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Let $\|\cdot\|$ be the spectral norm, i.e., largest singular value. The condition number of an invertible complex matrix $X$ is defined as $\kappa(X):=\|X\|\|X^{-1}\|$.

I am able to prove

Proposition Let $A, B$ be $n\times n$ positive definite matrices. If $X$ is an $n\times n$ invertible matrix such that $AXB$ is Hermitian, then \begin{eqnarray*} \|X^{-1}AXB\|\le \kappa(X) \|AB\|. \end{eqnarray*}

In particular, if moreover $X$ in the above proposition is unitary, then $\|X^{-1}AXB\|\le \|AB\|$.

I wonder if $\kappa(X)$ in the above proposition can always be replaced with $1$.

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  • $\begingroup$ Very interesting proposition: I am thinking to a proof, in particular the case that X is unitary. Did you wrote your proof some where(arxive, or some thing like this?)I am thinking for possible motivation for consideration of the assumption $AXB$ is hermitian?I am thinking for some geometric interpretation for the proposition in the unitary case. I am thinking also to the sharpnest of $\kappa (X)$ in your inequality. Very great post! $\endgroup$
    – Ali Taghavi
    Commented Mar 15, 2017 at 20:24
  • $\begingroup$ Does your argument work in arbitrary C* algebra? $\endgroup$
    – Ali Taghavi
    Commented Mar 15, 2017 at 20:31
  • $\begingroup$ Are there there matrices $A,B,X$ with $A, B$ are positive and X is unitary such that $AXB$ is Hermitian but $AB \neq BA$? $\endgroup$
    – Ali Taghavi
    Commented Mar 15, 2017 at 20:43
  • $\begingroup$ @AliTaghavi sure, for example you can find a unitary matrix $X$ such that $AXB$ is equal to the geometric mean of $A^2$ and $B^2$, usually denoted by $A^2\sharp B^2$. $\endgroup$
    – M. Lin
    Commented Mar 17, 2017 at 9:05
  • $\begingroup$ @AliTaghavi I emailed you a proof of the proposition. $\endgroup$
    – M. Lin
    Commented Mar 17, 2017 at 9:05

1 Answer 1

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No the norm of the left side can be very large.

For example $\left\| X^{-1}AXB \right\| $ is an unbounded function in $(x,y)$ where $A,X,B$ are the following matrices:

Put $A=B= \begin{pmatrix} 1&0\\0&2 \end{pmatrix}$ and $X^{-1}=\begin{pmatrix} -x^3&y\\y&x \end{pmatrix}$

where $(x,y) \in \mathbb{R}^{2} \setminus \{0\}$.

In this example $A,B$ are positive matrices and $AXB$ is equal to its transpose so is a Hermitian matrix.

This shows that the inequality $\left\| X^{-1}AXB \right\| \leq \left\| A B \right\| $ is not necessarily true. Because the left side is unbounded function in $(x,y)$ but the right side is a constant function.

The left side is unbounded because of the following argument:

Put $C=X^{-1}AXB$ then $C_{21}=xy/x^{4}+y^2$. Then $|C_{21}|$ tends to $+\infty$ when $y=x^2$ and $x \to 0$. So $\parallel C \parallel_{\infty} = Max |C_{ij}|$ goes to infinity. But the later norm is equivalent to the spectral norm.

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  • $\begingroup$ can you explain why $\parallel X^{-1}AXB \parallel$ is unbounded in your example? $\endgroup$
    – M. Lin
    Commented Mar 13, 2017 at 8:24
  • $\begingroup$ In this example, the left hand side is definitely not unbounded. $\endgroup$
    – Liam Baker
    Commented Mar 13, 2017 at 8:42
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    $\begingroup$ @M.Lin I explain the reason in the revised version of the answer. $\endgroup$
    – Ali Taghavi
    Commented Mar 13, 2017 at 8:54
  • $\begingroup$ @M.Lin you are well come. $\endgroup$
    – Ali Taghavi
    Commented Mar 13, 2017 at 9:53

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