Economic

Quadratic Forms

Dr Damien S. Eldridge

Australian National University

22 February 2021

D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 1 / 63

Reading Part 1

Anton, H (1987), Elementary linear algebra (fifth edition), John Wiley and Sons, USA: Chapter 6 and Chapter 7 (Section 3).

Chiang, AC (1984), Fundamental methods of mathematical economics (third edition), McGraw-Hill, Singapore: Chapter 11 (Section 3).

Debreu, G (1952), “Definite and semidefinite quadratic forms”, Econometrica 20(2), April, pp. 295–300.

Hicks, JR (1939), Value and capital: an inquiry into some fundamental principles of economic theory, Oxford University Press, Great Britain, February: Mathematical Appendix (pp. 303–328).

Hicks, JR (1946), Value and capital: an inquiry into some fundamental principles of economic theory (second edition), Oxford University Press, Great Britain: Mathematical Appendix (pp. 303–328).

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Reading Part 2

Mandy, DM (2013), “On second order conditions for equality constrained extremum problems”, Economics Letters 121(3), December, pp. 440–443.

Mandy, DM (2018), “Leading principal minors and semidefiniteness”, Economic Inquiry 56(2), April, pp. 1396–1398.

Mann, HB (1943), “Quadratic forms with linear constraints”. The American Mathematical Monthly 50(7), August to September, pp. 430–433.

Samuelson, PA (1947), Foundations of Economic Analysis, Harvard University Press, USA: Mathematical Appendix A (pp. 357–379).

Silbeberg, E, and W Suen (2001), The structure of economics: A mathematical analysis (third edition), Irwin McGraw-Hill, Singapore: Chapter 6.

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Reading Part 3

Simon, CP, and L Blume (1994), Mathematics for economists, WW Norton and Company, USA: Chapter 13 (Section 3), Chapter 16, Chapter 17, Chapter 19, Chapter 21, and Chapter 23.

Sundaram, RK (1996), A first course in optimization theory, Cambridge University Press, USA: Chapter 1 (Section 5).

Takayama, A (1993), Analytical methods in economics, The University of Michigan Press, USA: Chapter 1 (Section 4).

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What is a Quadratic Form? 1

Suppose that

A =

 

a11 a12 · · · a1n a21 a22 · · · a2n

... ...

. . . ...

an1 an2 · · · ann

 

is an (n×n) square matrix whose elements are all fixed parameters (constants) and

x =

 

x1 x2 ... xn

 

is an (n× 1) column vector whose elements are all variables. Consider the function f (x) = xT Ax.

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What is a Quadratic Form? 2

Note that

xT Ax = ( x1 x2 · · · xn

)  

a11 a12 · · · a1n a21 a22 · · · a2n

... ...

. . . ...

an1 an2 · · · ann

   

x1 x2 ... xn

 

= ( x1 x2 · · · xn

)  

∑ni=1 a1ixi ∑ni=1 a2ixi

...

∑ni=1 anixi

 

= n

∑ j=1

( xj

( n

∑ i=1

ajixi

))

Continued on the next slide.

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What is a Quadratic Form? 3

Continued from the previous slide.

xT Ax = n

∑ j=1

( xj

( n

∑ i=1

ajixi

))

= n

∑ j=1

( n

∑ i=1

ajixixj

)

= n

∑ j=1

n

∑ i=1

ajixixj

=

( n

∑ j=1

ajjx 2 j

) +

( ∑ j 6=i

n

∑ i=1

ajixixj

) .

Clearly f (x) = xT Ax is a quadratic function of the variables in the x vector. It is for this reason that xT Ax is known as a quadratic form.

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Symmetric and Non-Symmetric Matrices 1

Suppose that a square matrix A is not symmetric, so that aij 6= aji for at least one (i, j) pair for which i 6= j. We have already shown that

xT Ax =

( n

∑ j=1

ajjx 2 j

) +

( ∑ j 6=i

n

∑ i=1

ajixixj

) .

