Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $S$.

share|improve this question
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

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 $a_2=0$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.