Positive definite real quadratic forms

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  • #1
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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
===================================
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!:)
 

Answers and Replies

  • #2
CompuChip
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Indeed, it is possible for the product of two non-zero matrices to be zero. But if for a certain fixed matrix the product with any other matrix is zero, then the fixed matrix must be zero (you can then e.g. take the latter matrix the identity, and it follows).

Or, suppose that for any X nonzero there exists Y such that X = P Y, where P is the given orthogonal non-zero matrix.
Now assume that Y = 0. The product with the zero matrix is always zero, so X is zero. This is a contradiction.
 
  • #3
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Indeed, it is possible for the product of two non-zero matrices to be zero. But if for a certain fixed matrix the product with any other matrix is zero, then the fixed matrix must be zero (you can then e.g. take the latter matrix the identity, and it follows).

Thanks, but I don't understand this part that I quoted. Can you explain a bit more on it and how this helps in our situtation?
 
  • #4
CompuChip
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Let me prove it to you in this specific case:

You have
For any X not =0, X=PY
This implies that Y not =0
Here, P is a fixed matrix. That is, once I give you an A, you can in principle give me the corresponding matrix P and there's nothing we can do about it. So, there's no freedom in choosing P.
Now suppose we have any matrix X which is non-zero, and I want to write this as P Y for some matrix Y (which will depend on the entries of X, of course). If P is invertible, I can multiply by P^{-1} on both sides (from the left) and I will get
X = P Y <=> Y = P^{-1} X
The claim was, that this is non-zero if X is non-zero. So, let's suppose it is and derive a contradiction. If Y = 0 then the right hand side is zero, so P^{-1} X is zero. I can safely multiply by P again on both sides, as any matrix times zero equals zero. So then I get
P Y = P P^{-1} X = X = P 0 = 0
so X is equal to zero. This is the contradiction, because we assumed X was non-zero.


Actually, you can extend this argument to show that if A is a fixed non-zero invertible matrix then for any non-zero matrix, A X is also non-zero. (I think it will go wrong if A is non-invertible, but P is orthogonal so (det P)^2 = 1 hence that is not the case in your situation).

I hope I made it clearer now, not more obscure. I don't really know what you want, a heuristic argument or a rigorous proof, but as I think I gave the heuristics in the first post already, I now did the proof :smile:
If it's still not clear, ask again.
 
  • #5
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Thanks, it really helps!
 

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