Note that the second component of the right hand side of this expression includes only terms for which either j < i or j > i. If we collect like terms, we can rewrite this component as

∑ j 6=i

n

∑ i=1

ajixixj = ∑ j<i

n

∑ i=1

(aji + aij) xixj .

Suppose that we let bii = aii and bij = (aji +aij )

2 = bji .

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Symmetric and Non-Symmetric Matrices 2

This yields

xT Ax =

( n

∑ j=1

ajjx 2 j

) +

( ∑ j 6=i

n

∑ i=1

ajixixj

)

=

( n

∑ j=1

ajjx 2 j

) +

( ∑ j<i

n

∑ i=1

(aji + aij) xixj

)

=

( n

∑ j=1

bjjx 2 j

) +

( ∑ j<i

n

∑ i=1

( (aji + aij)

2 +

(aji + aij)

2

) xixj

)

=

( n

∑ j=1

bjjx 2 j

) +

( ∑ j<i

n

∑ i=1

(bji + bji) xixj

)

Continued on the next slide.

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Symmetric and Non-Symmetric Matrices 3

Continued from the previous slide.

xT Ax =

( n

∑ j=1

bjjx 2 j

) +

( ∑ j 6=i

n

∑ i=1

bjixixj

)

= xT Bx,

where

B =

 

b11 b12 · · · b1n b12 b22 · · · b2n

... ...

. . . ...

b1n b2n · · · bnn

 

is a symmetric matrix.

Thus any quadratic form xT Ax in which the matrix A is not symmetric can also be expressed as a quadratic form xT Bx in which the matrix B is symmetric.

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The Definiteness of a Matrix 1

The “definiteness” of a square matrix A is related to the sign of the quadratic form xT Ax when the x vector is not a null vector.

Trivially, when x is a null vector (that is, a vector of zeros), the quadratic form xT Ax must be equal to zero.

The matrix A is said to be “positive definite” if xT Ax > 0 for all x 6= 0. The matrix A is said to be “positive semi-definite” if xT Ax > 0 for all x 6= 0. The matrix A is said to be “negative semi-definite” if xT Ax 6 0 for all x 6= 0. The matrix A is said to be “negative definite” if xT Ax < 0 for all x 6= 0. The matrix A is said to be “indefinite” if xT Ax > 0 for at least one x 6= 0 and xT Ax < 0 for at least one x 6= 0.

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The Definiteness of a Matrix 2

Sometimes the definition of the various types of matrix definiteness can be used to establish the definiteness of a particular square matrix.

But often this is not a particularly convenient method for doing this.

When a square matrix is symmetric, there are two indirect approaches to determining its definiteness that are often much more convenient than attempting to directly employ the definition itself.

One of these indirect methods involves an examination of the “eigenvalues” of the matrix. The other of these indirect methods involves an examination of the “leading principal minors” of the matrix.

But what if a square matrix is not symmetric?

We can always use the technique discussed above to construct a symmetric square matrix that will have an identical definiteness to the original non-symmetric square matrix.

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Eigenvalues and Definiteness 1

Suppose that, in addition to being a square matrix, the matrix A is a symmetric matrix, so that AT = A.

The matrix A will be positive definite if and only if all of its eigenvalues are strictly positive (> 0). The matrix A will be positive semi-definite if and only if all of its eigenvalues are non-negative (> 0). The matrix A will be negative semi-definite if and only if all of its eigenvalues are non-positive (6 0). The matrix A will be negative definite if and only if all of its eigenvalues are strictly negative (< 0). The matrix A will be indefinite if and only if it has both at least one strictly positive eigenvalue and at least one strictly negative eigenvalue.

But what are the eigenvalues of a matrix and how do we find them?

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Eigenvalues and Definiteness 2

The characteristic matrix associated with a square matrix A is defined to be the square matrix (λI −A), where λ is a scalar variable and I is the identity matrix that has the same dimensions as A.

