definite matrices 13

8/7/2019 Definite Matrices 13

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Professor John NachbarEcon 4111February 19, 2008

Definite Matrices

1 Basic Definition

An N × N symmetric matrix A is positive definite iff for any v = 0, vAv > 0. Forexample, if

A =

a b

b c

then the statement is that for any v = (v1, v2) = 0,

vAv = [ v1 v2 ]

a b

b c

v1v2

= [ v1 v2 ]

av1 + bv2bv1 + cv2

= av21 + 2bv1v2 + cv2

2

> 0

(The expression av21

+ 2bv1v2 + cv22

is called a quadratic form .) An N ×N symmetricmatrix A is negative definite iff −A is positive definite. The definition of a positivesemi definite matrix relaxes > to ≥, and similarly for negative semidefiniteness.

If N = 1 then A is just a number and a number is positive definite iff it ispositive. For N > 1 the condition of being positive definite is somewhat subtle. Forexample, the matrix

A =

1 33 1

is not positive definite. If v = (1,−1) then vAv = −4. Loosely, a matrix is positivedefinite iff (a) it has a diagonal that is positive and (b) off diagonal terms are nottoo large in absolute value relative to the terms on the diagonal. I won’t formalizethis assertion but it should be plausible given the example. The canonical positivedefinite matrix is the identity matrix, where all the off diagonal terms are zero.

2 Inverses of definite matrices.

Theorem 1. If A is positive definite then A is invertible and A−1 is positive definite.

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Proof. If A is positive definite then vAv > 0 for all v = 0, hence Av = 0 for allv = 0, hence A has full rank, hence A is invertible.

For any invertible matrix A, (A−1) = (A)−1. To see this, note that [A(A−1)] =A−1A = I . Hence A(A−1) = I = I . Similarly, (A−1)A = I .

If A is symmetric and invertible then (A−1) = (A)−1 = A−1, hence A−1 issymmetric. It remains to show that A−1 is positive definite. Consider vA−1v forany v = 0. Then vA−1v = vA−1AA−1v = (A−1v)A(A−1v) > 0, where the secondequality comes from the fact that A−1 is symmetric and the last inequality followsfrom the fact that A is positive definite.

Likewise, if A is negative definite then A−1 exists and is negative definite.

3 Checking definiteness.

There is a mechanical check for definiteness of a matrix. It is messy but easy toimplement on a computer.

Given an N × N matrix A, a leading principal submatrix of A is a submatrixformed by deleting all but the first n rows and columns. Thus, if

A =

a11 a12 a13

a21 a22 a23a31 a32 a33

,

then the leading principal submatrices are

a11 ,

a11 a12a21 a22

,

and A itself. A leading principal minor is the determinant of a leading principalsubmatrix.

Theorem 2. A matrix is positive definite iff all of the leading principal minors arepositive.

Proof. Omitted.

Since A is negative definite iff −A is positive definite, and since multiplying ann × n matrix by -1 multiplies the determinant by (−1)n, Theorem 2 implies that amatrix is negative definite iff the principal minors alternate in sign, with the signnegative iff the number of columns in the submatrix of A is odd.

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4 Definiteness on subspaces.

Let B be N ×K where K ≤ N . Consider the set

T = {v ∈ RN : Bv = 0}.

That is, T consists of all vectors that are at right angles to the columns of B. T isa vector subspace in RN : T is a line, plane, or hyperplane though the origin. If thecolumns of B are linearly independent then T has dimension N −K .

If vAv > 0 for all v ∈ T , v = 0, then A is positive definite on T . Definitenesson a subspace shows up in the second order conditions for constrained optimiza-tion. There is a mechanical check for this that parallels that for (full) definiteness.Consider the (N + K ) × (N + K ) bordered matrix ,

A B

B 0

.

A border preserving leading principal submatrix of this bordered matrix is the matrixformed by deleting all but the first n ≥ K rows and columns of A, while retainingthe associated components of the border. For example, if the bordered matrix is,

a11 a12 a13 b11 b12a21 a22 a23 b21 b22a31 a32 a33 b31 b32b11 b21 b31 0 0b12 b22 b32 0 0

,

then the border preserving leading principal submatrices are

a11 a12 b11 b12a21 a22 b21 b22b11 b21 0 0b12 b22 0 0

and the original bordered matrix. The border preserving leading principal minorsare the determinants of these submatrices.

Theorem 3. Let B by an N ×K matrix, with K ≤ N . Let T = {v ∈ RN : Bv = 0}.Assume that B has full rank ( K ). A symmetric matrix A is positive definite on T

iff the border preserving leading principal minors of the associated bordered matrix all have the same sign, with the sign positive iff K is even.

Proof . Omitted.

Thus, if K = 1, which will be the typical case in our applications, all the borderpreserving principal minors must be negative.

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For negative definite matrices, one gets the same result, except that the minorsalternate in sign, with the sign negative iff the number of columns in the submatrixof A is odd. In particular, if K = 1, then the smallest submatrix has one column

and the sign of the corresponding minor is negative. The sign of the minor for thenext smallest matrix, with two columns, is positive, and so on.

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definite matrices 13

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