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I have a question regarding the real part of some matrix A, defined as $$ Re\{A\} = \frac{1}{2}\left(A + A^\dagger \right).$$ Where $A^\dagger$ denotes the Hermitian conjugate. One can also assume that the real part is positive semi-definite, i.e. $Re\{A\} \geq 0$ .
Suppose I were to apply a similarity transformation with $S > 0$, and $S$ Hermitian to A as $S^{-1}AS$, what can be said about the real part of this similar matrix? Are there any bounds known of the form $$ Re\{S A S^{-1}\} \leq c(S) Re\{A\} $$ for some constant $c(S)$ that can depend on the matrix S? I would guess the constant $c(S)$ is somehow related to the largest eigenvalue of $S$.

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When is a matrix $A$ smaller than a matrix $B$? Do you mean a norm of $\text{Re}\{M\}$? –  draks ... Apr 3 '12 at 9:04
    
Hi draks, Sorry I should have been more clear. No I meant in terms of the partial order for matrices, i.e. matrix $A \geq B$ if the difference between $A - B \geq 0$ is positive semi-definite. So to state the problem differently what is the smallest constant $c(S)$ so that for all vectors $\psi$ we have that $(\psi ,(c(S)Re\{A\} - Re\{SAS^{-1}\})\psi) \geq 0$. –  DrMabuse Apr 3 '12 at 12:04
    
It doesn't really have to be the smallest, any "reasonable" upper bound to $c(S)$ would do. –  DrMabuse Apr 3 '12 at 12:11
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1 Answer 1

There is little hope here, unless I misunderstood your purpose, even for positive Hermitian matrices.

Assume that $A=\begin{pmatrix}a_1 & 0\\ 0&a_2\end{pmatrix}$ for some positive real numbers $a_1$ and $a_2$ and that $S=\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$. Then $SAS^{-1}=\begin{pmatrix}a_2 & 0\\ 0&a_1\end{pmatrix}$ hence $\text{Re}(A)=A$ and $\text{Re}(SAS^{-1})=SAS^{-1}$ but the smallest $c$ such that $SAS^{-1}\leqslant c\cdot A$ in the sense of the Hermitian matrices is $c=\max\{a_1/a_2,a_2/a_1\}$ hence there can exist no finite $c=c(S)$ independent on $A$ such that the upper bound you are interested in holds for every $A$.

If non invertible matrices are allowed things are even simpler: consider the example above with $a_1=1$ and $a_2=0$.

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