Sometimes the characteristic matrix is defined to be (A− λI). The characteristic polynomial associated with the matrix A is defined to be det (λI −A).

Sometimes the characteristic polynomial associated with the matrix A is defined to be det (A− λI). If A is an (n×n) matrix for some n ∈ N, then the characteristic polynomial will be an nth degree polynomial function of the scalar variable λ.

The characteristic equation associated with the matrix A is defined to be det (λI −A) = 0.

Sometimes the characteristic equation associated with the matrix A is defined to be det (A− λI) = 0.

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Eigenvalues and Definiteness 3

The eigenvalues (λ) of a square matrix (A) are the solutions to the characteristic equation.

Both versions of the characteristic equation will yield the same set of solution values for λ.

If A is a symmetric matrix, then all of its eigenvalues will be real numbers.

This is Part (a) of Theorem 23.16 in Simon and Blume (1994, p. 621).

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Principal Minors and Definiteness 1

Once again, suppose that A is a symmetric square matrix.

Symmetry requires that aij = aji for all i 6= j. In other words, if A is an an (n×n) square matrix, then it takes the form

A =

 

a11 a12 · · · a1n a12 a22 · · · a2n

... ...

. . . ...

a1n a2n · · · ann

  .

The leading principal sub-matrices for A are given by

A1 = (a11) ,

A2 =

( a11 a12 a12 a22

) ,

Continued on the next slide.

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Principal Minors and Definiteness 2

Continued from the previous slide.

A3 =

  a11 a12 a13a12 a22 a23

a13 a23 a33

  ,

and so on and so forth up until

An = A =

 

a11 a12 · · · a1n a12 a22 · · · a2n

... ...

. . . ...

a1n a2n · · · ann

  .

The leading principal minors for A are given by det (A1), det (A2), det (A3), and so on and so forth up until det (An) = det (A).

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Principal Minors and Definiteness 3

The matrix A will be positive definite if and only if det (Ai) > 0 for all i ∈{1, 2, · · · , n}. The matrix A will be negative definite if and only if both det (Ai) < for all odd i and det (Ai) > 0 for all even i.

Using the leading principle minors of a matrix to determine whether or not it is either positive semi-definite or negative semi-definite is slightly more complicated.

Unfortunately, we cannot just directly modify the strict inequalities in the positive definite and negative definite tests to incorporate weak inequalities. Instead, we need the pattern implied by doing just that to hold for all possible permutations of the entries in the matrix. This makes this approach to determining positive semi-definiteness or negative semi-definiteness somewhat cumbersome. Further details can be found in Sundaram (1996, pp. 50–55) and Mandy (2018, pp. 1396–1398).

In any other circumstance, the matrix A is indefinite.

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Example 1 Part 1

Consider the matrix

H =

 

α x

0 0

0 β y

0

0 0 γ z

  .

where x > 0, y > 0, z > 0, α > 0, β > 0, γ > 0 and (α + β + γ) = 1.

Note that H is a symmetric matrix.

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Example 1 Part 2

First, let us see if we can determine the definiteness of the matrix H by using the leading principal minors approach.

Note that the first leading principle minor of H is

det (H1) = det ((α

x

)) =

α

x > 0

because x > 0 and α > 0.

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Example 1 Part 3

Note that the second leading principle minor of H is

det (H2) = det

   

α x

0

0 β y

   

= (α x

)( β y

) − (0) (0)

= αβ

xy − 0

= αβ

xy

> 0 (because x > 0, y > 0, α > 0 and β > 0).

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Example 1 Part 4

Note that the third (and final) leading principle minor of H is simply the determinant of the matrix H itself.

Upon employing a cofactor expansion along the first row of matrix H, we obtain

det (H3) = det (H)

= (α x

) (−1)1+1 det

   

β y

0

0 γ z

    + 0 + 0

= (α x

) (−1)2

[( β

y

)(γ z

) − (0) (0)

]

= (α x

) (1)

[ βγ

yz − 0 ]

.

