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_math_help.zipTest 1.pdf
Solutions
Name:
Student ID No.:
Discussion Section:
Math 20F Midterm I(ver. a) Winter 2016
Problem Score
1 /24
2 /24
3 /26
4 /26
Total /100
1. (24 Points.) The following are True/False questions. For this problem only, you do not have to show any work. There will be no partial credit given for this problem. For this problem:
• A correct answer gives 4 points.
• An incorrect answer gives 0 points.
• If you leave the space blank, you receive 2 points.
F (a) If T : Rn → Rm and T (~0) =~0, then T is a linear transformation.
T (b) Assume A is an m×n matrix. If A~x =~b has a solution for every vector~b then the columns of A span Rm.
T (c) If the Row Echelon form of A has a pivot in every row then A~x =~b has a solution for every vector ~b.
F (d) Whenever ~x1 and ~x2 are two solutions to A~x =~b, then ~x1 +~x2 is also a solution to A~x =~b.
T (e) If the homogeneous problem: A~x =~0 has only the trivial solution, then the problem A~x =~b either has no solutions or has exactly one solution.
T (f) If the Row Echelon form of the matrix A has a column without a pivot, then the problem A~x =~0 has many solutions.
2. Each statement below is either true (in all cases) or false (in at least one case). If false, construct a counterexample. If true, give a justification using the definition of “linearly independent.”
(a) (8 Points.) If v1,...,v4 are in R5 and {v1,v2,v3} is linearly dependent then {v1,v2,v3,v4} is linearly dependent.
(b) (8 Points.) If v1,...,v4 are in R5 and v1 is not a linear combination of {v2,v3,v4}, then {v1,v2,v3,v4} is linearly independent.
(c) (8 Points.) If v1,...,v4 are linearly independent vectors in R5, then {v1,v2,v3} is also linearly independent.
(a) True. {v1,v2,v3} linearly dependent means there exists weights a1, a2, and a3 not all zero such that
a1v1 + a2v2 + a3v3 = 0.
This means a1v1 + a2v2 + a3v3 + a4v4 = 0,
for a4 = 0 and the same a1, a2, a3 as above. Since not all of a1, a2, and a3 are zero, that means not all of of a1, a2, a3, a4 are zero. And that means {v1,v2,v3,v4} are linearly dependent.
(b) False. Here’s a simple counterexample: choose v1 = e1 and v2 = v3 = v4 = e2. Then v1 is not a linear combination of {v1,v2,v3}, yet {v1,v2,v3,v4} are linearly dependent (because, for example, 0v1 + 0v2 + v3 −v4 = 0).
(c) True. To show that {v1,v2,v3} is linearly independent, we must show that if a1v1 + a2v2 + a3v3 = 0, then a1 = a2 = a3 = 0 (“only the trivial solution...”). So assume a1v1 +a2v2 +a3v3 = 0. Then a1v1 +a2v2 +a3v3 +0v4 = 0 must also be true (we simply added 0 to the left hand side). But because {v1,v2,v3,v4} are linearly independent, that means all the weights in this expression must be zero. In particular, a1 = a2 = a3 = 0. Which is what we wanted.
3. (25 Points.) Find the general flow pattern of the network shown in the figure.
x3 x4
x1
x5
x2
The flow into each corner must equal the flow out. This gives us the system of equations:
x1+ = x5,
x4 + x5 = x3,
x3 = x2,
x2 = x1 + x4,
which is equivalent to the homogeneous system:
x1 −x5 = 0,
−x3 + x4 + x5 = 0,
−x2 + x3 = 0,
−x1 + x2 −x4 = 0.
Solving this system gives x1 = x5, x2 = x4 + x5, x3 = x4 + x5, x4 = free, and x5 = free.
4. Each of the following two parts are about linear transformations (but are otherwise not related).
(a) (13 Points.) Consider the transformation T : R3 → R2 defined by
T
x1
x2
x3
=
x1x2
x3
.
Is T a linear transformation? You must justify your answer using the definition of linear transformation.
