to 4.
Step 4: Substituting these results into the expression for B=A2−4A+ 4I,
we get: Eigenvalues of B: 1−4+4 = 1, 4−(−8)+4 = 16, and 16−(−16)+4 = 36.
Therefore, the eigenvalues of Bare 1, 16, and 36.
Question 34
Question
Let Abe a 3 ×3 matrix with eigenvalues λ1= 2, λ2=−1, and λ3= 4. Find
the eigenvectors corresponding to each eigenvalue.
Solution
Step 1: To find the eigenvector corresponding to λ1= 2, we solve the equation
(A−2I)v1=0, where v1is the eigenvector corresponding to λ1and Iis the
3×3 identity matrix. Step 2: Subtract 2Ifrom Ato get:
−121
1 1 1
2 2 2
Step 3: Applying row operations, we get the row echelon form of the matrix as:
111
010
000
Step 4: The system of equations corresponding to the row echelon form is:
v11 +v12 +v13 = 0
v22 = 0
Step 5: Thus, the eigenvector corresponding to λ1= 2 is
−1
0
1
after fixing a
free variable.
Step 6: Repeat steps 1-5 for λ2=−1 and λ3= 4 to find the eigenvectors
corresponding to these eigenvalues.
Step 7: The eigenvectors corresponding to λ2=−1 and λ3= 4 are
1
−2
1
and
1
1
0
respectively.
31
Question 35
Question
Let Abe a 3 ×3 matrix with eigenvalues −2,1,and 3. Suppose B= 3A2−4A.
Find the eigenvalues of B.
Solution
Step 1: First, let’s find the eigenvalues of Busing the given eigenvalues of A.
Let λbe an eigenvalue of Awith corresponding eigenvector v. Then, we have
Av=λv.
Step 2: Now, we can find Bvin terms of Av.
Bv= (3A2−4A)v= 3A2v−4Av.
Step 3: Since Av=λv, we can simplify Bv.
Bv= 3A2v−4Av= 3λ2v−4λv.
Step 4: Therefore, Bv= (3λ2−4λ)v. This shows that if λis an eigenvalue
of A, then 3λ2−4λis an eigenvalue of B.
Step 5: Using the eigenvalues of A(−2,1,and 3), let’s find the eigenvalues
of B. For λ=−2: 3λ2−4λ= 3(−2)2−4(−2) = 12 + 8 = 20. For λ= 1:
3λ2−4λ= 3(1)2−4(1) = 3 −4 = −1. For λ= 3: 3λ2−4λ= 3(3)2−4(3) =
27 −12 = 15.
Step 6: Therefore, the eigenvalues of Bare 20,−1,and 15.
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