Continued on the next slide.

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Example 1 Part 5

Continued from the previous slide.

det (H3) = det (H)

= (α x

) (1)

[ βγ

yz − 0 ]

= (α x

)( βγ yz

)

= αβγ

xyz

> 0

because x > 0, y > 0, z > 0, α > 0, β > 0, and γ > 0.

Since det (H1) > 0, det (H2) > 0 and det (H3) > 0, we can conclude that the matrix H is positive definite.

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Example 1 Part 6

Now let us now establish that H is positive definite by using the eigenvalue approach.

The characteristic matrix associated with H is

λI −H = λ

 

1 0 0

0 1 0

0 0 1

 −

 

α x

0 0

0 β y

0

0 0 γ z

 

=

 

λ 0 0

0 λ 0

0 0 λ

 −

 

α x

0 0

0 β y

0

0 0 γ z

  .

Continued on the next slide. D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 24 / 63

Example 1 Part 7

Continued from the previous slide.

λI −H =

 

λ 0 0

0 λ 0

0 0 λ

 −

 

α x

0 0

0 β y

0

0 0 γ z

 

=

 

λ − α x

0 0

0 λ − β y

0

0 0 λ − γ z

  .

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Example 1 Part 8

It is straight-forward to establish that the characteristic polynomial for H is

det (λI −H) = (

λ − α

x

)( λ −

β

y

)( λ −

γ

z

) .

You should establish the validity of this claim as a form of revision of the calculation of determinants.

Thus the characteristic equation associated with H is

det (λI −H) = (

λ − α

x

)( λ −

β

y

)( λ −

γ

z

) = 0.

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Example 1 Part 9

Clearly the eigenvalues for H are λ1 = α x

, λ2 = β y

, and λ3 = γ z

.

Since x > 0, y > 0, z > 0, α > 0, β > 0, and γ > 0, we know that λ1 > 0, λ2 > 0, and λ3 > 0.

Thus we can conclude that H is a positive definite matrix.

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Example 2

Consider the matrix

A =

( 2 3 3 7

) .

Note that A is a symmetric matrix.

The leading principal sub-matrices for this matrix are A1 = (2) and A2 = A.

As such, the leading principle minors for this matrix are

det (A1) = 2 > 0

and

det (A2) = det (A) = (2) (7)− (3) (3) = 14 − 9 = 5 > 0.

Thus we can conclude that A is a positive definite matrix.

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Example 3 Part 1

Consider the matrix

A =

( 2 4 4 7

) .

Note that A is a symmetric matrix.

The leading principal sub-matrices for this matrix are A1 = (2) and A2 = A.

As such, the leading principle minors for this matrix are

det (A1) = 2 > 0

and

det (A2) = det (A) = (2) (7)− (4) (4) = 14 − 16 = −2 < 0.

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Example 3 Part 2

Since det (A1) > 0 and det (A2) = det (A) < 0, we know that A is neither negative definite nor positive definite.

But is it negative semi-definite, positive semi-definite, or indefinite?

Let us find the eigenvalues of A.

The characteristic matrix for A is

λI −A = λ (

1 0 0 1

) − (

2 4 4 7

)

=

( λ 0 0 λ

) − (

2 4 4 7

)

=

( λ − 2 −4 −4 λ − 7

) .

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Example 3 Part 3

This means that the characteristic polynomial for A is

det (λI −A) = (λ − 2) (λ − 7)− (−4) (−4)

= λ2 − 7λ − 2λ + 14 − 16

= λ2 − 9λ − 2.

Thus the characteristic equation for A is

det (λI −A) = λ2 − 9λ − 2 = 0.

Note that the characteristic equation for A is a quadratic equation in the variable λ.

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Example 3 Part 4

Upon applying the quadratic formula to the characteristic equation

for A, we find that the eigenvalues for the matrix A are λ1 = 9+ √ 89

2

and λ2 = 9− √ 89

2 .

