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(For this queation, all matrices are real).

According to the ancient paper which I cannot really read fully, since it is in german, any square matrix can be written as a product of two symmetric matrices (one of which is non-singular). If we strengthen the conditions, such that one of the factors must be positive-(semi)definite, what can we say? Any way of characterizing square matrices which can be written as a product of a symmetric and a symmetric positive-semidefinit matrix?

If $A$ is symmetric positive-definite and $B$ is symmetric, then the product $AB$ is similar to a symmetric matrix, so has real eigenvalues. So if any square matrix could be written such, all square matrices would have real eigenvalues, which is absurd. So there must be some restriction.

And, any more recent references for this problem?

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up vote 3 down vote accepted

The only restriction is to be diagonalizable with real eigenvalue. For if $M=PDP^{-1}$ with $P,D$ real and $D$ diagonal, then $M=AB$ with $B=P^{-T}DP^{-1}$ symmetric and $A=PP^T$ is PSD. And conversely, such a product is similar to the symmetric matrix $A^{1/2}BA^{1/2}$, hence is diagonalizable with real eigenvalues.

About the fact that every square real matrix is the product of two Hermitian matrices (complex counterpart of what you mention), see Factorization of a real matrix into Hermitian x Hermitian. Is it stable ? .

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Not quite: see my response for a non-diagonalizable example. – Robert Israel Oct 12 '12 at 6:10

I don't have an answer, but slightly more restrictions.

If $A$ is symmetric positive semidefinite and $B$ is symmetric, $AB = A^{1/2} A^{1/2} B$ has the same characteristic polynomial as the symmetric matrix $C = A^{1/2} B A^{1/2}$, and in particular has real eigenvalues. If $u$ is an eigenvector of $C$ for nonzero eigenvalue $\lambda$ then $A^{1/2} u \ne 0$ and $AB A^{1/2} u = \lambda A^{1/2} u$, so $A^{1/2}$ maps the eigenspaces of $C$ for nonzero eigenvalues to the corresponding eigenspaces of $AB$. In particular, the geometric and algebraic multiplicities of a nonzero eigenvalue of $AB$ must be equal. However, this may not be the case for eigenvalue $0$, e.g. $$ \pmatrix{0 & 1\cr 0 & 0\cr} = \pmatrix{1 & 0\cr 0 & 0\cr} \pmatrix{0 & 1\cr 1 & 0\cr}$$


On the other hand, $AB$ can't have a Jordan block of size greater than $2$ for eigenvalue $0$. In fact, suppose $(AB)^3 v = 0$.
Since $\text{Ker}(A) = \text{Ker}(A^{1/2})$, this says $ A^{1/2} B A B A B v = (A^{1/2} B A^{1/2})^2 A^{1/2} B v = 0$, and since $\text{Ker}((A^{1/2} B A^{1/2})^2) = \text{Ker}(A^{1/2} B A^{1/2})$ we have $(A^{1/2} B A^{1/2})A^{1/2} B v = A^{1/2} B A B v = 0$ and thus $(AB)^2 v = 0$.

Since being of the form $AB$ with $A$ and $B$ symmetric and $A$ positive definite is a similarity invariant, i.e. for any matrix $AB$ of this form and any invertible $S$, $SABS^{-1} = (SAS^T)((S^{-1})^T B S^{-1})$, we see that a necessary and sufficient condition is that the eigenvalues are real and the only Jordan blocks of size greater than $1$ are $\pmatrix{0 & 1\cr 0 & 0\cr}$.

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Robert, your example does not contradict my answer, because your factor is only semi-positive definite, not definite. – Denis Serre Oct 12 '12 at 7:59
The original question seems to ask about positive semi-definite. Also you mention PSD in your answer. – Robert Israel Oct 12 '12 at 16:01

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