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Consider the matrix

$$A_2:= \begin{pmatrix} a & b_1 \\ b_2 & a\end{pmatrix}.$$

Let $\sigma_2 = \begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$, then

$$\sigma_2 A_2 \sigma_2 = \begin{pmatrix} a & -b_2 \\ -b_1 & a\end{pmatrix}.$$

I wonder if I have the matrix

$$A_3:= \begin{pmatrix} a & b_1 & 0 \\ b_2 & a & c_1 \\ 0 & c_2 & a \end{pmatrix}$$ if there is an analogous matrix

$\sigma_3$ that works for all possible choices of coefficients in $A_3$ such that

$$\sigma_3 A_3 \sigma_3^{-1} = \begin{pmatrix} a & -b_2 & 0 \\ -b_1 & a & -c_2 \\ 0 & -c_1 & a \end{pmatrix}.$$

That there exists one such matrix for each set of coefficients is clear, since the eigenvalues of the two $3x3$ matrices are the same. I am looking for one that works for all choices.

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  • $\begingroup$ Do you want $\sigma_3$ to be a unitary matrix? $\endgroup$ Commented Apr 17, 2023 at 16:10
  • $\begingroup$ @IosifPinelis actually invertible is fine, but yes, initially I had unitary in mind. Thanks a lot for clarifying this. $\endgroup$ Commented Apr 17, 2023 at 16:12
  • $\begingroup$ You have $\sigma_2 A_2 \sigma_2$ but $\sigma_3 A_3 \sigma_3^{-1}$. Do you want $\sigma_3 A_3 \sigma_3$? $\endgroup$ Commented Apr 17, 2023 at 16:39
  • $\begingroup$ @RobertIsrael Notice that $\sigma_2$ is self-involutive, i.e. $\sigma^2_2=1,$ which is why I am happy with the statement as written in the question. $\endgroup$ Commented Apr 17, 2023 at 16:43
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    $\begingroup$ $\sigma_3 A_3 \sigma_3^{-1}$ won't work: you can explicitly solve the linear system $\sigma_3 A_3 = B_3 \sigma_3$ where $B_3$ is your other $3 \times 3$ matrix and $\sigma_3$ is a $3 \times 3$ matrix of variables, and see that all nontrivial solutions depend on $c_i$ and $b_i$. $\endgroup$ Commented Apr 17, 2023 at 16:48

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The answer is no. Indeed, let $T:=\sigma_3=(t_{ij}\colon i,j\in\{1,2,3\})$, $A:=A_3$, and $$B:=\begin{pmatrix} a & -b_2 & 0 \\ -b_1 & a & -c_2 \\ 0 & -c_1 & a \end{pmatrix}.$$ Then the equality in question would imply that $$TA=BT \tag{1} \label{1}$$ for all $a,b_1,b_2,c_1,c_2$. Solving system \eqref{1} of linear equations for the $t_{ij}$'s, we get $$b_1 t_{11}+c_2 t_{13}+b_2 t_{22}=0,\quad c_1 t_{12}+b_2 t_{23}=0$$ for all $b_1,b_2,c_1,c_2$. It follows that $$t_{11}=t_{13}=t_{22}=0,\quad t_{12}=t_{23}=0,$$ so that $t_{11}=t_{13}=t_{12}=0$ and hence $\det T=0$, which precludes the desired identity $TAT^{-1}=B$. $\quad\Box$

In fact, one can see that, if \eqref{1} holds for all $a,b_1,b_2,c_1,c_2$, then necessarily $T=0$.

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