Since 92 = 81 < 89 < 100 = 102, we know that 9 < √

89 < 10.

This means that λ1 = 9+ √ 89

2 > 0 and λ2 =

9− √ 89

2 < 0.

Thus we can conclude that the A is an indefinite matrix.

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Example 4

Consider the matrix

A =

( 0 0 0 7

) .

Note that xT Ax = 7x22 > 0 for all (x1, x2) T 6= (0, 0)T .

Thus this matrix is positive semi-definite.

This follows directly from the definition of positive semi-definiteness of a matrix. It is not positive definite because xT Ax = 7x22 = 0 when

(x1, x2) T = (1, 0) and (1, 0) 6= (0, 0).

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Example 5

Consider the matrix

A =

( 0 0 0 −7

) .

Note that xT Ax = −7x22 6 0 for all (x1, x2)T 6= (0, 0)T . Thus this matrix is negative semi-definite.

This follows directly from the definition of negative semi-definiteness of a matrix. It is not negative definite because xT Ax = 7x22 = 0 when

(x1, x2) T = (1, 0) and (1, 0) 6= (0, 0).)

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Example 6 Part 1

Consider the matrix

A =

( 2 2 2 −1

) .

Note that A is a symmetric matrix.

The leading principal sub-matrices for this matrix are A1 = (2) and A2 = A.

As such, the leading principle minors for this matrix are

det (A1) = 2 > 0

and

det (A2) = det (A) = (2) (−1)− (2) (2) = −2 − 4 = −6 < 0.

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Example 6 Part 2

Since det (A1) > 0 and det (A2) = det (A) < 0, we know that A is neither negative definite nor positive definite.

But is it negative semi-definite, positive semi-definite, or indefinite?

Let us find the eigenvalues of A.

The characteristic matrix for A is

λI −A = λ (

1 0 0 1

) − (

2 2 2 −1

)

=

( λ 0 0 λ

) − (

2 2 2 −1

)

=

( λ − 2 −2 −2 λ + 1

) .

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Example 6 Part 3

This means that the characteristic polynomial for A is

det (λI −A) = (λ − 2) (λ + 1)− (−2) (−2)

= λ2 + λ − 2λ − 2 − 4

= λ2 − λ − 6

= (λ + 2) (λ − 3) .

Thus the characteristic equation for A is

det (λI −A) = (λ + 2) (λ − 3) = 0.

This means that the eigenvalues for the matrix A are λ1 = −2 and λ2 = 3.

Since λ1 < 0 and λ2 > 0, we can conclude that A is an indefinite matrix. D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 37 / 63

Example 7 Part 1

Consider the matrix

A =

  1 2 02 4 5

0 5 6

  .

Note that A is a symmetric matrix.

The leading principal sub-matrices for this matrix are A1 = (1),

A2 =

( 1 2 2 4

) and A3 = A.

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Example 7 Part 2

As such, the leading principle minors for this matrix are det (A1) = 1,

det (A2) = (1) (4)− (2) (2) = 4 − 4 = 0

and

det (A3) = det (A)

= (1) (−1)1+1 det ((

4 5 5 6

)) + (2) (−1)1+2 det

(( 2 5 0 6

)) + (0) (−1)1+3 det

(( 2 4 0 5

)) = (1) (−1)2 {(4) (6)− (5) (5)}

+ (2) (−1)3 {(2) (6)− (5) (0)}+ 0

Continued on the next slide. D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 39 / 63

Example 7 Part 3

Continued from the previous slide.

= (1) (−1)2 {(4) (6)− (5) (5)} + (2) (−1)3 {(2) (6)− (5) (0)}+ 0

= (1) (1){24 − 25}+ (2) (−1){12 − 0} = (1) (1) (1)− (2) (−1) (12) = 1 − (−24) = 1 + 24

= 25.

Since det (A1) > 0, det (A2) = 0, and det (A3) = det (A) < 0, we know that A is neither negative definite nor positive definite.