(b) (13 Points.) Assume U : R2 → R3 is a linear transformation where
U
3
0
=
4
3
0
and U
1
2
=
0
2
1
.
Find U
14
4
.
(a) T is not a linear transformation. If it were, it would have to satisfy:
T
c
x1
x2
x3
= cT
x1
x2
x3
.
But the left hand side is equal to
c2x1x2
cx3
, while the left hand side is equal to c
x1x2
x3
. For these two
vectors to be equal, we’d need cx1x2 = c 2x1x2 for all possible choices of c, x1, and x2. Clearly this isn’t true
whenever c 6= 1 and x1 and x2 not both zero: for example, with c = 2, x1 = 1, and x2 = 1 (x3 can be anything), we would not have equality. In short, we have
T
2
1
1
0
= T
2
2
0
=
4
0
,
while:
2T
1
1
0
= 2
1
0
=
2
0
.
(Note that T (u + v) = T (u)+ T (v) is generally not true either.)
(b) We first write
14
4
as a linear combination of
3
0
and
1
2
. This involves solving the system of equa-
tions 14
4
= c1
3
0
+ c2
1
2
.
The solution is: 14
4
= 4
3
0
+ 2
1
2
.
Therefore, because U is a linear transformation, we have:
U
14
4
= U
4
3
0
+ 2
1
2
= 4U
3
0
+2U
1
2
= 4
4
3
0
+2
0
2
1
=
16
16
2
.
Test 2.pdf
Solutions
Name:
Student ID No.:
Discussion Section:
Math 20F Midterm II(ver. a) Winter 2016
Problem Score
1 /24
2 /25
3 /27
4 /24
Total /100
1. (24 Points.) The following are True/False questions. For this problem only, you do not have to show any work. There will be no partial credit given for this problem. For this problem:
• A correct answer gives 4 points.
• An incorrect answer gives 0 points.
• If you leave the space blank, you receive 2 points.
T (a) If a vector space V has p linearly independent vectors, then it’s impossible for a set of fewer than p vectors to span V .
F (b) If A is an m×n matrix whose columns are linearly dependent, then the column space of A must have dimension less than m.
F (c) Let W be the subset of R2 consisting of all vectors
x
y
such that xy ≤ 0. W is a subspace of
R2.
T (d) Let A be an n×n matrix. If there is a ~b such that A~x =~b has no solutions, then A cannot be one-to-one.
T (e) Let A be a matrix. Any column of A that does not have a pivot (when A is put into Row Echelon form) can be written as a linear combination of the columns of A that do have pivots.
F (f) Let C(R) be the vector space of all continuous functions defined on the real line. Let H be the subset consisting of all continuous functions f (x) such that f (3)≥ 0. H is a subspace of C(R).
2. (25 Points.) Given below is the LU -factorization of a matrix A.
A =
1 0 0
2 1 0
−1 −3 1
4 3 2
0 3 −7
0 0 5
.
Solve A~x =~b, where
~b =
9
25
−10
.
In order to receive credit, you must use the LU -factorization to solve this problem.
First solve:
1 0 0
2 1 0
−1 −3 1
y1
y2
y3
=
9
25
−10
.
The solution is:
y1
y2
y3
=
9
7
20
.
Plug this into:
4 3 2
0 3 −7
0 0 5
x1
x2
x3
=
y1
y2
y3
,
and solve again for the final answer:
x1
x2
x3
=
−172
35 3
4
.
3. The following are all proof-type problems. They are otherwise unrelated.
(a) (9 Points.) Let T be a linear transformation from a vector space V to a vector space W . Show that the kernel of T is a subspace of V . You must note every place where you use the fact that T is a linear transformation in order to receive full credit.
(b) (9 Points.) Show that the vector space C(R) of all continuous functions defined on the real line is infinite-dimensional. Write down the statements of any Theorems you use.
(c) (9 Points.) Assume A is a 6×4 matrix and B is a 4×6 matrix. Show that AB is not invertible.
(a) (1) Whenever ~u and ~v are in the kernel of T , we have
T (~u +~v) = T (~u)+ T (~v) =~0 +~0 =~0.
