Finding Error in Proof Showing Product of Symmetric Positive Definite Matrices is Positive Definite?

Question

Sorry if this is a duplicate question, but I haven't been able to find this exact question. Is the product of symmetric positive semidefinite matrices positive definite? is close, but for reasons I point out in a comment below, I don't think it provides a clear answer to the questions.

Edit 1: Update! I better understood Robert Israel's answer at the other post. Let me rephrase my question. I have a "proof" below that shows the product of two real, symmetric, positive definite matrices $A, B$ is also positive definite. But this is apparently false. Can someone help me find the error.

Consider real symmetric positive definite matrices $A, B$. This is defined as $A = A^T, B=B^T$ and $\forall x \neq 0, x^T A x >0$ and $\forall x \neq 0, x^T B x > 0$. Choose an eigenvalue-eigenvector pair $\lambda, v$ of the product $AB$:

$$ AB v = \lambda v$$

Left-multiply both sides by $v^T B^T$.

$$ v^T B^T A B v = \lambda v^T B^T v = \lambda v^T B v$$

Since $B$ is positive definite, we know that $v^T B v > 0$. We also know that $B^T A B$ is positive definite. Rearranging, we see that

$$\lambda = \frac{v^T B^T A B v}{v^T B v}$$

Since both the numerator and denominator are positive, $\lambda$ must also be positive. Since this holds for all eigenvalues, the product $AB$ must also be positive definite.

Had to correct a few typing errors! Should be good to go now. — Rylan Schaeffer, Mar 08 '20 at 18:40
Does this answer your question? Is the product of symmetric positive semidefinite matrices positive definite?. See the answer by Plainview. — Dietrich Burde, Mar 08 '20 at 18:42
Close, but not quite. Arseny Nerinovsky's answer cites a paper without explanation, Jog points out that the paper makes specific assumptions, and Robert Israel's answer confuses me since he distinguishes between two different definitions of positive definiteness that I thought were equivalent. — Rylan Schaeffer, Mar 08 '20 at 18:47
@DietrichBurde I better understood the other post. I rephrased the question to ask where my "proof" goes awry. — Rylan Schaeffer, Mar 08 '20 at 18:53

score 1 · Answer 1 · answered Mar 24 '20 at 23:17

1

I would hope somebody would correct me if I'm wrong, but I see no mistake in your proof. It is clearly laid out and each step follows logically. In fact, the exchange that Dietrich Burde linked in the comment seems to back up your assertion as well.

answered Mar 24 '20 at 23:17

PidgeonAl

21

I think one problem might be that I assume $v$ is a non-zero vector. But this might not be true. – Rylan Schaeffer Mar 24 '20 at 23:18
1

Eigenvectors are always non-zero (here zero means the zero vector), so when you write out the eigenvalue-eigenvector pairs of AB, they are always implicitly nonzero. To say that there is an eigenvalue with an eigenvector that is just the zero vector has no meaning. – PidgeonAl Mar 24 '20 at 23:31

Finding Error in Proof Showing Product of Symmetric Positive Definite Matrices is Positive Definite?

1 Answers1