But is it negative semi-definite, positive semi-definite, or indefinite?

Let us find the eigenvalues of A.

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Example 7 Part 4

The characteristic matrix for A is

λI −A = λ

  1 0 00 1 0

0 0 1

 −

  1 2 02 4 5

0 5 6

 

=

  λ 0 00 λ 0

0 0 λ

 −

  1 2 02 4 5

0 5 6

 

=

  λ − 1 −2 0−2 λ − 4 −5

0 −5 λ − 6

  .

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Example 7 Part 5

The characteristic polynomial for A is

det(λI −A) = (λ − 1) (−1)1+1 det ((

λ − 4 −5 −5 λ − 6

))

+ (−2) (−1)1+2 det (( −2 −5 0 λ − 6

)) + 0

= (λ − 1)(−1)2 [(λ − 4)(λ − 4)− 25]

+(−2)(−1)3 [−2(λ − 6)− 0]

= (λ − 1)(1) [ λ2 − 10λ + 24 − 25

] −2(−1) [−2λ + 12]

Continued on the next slide. D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 42 / 63

Example 7 Part 6

Continued from the previous slide.

det(λI −A) = (λ − 1)(1) [ λ2 − 10λ + 24 − 25

] −2(−1) [−2λ + 12]

= (λ − 1) [ λ2 − 10λ − 1

] + 2 [−2λ + 12]

= λ3 − 10λ2 − λ − [ λ2 − 10λ − 1

] − 4λ + 24

= λ3 − 10λ2 − λ − λ2 + 10λ + 1 − 4λ + 24

= λ3 − 11λ2 + 5λ + 25.

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Example 7 Part 7

The characteristic equation for the matrix A is

det(λI −A) = λ3 − 11λ2 + 5λ + 25 = 0.

Note that this is a cubic equation in the variable λ.

While there do not appear to be any “obvious” factorisations of this characteristic polynomial, it is possible to obtain numerical approximations to the eigenvalues of A (that is, the solutions to the characteristic equation) by using Microsoft Excel.

Upon doing this we find that λ1 ≈−1.2395, λ2 ≈ 1.9627, and λ3 ≈ 10.277. A graph of the characteristic polynomial can be found on the next slide.

Since the matrix A has both positive and negative eigenvalues, we can conclude that it is indefinite.

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Example 7 Part 8

-500

-400

-300

-200

-100

0

100

200

300

400

500

-6 -4 -2 0 2 4 6 8 10 12 14

f(Lambda)

f(Lambda)

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Example 8 Part 1

Consider the matrix

A =

  −1 1 01 −1 0

0 0 −2

  .

Note that A is a symmetric matrix.

The characteristic matrix for A is

(λI −A) =

  λ + 1 −1 0−1 λ + 1 0

0 0 λ + 2

  .

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Example 8 Part 2

The characteristic polynomial for A is

det(λI −A) = 0 + 0 + (λ + 2) (−1)3+3 det ((

λ + 1 −1 −1 λ + 1

)) = 0 + 0 + (λ + 2)(−1)6

[ (λ + 1)2 − 1

] = (λ + 2)(1)((λ + 1) + 1)((λ + 1)− 1)

= (λ + 2)(1)(λ + 2)(λ)

= (λ + 2)2λ.

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Example 8 Part 3

The characteristic equation for the matrix A is

det(λI −A) = (λ + 2)2λ = 0.

Clearly the eigenvalues for the matrix A are λ1 = −2, λ2 = −2, and λ3 = 0.

Since the matrix A has both negative eigenvalues and a zero eigenvalue , we can conclude that it is negative semi-definite.

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Example 9 Part 1

Consider the matrix

A =

( −1 −2 4 3

) .

Note that the matrix A is not symmetric.

