applied linear algebra: We consider quadratic forms. Let A be an n×n matrix and

Question

applied linear algebra:

We consider quadratic forms. Let A be an n×n matrix and let

the function of n variables defined by q(x1, x2, · · · , xn) = XT AX. This is called a

quadratic form.

a) Show that we may assume that the matrix A in the above definition is symmetric by proving the following two facts. First, show that (A+AT)/2 is a symmetric matrixe. Second, show that XT (A+AT/2)X=XTAX.

b) Show that if A is positive definite (i.e., A has an LDU factorization such that the diagonal elements of D are positive numbers) then for every vector X ?= 0, XT AX > 0.

Assuming that X0 is a critical point of f, i.e., ?f(X0) = 0, show that X0 is a local maximum of f if H(X0) is positive definite and that X0 is a local minimum of f if H(X0) is negative definite (i.e., if ?H(X0) is positive definite).

Explanation / Answer

Quadratic form is defined here q(x1,x2,.....,xn)=XTAX

part(a) first one thing must clear that a matrix A is symmetric if AT=A

(i) we have given that A is symmetric

RTP:- (A+AT)/2 is symmetric

proof:- taking transpose

then [(A+AT)/2]T=[((AT)+(AT)T)/2]=[(A+AT)/2] as we have result AT=A

HENCE [(A+AT)/2] is symmetric

(ii) RTP:- XT[(A+AT)/2]X=XTAX

take LHS XT[(A+AT)/2]X=XT[(A+A)/2]X (SINCE A is symmetric hence AT=A)

so XT[2A/A]X=XTAX hence proved

part(B) given A is positive definite

RTP:- XTAX>0 for every non zero vector X

proof :- since A is positive definite and symmetric , hence it is similar to a diagonal matrix with positive diagonal entries

hence A=PTDP where D is diagonal element with positive entries

now XTAX=XT(PTDP)X=YTAX>0 where Y=XP

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