# Every matrix is a product of two symmetric matrices

Let $$\mathbb{F}$$ be a field with char $$\mathbb{F} \neq 2$$. Let $$A \in M_n(\mathbb{F})$$. Does there exist symmetric matrices $$B,C \in M_n(\mathbb{F})$$ such that $$A=BC$$?

The answer is yes when $$\mathbb{F} = \mathbb{C}$$ or $$\mathbb{R}$$. But how about other fields?

• I'm wondering who has voted to close this. Questions do not get more interesting than this. Feb 27 at 17:12
• @BrauerSuzuki I did not vote to close but to me the issue with lack of context makes sense. Useful context would be for instance where a proof for $\mathbb{C}$ or $\mathbb{R}$ fails for a general field. Mar 6 at 10:43
• @ronno no, that would just bloat the question. As written, it provides enough context. Mar 6 at 14:24
• @ronno I thought this site is for people to ask questions and not to educate everybody. I don't even read questions when they are longer than 10 lines. Mar 6 at 17:21
• @BrauerSuzuki I did not have educating everybody in mind. "How does the proof work over $\mathbb{R}$" is probably the first thing I would ask if a colleague were to ask me this question. But I don't know if that would actually be helpful and there seem to already be good answers to the question so I did not vote to close. You are of course free to have your own standards for reading and voting on questions. If you want to discuss the purpose of the site, that's better suited for meta and/or chat. Mar 6 at 17:39

This is true over all fields, including those of characteristic two. Moreover, one of the two symmetric matrices can be taken to be nonsingular. Let $$A=P^{-1}CP$$ where $$C$$ be the Frobenius normal form (a.k.a. rational canonial form) of $$A$$. Suppose for the moment that $$C=SC^TS^{-1}$$ for some symmetric matrix $$S$$. Then \begin{align*} A &=P^{-1}CP\\ &=P^{-1}SC^TS^{-1}P\\ &=P^{-1}S(PAP^{-1})^TS^{-1}P\\ &=\underbrace{P^{-1}S(P^{-1})^T}_{S_1}\,A^T\,\underbrace{P^TS^{-1}P}_{S_1^{-1}}\\ &=S_1A^TS_1^{-1} \end{align*} for some symmetric matrix $$S_1$$. Therefore $$A=S_1S_2$$ where $$S_2=A^TS_1^{-1}$$ is also symmetric because $$S_2^T=S_1^{-1}A=A^TS_1^{-1}=S_2$$. Alternatively, you may write $$A=H_1H_2$$ where $$H_1=S_1A^T$$ is symmetric and $$H_2=S_1^{-1}$$ is symmetric and nonsingular.
So, it suffices to prove that $$C=SC^TS^{-1}$$ for some symmetric matrix $$S_1$$. In turn, by considering the problem blockwise, it suffices to consider the special case where $$C$$ is a companion matrix (with ones on the first subdiagonal) for a polynomial $$f(x)=x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$$. In this case, one may take $$S=\pmatrix{c_1&c_2&c_3&\cdots&c_{n-1}&1\\ c_2&c_3&\cdots&c_{n-1}&1\\ c_3&\vdots\\ \vdots&c_{n-1}&1\\ c_{n-1}&1\\ 1}.$$ Reference: Olga Taussky, The role of symmetric matrices in the study of general matrices, LAA 5:147-154 (1972). From the citations appeared in Taussky’s paper, apparently this result was first mentioned, if not proved, by Frobenius himself.