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 $1$.



A matrix norm inequality II
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.