Before proceeding, let us construct a symmetric matrix B such that xT Ax = xT Bx for all x ∈ R2. We will do this by setting b11 = a11 = −1, b22 = a22 = 3, and b12 = b21 =

( 1 2

) (a12 + a21) =

(−2+4) 2 =

2 2 = 1.

The resulting matrix B is

B =

( −1 1 1 3

) .

Note that the matrix B is symmetric.

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Example 9 Part 2

The leading principal sub-matrices for this matrix are B1 = (−1) and B2 = B.

As such, the leading principle minors for this matrix are

det (B1) = −1 < 0

and

det (B2) = det (B) = (−1) (3)− (1) (1) = −3 − 1 = −4 < 0.

Since det (B1) < 0 and det (B2) = det (B) < 0, we know that B (and hence A) is neither negative definite nor positive definite.

But is it negative semi-definite, positive semi-definite, or indefinite?

Let us find the eigenvalues of B.

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Example 9 Part 3

The characteristic matrix for B is

λI −B = λ (

1 0 0 1

) − ( −1 1 1 3

)

=

( λ 0 0 λ

) − ( −1 1 1 3

)

=

( λ + 1 −1 −1 λ − 3

) .

This means that the characteristic polynomial for B is

det (λI −B) = (λ + 1) (λ − 3)− (−1) (−1)

= λ2 − 3λ + λ − 3 − 1

= λ2 − 2λ − 4.

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Example 9 Part 4

Thus the characteristic equation for B is

det (λI −B) = λ2 − 2λ − 4 = 0.

Note that the characteristic equation for B is a quadratic equation in the variable λ.

Upon applying the quadratic formula to the characteristic equation

for B, we find that the eigenvalues for the matrix B are λ1 = 2+ √ 20

2

and λ2 = 2− √ 20

2 .

Since 42 = 16 < 20 < 25 = 52, we know that 4 < √

20 < 5.

This means that λ1 = 2+ √ 20

2 > 0 and λ2 =

2− √ 20

2 < 0.

Thus we can conclude that the B (and hence A) is an indefinite matrix.

D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 52 / 63

Example 10 Part 1

Consider the matrix

A =

( −α x2

0

0 −(1−α)

y2

) ,

where x > 0, y > 0, and 0 < α < 1.

The leading principal sub-matrices for this matrix are A1 = (−α x2

) and

A2 = A.

As such, the leading principle minors for this matrix are

det (A1) = −α x2

< 0

and

det (A2) = det (A)

Continued on the next slide. D. S. Eldridge (ANU) Quadratic Forms 22 February 2021 53 / 63

Example 10 Part 2

Continued from the previous slide.

det (A2) = det (A)

=

( −α x2

)( −(1 − α)

y2

) − (0) (0)

= α(1 − α) x2y2

− 0

= α(1 − α) x2y2

> 0.

Since det (A1) < 0 and det (A2) = det (A) > 0, we can conclude that A is a negative definite matrix.

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Quadratic Forms On Sets of Linear Constraints 1

Up until now, we have been allowing the x vector to vary over all of Rn (for some n ∈ N) and examining the sign of the quadratic form xT Ax for all such x 6= 0.

In other words, we have been examining the sign of an “unconstrained” quadratic form in x.

In economics, however, we are often faced with situations in which the x vector will be subject to one or more constraints.

Is it possible to determine the sign of a quadratic form for all x vectors that satisfy one or more constraints in addition to the standard constraint that x 6= 0? If so, how can this be done?

In the remainder of these notes, we (at least partially) address these questions for the case in which the x vector is subject to one or more linear constraints (in addition to the standard constraint that x 6= 0).

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Quadratic Forms On Sets of Linear Constraints 2

Let x ∈ Rn be a vector of n real-valued variables, A be an (n×n) matrix of constant real-valued coefficients, C be an (m×n) matrix of constant real-valued coefficients, m ∈ N, n ∈ N, and m < n. Note that xT Ax is a quadratic form in the x vector and Cx is a linear form in the x vector.

