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.



Matrix transformation that always works?
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 $TAT^{-1}=B$. $\quad\Box$




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In fact, one can see that, if \eqref{1} holds for all $a,b\_1,b\_2,c\_1,c\_2$, then necessarily $T=0$.


