If $\sigma:M_3\rightarrow S_3$ is a linear map such that $\sigma(PMP^{-1})=P\sigma(M)P^{-1}$, then $\sigma(M)=\ldots$ In page 32 of the book A Mathematical Introduction to Fluid Mechanics, by Alexandre Chorin and Jerrold E. Marsden, it is used and commented the following property: let $M_3$ be the space of $3\times 3$ matrices and $S_3$ be the subspace of symmetric matrices. Let $\sigma:M_3\rightarrow S_3$ be a linear map, with the property that $\sigma(PMP^{-1})=P\sigma(M)P^{-1}$ for each orthogonal matrix $P$ and each $M\in M_3$. Then there exist constants $\lambda$ and $\mu$ such that $\sigma(M)=\lambda\text{Trace}(M)I_3+\mu(M+M^T)$. I do not understand the brief proof from the book. Could you provide a detailed proof?
 A: Presumably the matrices are real. Let $R_t$ denotes the $2\times2$ rotation matrix for angle $t$ and let $Q_t=(R_t\oplus1)\in SO(3)$.
We first show that $\sigma(K)=0$ when $K$ is skew-symmetric. We may consider only the case $K=Q_{\pi/2}$, because every $3\times3$ skew-symmetric matrix is orthogonally similar to a scalar multiple of this $K$. Let $S=\sigma(K)$. Then $Q_tKQ_t^\top=K$ and $Q_tSQ_t^\top=S$ for every $t$. Therefore $S=aI_2\oplus b$ for some scalars $a$ and $b$. However, for $D=\operatorname{diag}(1,-1,1)$, we also have $DSD=-S$ because $DKD=-K$. Hence $a=b=0$ and $S=\sigma(K)=0$.
Next, we show that $\sigma(Z)$ is a scalar multiple of $Z$ when $Z$ is a traceless symmetric matrix. Let us consider the case $Z=\operatorname{diag}(1,-1,0)$ first. Let $S=\sigma(Z)$. Then $\Lambda Z\Lambda=\Lambda$ and $\Lambda S\Lambda=S$ for every diagonal orthogonal matrix $\Lambda=\operatorname{diag}(\pm1,\pm1,\pm1)$. Hence $S$ is a diagonal matrix. Yet, we also have $Q_{\pi/2}ZQ_{\pi/2}^\top=-Z$ and $Q_{\pi/2}SQ_{\pi/2}^\top=-S$. Thus $S$ is a scalar multiple of $Z$. Let $S=\sigma(Z)=2\mu Z$.
It follows that $\sigma(Z)=2\mu Z$ for every traceless symmetric matrix $Z$, because $\sigma$ is a linear map that preserves orthogonal similarity and every traceless symmetric matrix $Z$ can be written as a finite weighted sum of the form $\sum_i c_i U_i\operatorname{diag}(1,-1,0)U_i^\top$ where each $U_i$ is an orthogonal matrix.
Finally, the given properties of $\sigma$ clearly imply that $\sigma(I_3)=\gamma I_3$ for some scalar $\gamma$. Since every matrix $M\in M_3(\mathbb R)$ can be written as $\frac13\operatorname{tr}(M)I_3+K+Z$, where $K=\frac12(M-M^\top)$ is skew-symmetric and $Z=\frac12(M+M^\top)-\frac13\operatorname{tr}(M)I_3$ is traceless and symmetric, we get
\begin{align}
\sigma(M)
&=\frac13\operatorname{tr}(M)\sigma(I_3)+\sigma(K)+\sigma(Z)\\
&=\frac\gamma3\operatorname{tr}(M)I_3+0+2\mu Z\\
&=\frac{\gamma-2\mu}3\operatorname{tr}(M)I_3+\mu(M+M^\top).
\end{align}
