# Is $\begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix}$ similar to $\begin{bmatrix} 0 & CAC^{-1} \\ C^{-1}BC & 0 \end{bmatrix}$ by some transformation?

Consider a matrix with two entries being some operator or matrix $$D=\begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix}.$$ I want to construct another $$2\times2$$ matrix $$S$$ such that $$SDS^{-1} = \begin{bmatrix} 0 & CAC^{-1} \\ C^{-1}BC & 0 \end{bmatrix}$$ where $$C$$ is some operator or matrix.

Is it possible or not?

## 1 Answer

No, it is not possible. The matrices $$D$$ and $$\begin{bmatrix} 0 & CAC^{-1} \\ C^{-1}BC & 0 \end{bmatrix}$$ are not similar in general; they do not even have the same characteristic polynomial in general.

For a counterexample, set $$A=\left( \begin{array}{cc} a & 0\\ 0 & b \end{array} \right)$$ and $$B=\left( \begin{array}{cc} c & 0\\ 0 & d \end{array} \right)$$ and $$C=\left( \begin{array}{cc} 2 & 1\\ 1 & 0 \end{array} \right)$$.

• Thanks for the answer! Do you think it possible with any additional condition? Actually, I am thinking about the case when $A,B$ are simple functions of $x$ and $C=\exp(−\mathrm{i}a\frac{d}{dx})$ is a differential operator with constant a. – xiaohuamao Feb 4 at 23:31
• @xiaohuamao: Your $C$ looks more like a shift operator than a differential operator :) But no, I'M afraid I don't know any conditions that could help. – darij grinberg Feb 4 at 23:37
• Yes, it's the shift operator. Thanks for your opinion. I'll wait for a little while and accept your answer if no other. – xiaohuamao Feb 4 at 23:46