It is known (see Theorem 4.1.7 in R. Horn & C. Johnson) that every matrix $A\in M\_n(\mathbb R)$ (*real* entries) can be written as the product $HK$ of two Hermitian matrices (*complex* entries). Of course, the pair $(H,K)$ is far from being unique, because the real dimension of $\mathbb H\_n\times\mathbb H\_n$ is $2n^2$, much larger than $n^2=\dim M\_n(\mathbb R)$. The question is whether this factorization can be done in a stable manner:



> 
> Does there exist a finite constant $c\_n$ such that, for every $A\in M\_n(\mathbb R)$, the pair $(H,K)\in\mathbb H\_n\times\mathbb H\_n$ can be chosen so that $A=HK$ and $\|H\|\cdot\|K\|\le c\_n\|A\|$ ?
> 
> 
> 


Of course the answer does not depend on the choice of the matrix norm. Only the constant does.


**Edit**. I must mention, to my shame, that at the beginning of Chapter 6 of my book on matrices (Springer-Verlag, GTM 216), I pretend that $\mathbb H\_n\times\mathbb H\_n$ equals $M\_n(\mathbb C)$; without proof of course. Thanks to Jean Gallier, who pointed it out.



Factorization of a real matrix into Hermitian x Hermitian. Is it stable ?
Surprisingly (at least to me) the answer is no when $n\ge 3$. This was proved by Yves Benoist and me after I mentioned the problem in a talk at MSRI and Yves came up with a great idea. 


It is enough to show that there is no uniform bound when $n=3$.
Here is an elementary argument that involves little computation. I don't pay attention to getting exact constants. 


For small positive $\epsilon$ define
\begin{equation}
 x\_1 = e\_1 \quad \quad 
 x\_2 = e\_1 + \epsilon e\_2 \quad \quad 
 x\_3 = e\_1 + \epsilon e\_3 
\end{equation}`
For certain distinct non zero real $\lambda\_i$, which will depend on $\epsilon$, let $T\in M\_n(\mathbb{R})$ be defined by $Tx\_i = \lambda\_i x\_i$. The dual basis to $(x\_i)$ is defined by
\begin{equation}
 f\_1=e\_1 - {1\over\epsilon} (e\_2 + e\_3) \quad \quad 
 f\_2={1\over\epsilon} e\_2 \quad \quad 
 f\_3={1\over\epsilon} e\_3
\end{equation}


So the transpose and adjoint of $T$ is defined by $T^\*f\_i = \lambda\_i f\_i$. If $S$ implements a similarity between $T$ and $T^\*$, it is more or less clear that $\|S\|\cdot \|S^{-1}\| \to \infty$ as $\epsilon \to 0$ uniformly over all permissible choices of $\lambda\_i$. (For me the easy way to see this is to semi-normalize the $f\_i$ as
\begin{equation}
 \tilde{f}\_1=\epsilon e\_1 - (e\_2 + e\_3) \quad \quad 
 \tilde{f}\_2= e\_2 \quad \quad 
 \tilde{f}\_3= e\_3
\end{equation}
The similarity $S$ must be given by $Sx\_i = a\_i \tilde{f}\_i$ for some non zero $a\_i$ since the $\lambda\_i$ are distinct, and if $\|S\| \vee \|S^{-1}\| \le (1/3) C$, then all $|a\_i|$ and all $1/|a\_i|$ are less than $C$. But $\|x\_2-x\_3\|=\sqrt{2}\epsilon$ and 
$\|a\_2 \tilde{f}\_2 - a\_3 \tilde{f}\_3 \| =\sqrt{|a\_2|^2+|a\_3|^2} > \sqrt{2} /C$, which forces 
$\|S\| >1/(C\epsilon)$.)


Next a trivial but (it seems) important point. For fixed $\epsilon$, you can choose the $\lambda\_i$ close enough to one so that $T$ is as close to the identity as you want. So we can choose $\lambda\_i$ so that $\|T\|=1$ and $\|T^{-1}\|< 1+\epsilon$; denote such a $T$ by $T\_\epsilon$.


Write $T\_\epsilon = H\_\epsilon K\_\epsilon$ with $H\_\epsilon$, $K\_\epsilon$ (complex) Hermitian and $\|H\_\epsilon\|=1$. We want to see that $\|K\_\epsilon\| \to \infty$ as $\epsilon \to 0$. Notice that $H\_\epsilon$ and $K\_\epsilon$ are non singular since no $\lambda\_i$ is zero. So we have
 $H\_\epsilon^{-1}T\_\epsilon H\_\epsilon = K\_\epsilon H\_\epsilon = T\_\epsilon^\*$ and hence, by the first part of the proof, $\|H\_\epsilon^{-1}\| \to \infty$ as $\epsilon \to 0$. But $H\_\epsilon^{-1} = K\_\epsilon T\_\epsilon^{-1}$, so 
 $$\|H\_\epsilon^{-1}\| \le \|K\_\epsilon\| \|T\_\epsilon^{-1}\| \le (1+\epsilon) \|K\_\epsilon\|.
 $$