We have already seen that if A is not a symmetric matrix, then there exists some other matrix B such that xT Ax = xT Bx for all x ∈ Rn.

As such, we can assume, without loss of generality, that A is a symmetric (n×n) matrix. We will make this assumption for the remainder of these notes.

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Quadratic Forms On Sets of Linear Constraints 3

The linear form Cx will represent the linear constraints that we are imposing on the x vector in addition to the standard restriction that x 6= 0.

To be precise, the set of linear constraints that are being imposed are given by the matrix equation Cx = 0, where 0 is an (m× 1) null vector (that is, a vector consisting entirely of entries that are zeros). It is important that m < n because we do not want the x vector to be completely determined by the additional constraints that are being imposed.

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Quadratic Forms On Sets of Linear Constraints 4

We can use the matrices A and C to form a “bordered matrix” D that takes the partitioned form

D =

  0 C

CT A

  ,

where 0 is an (m×m) null matrix (that is, a matrix consisting entirely of entries that are zeros).

Note that D is a symmetric square matrix.

It is an ((m + n)× (m + n)) matrix. It is symmetric because we have assumed that A is a symmetric matrix.

The theorem on the following two slides is contained within (that is, is part of) Theorem 16.4 in Simon and Blume (1994, p. 389).

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Quadratic Forms On Sets of Linear Constraints 5

Theorem: Consider the quadratic form Q (x) = xT Ax subject to the set of linear constraints Cx = 0(m×1) and x 6= 0(n×1), where x ∈ Rn is an (m× 1) vector of real-valued variables, A is an (n×n) symmetric matrix of constant real-valued coefficients, C is an (m×n) matrix of constant real-valued coefficients, m ∈ N, n ∈ N, and m < n. Form the ((m + n)× (m + n)). symmetric “bordered matrix” (in partitioned form)

D =

  0(m×m) C

CT A

  .

Continued on the next slide.

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Quadratic Forms On Sets of Linear Constraints 6

Continued from the previous slide. 1 The quadratic form Q (x) is negative definite on the constraint set if

det (D) has the same sign as (−1)n and if the last (n−m) leading principal minors alternate in sign.

2 The quadratic form Q (x) is positive definite on the constraint set if the last (n−m) leading principal minors all have the same sign as (−1)m.

3 The quadratic form Q (x) is indefinite on the constraint set if both condition (1) and condition (2) are violated by non-zero leading principal minors.

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Example 11 Part 1

This is Example 16.7 from Simon and Blume (1994, pp. 389–390).

Consider the case where A is the symmetric (4 × 4) matrix

A =

 

1 0 0 −1 0 −1 2 0 0 2 1 0 −1 0 0 1

  ,

and C is the (2 × 4) matrix

A =

( 0 1 1 1 1 −9 0 1

) .

Note that in this case there are four variables (that is, n = 4) and two linear constraints other than the constraint that x 6= 0 (that is, m = 2).

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Example 11 Part 2

The symmetric bordered matrix for this example is the (6 × 6) matrix

D =

  0(2×2) C

CT A

  =

 

0 0 0 1 1 1 0 0 1 −9 0 1 0 1 1 0 0 −1 1 −9 0 −1 2 0 1 0 0 2 1 0 1 1 −1 0 0 1

  .

Since n = 4 and m = 2 in this example, we need to check the signs of the last (n−m) = (4 − 2) = 2 leading principal minors.

The last leading principle minor is det (D6) = det (D). The second-last leading principle minor is det (D5).

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Example 11 Part 3

According to Simon and Blume (1994, p. 390), det (D6) = det (D) = 24 > 0 and det (D5) = 77 > 0.

You should confirm these values for the last two leading principle minors of D for yourself.

Note that (−1)m = (−1)2 = 1 > 0. Thus we have

sign (det (D6)) = sign (det (D5)) = sign ((−1) m ) = sign

( (−1)2

) .

This means that the matrix A is positive definite on the set of constraints given by both Cx = 0 and x 6= 0.

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