(adsbygoogle = window.adsbygoogle || []).push({}); Q: Suppose q(X)=(X^T)AX where A is symmetric. Prove that if all eigenvalues of A are positive, then q is positive definite (i.e. q(X)>0 for all X not =0).

Proof:

Since A is symmetric, by principal axis theorem, there exists an orthogonal matrix P such that (P^T)AP=diag{c1,c2,...,cn} is diagonal, where c1,...,cn are eigenvalues of A.

Suppose ci>0 for all i=1,...,n

For any X not =0, X=PY

This implies that Y not =0

=> q=(X^T)AX=[(PY)^T]A(PY)=c1(y1)^2+...+cn(yn)^2 > 0 for all X not =0 since ci>0 and Y not =0

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Now, I don't understand the red part. As far as I know, it's definitely possible for the product of two nonzero matrices to be the zero matrix. Then, how does X not =0, X=PY imply Y not=0 ?

Can someone please explain? Thanks!:)

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# Positive definite real quadratic forms

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