So ~u +~v is also in the kernel of T (“the kernel is closed under addition”). Note that in the first equality we used the fact that T is a linear transformation.
(2) If c ∈ R and ~u is in the kernel of T , then
T (c~u) = cT (~u) = c~0 =~0.
So c~u is in the kernel of T (“the kernel is closed under scalar multiplication”). Note that in the first equality we used the fact that T is a linear transformation.
(1) and (2) together tells us the kernel of T is a subspace of V .
(b) You can write this one up in many different ways, but the general idea is that the vector space of polynomials is infinite dimensional and is a subspace of C(R), therefore C(R) must be infinite dimensional as well (a subspace can’t be “bigger” than the vector space it lives in).
Here’s one way to write it up: If C(R) were finite dimensional (of dimension n), then any set of vectors with more than n vectors must be linearly dependent. (“If V is a finite dimensional vector space, then any set of more than dimV vectors must be linearly dependent”). However, {1,t,t 2,...,t n} is a linearly independent set in C(R) of n + 1 elements. Therefore C(R) can’t be finite dimensional. How do we know the above set is linearly independent? Assume
a0 + a1t +···+ ant n = 0.
For our set to be linearly independent, we need to show each ai = 0. Take the biggest k such that ak 6= 0 (if every ai = 0, then we’re done). Divide through by akt
k , giving us:
a0 ak
1 t k
+ a1 ak
1 t k−1
+···+ ak−1 ak
1 t + 1 = 0.
Choose t so large that every term in the sum above (besides the last term) has absolute value less than 1k (think about why this is possible). Then for that t, the left hand side can’t be equal to zero. So it must be that all the coefficients ai are equal to zero.
Note that you did not have to be so detailed on the exam.
(c) There are many ways to do this one. Here’s one: Think of AB as the composition of two linear transformations: B : R6 → R4, followed by A : R4 → R6. With this point of view, it’s easy to see that if B is not one-to-one, then the composition can’t be either. So the problem boils down to showing that B is not one-to-one. But B has more columns than rows so it can’t be one-to-one.
Here it is in more detail: We’ll show AB is not one-to-one. Because B is a linear transformation from a “bigger” vector space to a “smaller” one, it can’t be one-to-one: More precisely, you can say that its Row Echelon form must have at least 2 columns without pivots so there are many solutions to B~x =~0, so it’s not one-to-one. So there are two different vectors ~v1 and ~v2 such that B~v1 = B~v2. Applying A to both sides, we get AB~v1 = AB~v2. Because ~v1 6= ~v2, AB is not one-to-one, so it can’t be invertible.
Note that you can also do this problem by showing AB is not onto, which boils down to showing A is not onto, which boils down to the fact that A has more rows than columns.
4. The following problems are computational. They are otherwise unrelated.
(a) (12 Points.) Let T : R2 → R2 be a linear transformation that first reflects through the x1-axis, then rotates π4 radians counterclockwise. Find the standard matrix of this linear transforma- tion.
(b) (12 Points.) Determine whether the polynomials 1 + 2x + x2 + x3, −1 + 8x−3x2 + x3, and 2−x + 3x2 + x3. are linearly independent (in the vector space P3).
(a) ~e1 first goes (under reflection) to itself, and then to
√22√
2 2
. ~e2 first goes (under reflection) to −~e2, and then to
√22 − √
2 2
. So the standard matrix is:
√22 √22√
2 2 −
√ 2
2
(a) This is most easily done using coordinates. Using the basis {1,x,x2,x3} we know that our polynomials are linearly independent if and only if the vectors:
1
2
1
1
,
−1
8
−3
1
,
2
−1
3
1
are linearly independent. Putting these into a matrix and row reducing, we get:
1 −1 2
2 8 −1
1 −3 3
1 1 1
→
1 −1 2
0 10 −5
0 −2 1
0 2 −1
→
1 −1 2
0 10 −5
0 0 0
0 0 0
.
There is a column without a pivot, so the homogeneous problem has many solutions, so the vectors (and hence the polynomials) are not linearly independent.
