TITLE: Are Jordan "Formable" matrices closed under multiplication?
QUESTION [5 upvotes]: After learning about Jordan Canonical Form, I began thinking about if you have two matrices $A, B \in M_n(\mathbb{R}),$ whether or not their product $AB$ will also have a Jordan Canonical Form. Trivially, if we consider polynomials in $ M_n(\mathbb{C})$ then our polynomial will always have roots in the field. So, from what I understand, this is equivalent to asking whether or not the characteristic polynomial of $AB$ will have real roots.
I began playing around with the $2 \times 2$ case and if we have that $AB$ has a negative determinant then we can't have that the eigenvalues of $AB$ are in $\mathbb{C},$ since they must multiply to the determinant, but since the complex eigenvalues always come in a conjugate pair, that would mean that their product is always positive. I am not sure how I can break down the case if the determinant is positive. 
Finally, my own intuition says matrices which have Jordan form will be closed under multiplication. This is because if we think about what matrices with strictly, real eigenvalues do, they simply reflect and stretch space. So I don't believe it is possible to find two matrices whose product matrix will be a rotation in the plane. 
Also, I couldn't find this question asked elsewhere, but if it has an answer, I'd be happy to read it.

REPLY [5 votes]: Every square matrix $A$ over any field is similar to its own transpose via a non-singular symmetric matrix $S$ (see Taussky and Zassenhaus, On the similarity transformation between a matrix and its transpose).
If $A=SA^TS^{-1}$ for a symmetric $S$, then $SA^T=AS=(SA^T)^T$, i.e. $SA^T$ is symmetric. Similarly for $A^TS^{-1}$. Consequently, every square matrix can be written as a product of two symmetric matrices  (because $A=(SA^T)S^{-1}=S(A^TS^{-1})$) and one of them can be chosen to be non-singular.
It follows immediately that the answer to your question is negative unless $n=1$.