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.

  • $\begingroup$ Do you think we can get away with $c_n \sim \log n$ for the operator-2 norm? or perhaps even with a fixed constant, or is it just wishful thinking on my part? $\endgroup$
    – Suvrit
    Mar 31, 2011 at 20:00
  • $\begingroup$ Dear Suvrit, I see that you have in mind an MO question about commutators. A famous one indeed. $\endgroup$ Apr 1, 2011 at 5:13
  • $\begingroup$ Dear Denis, I actually did think about commutators in the beginning (that's where the $\log n$ came from), but later while thinking in terms of singular value inequalities, I thought maybe a constant works. But it seems that my intuition is incorrect. Very nice question, and I hope you soon find a nice solution. $\endgroup$
    – Suvrit
    Apr 1, 2011 at 12:23
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    $\begingroup$ I, on the other hand, wish you had not brought it up as it is giving me a headache. :) But I do hope you find a solution quickly; better you than I. $\endgroup$ Apr 1, 2011 at 15:22
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    $\begingroup$ Re: Edit. No proof given; no shame. I have made much wilder and wronger claims (usually with no proofs and, luckily, always minor) in papers. $\endgroup$ Apr 2, 2011 at 16:28

2 Answers 2


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 $$\|H_\epsilon^{-1}\| \le \|K_\epsilon\| \|T_\epsilon^{-1}\| \le (1+\epsilon) \|K_\epsilon\|. $$

  • $\begingroup$ Thanks Bill. I see the main point: if $T$ has real eigenvalues, it is similar to $T^*$. In addition, the similarity can be written thanks to a Hermitian change of basis. There remains to choose a $T$ with such an eigenbasis that the similarity is ill-conditioned. $\endgroup$ Sep 6, 2011 at 20:59
  • $\begingroup$ Right, Denis, except that there is the additional strange point that you apparently need to keep $\|T^{-1}\|$ bounded. $\endgroup$ Sep 7, 2011 at 0:45

Let $H$ be the Hermitians in $M_n$ and for norm one $B$ in $H$ define $T_B$ from $H$ to $M_n$ by $T_B(A)=AB$. Each $T_B$ is open onto its image and the assignment $B \mapsto T_B$ is continuous (by direct checking or because it is linear). From this it is easy to check that the degree of openness of $T_B$ is bounded away from zero for $B$ a norm one Hermitian.

From this it follows easily that such a $c_n$ exists.

(For the purpose of this post: if $T(Ball X) $ contains $a Ball (TX)$ say that the degree of openness of $T$ is at least $a$.)

Getting a good estimate for $c_n$ looks like a nice problem.

EDIT: As Denis so kindly pointed out, this answer is utter nonsense.

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    $\begingroup$ I don't think that the continuity of $B\mapsto T_B$ is enough. The fact that the range of $T_B$ does not have a constant dimension (hence a ``lack of continuity'' certainly creates a problem. $\endgroup$ Mar 31, 2011 at 13:54
  • $\begingroup$ I agree, Denis. Sorry. $\endgroup$ Mar 31, 2011 at 14:12
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    $\begingroup$ I recommend an edit to this question to indicate to non-cognoscenti (like me) that it does not answer the question. (Some people do not read as far as the comments, and it is best to herald this answer with such an edit so the reader can make an informed decision as to whether to read further. Deleting the answer (while within your right, Bill) might rob someone else of the opportunity to learn from this line of thought, however mistaken it might be. Gerhard "Ask Me About Keeping Mistakes" Paseman, 2011.09.06 $\endgroup$ Sep 6, 2011 at 21:02
  • $\begingroup$ Excuse, I mean "I recommend an edit to this answer..." . Gerhard "Ask Me About Making Mistakes" Paseman, 2011.09.06 $\endgroup$ Sep 6, 2011 at 21:03
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    $\begingroup$ @Gerhard: done. $\endgroup$ Sep 6, 2011 at 23:31

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