If $M:=i\mathbf{\sigma}\cdot \mathbf{v}$, how do I see that $SMS^{-1}=i\mathbf{\sigma}\cdot \mathbf{v}'$

Question

Let $\mathbf{\sigma}=\sigma_1+\sigma_2+\sigma_3$, where the $\sigma_i$ are the Pauli matrices and define:

$$M:=i\mathbf{\sigma}\cdot \mathbf{v}$$

The claim is that if I change $M$ through a similarity change $M'=SMS^{-1}$, then I can find another vector $\mathbf{v}'$ such that $M'=i\mathbf{\sigma}\cdot \mathbf{v}'$. I checked it by explicitly finding the coefficients $\{f^i_1,f^i_2,f^i_3\}$ for each $\sigma_i$ such that

$$S\sigma_iS^{-1} = \sum_j f^i_j\sigma_j$$

For example, if $S=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, then

$$S\sigma_1S^{-1} = (bd-ac)\sigma_1 - i(bd-ac)\sigma_2+ (ab+bc)\sigma_3.$$

My question is: could I have convinced myself that I can find $\mathbf{v}'$ without actually going through the calculation?

Answer 1

Let $M_2$ denote the set of complex matrices of order 2. Notice that $\sigma_1,\sigma_2,\sigma_3$ is a basis for the subspace $W=\{M\in M_2,\ tr(M)=0\}$

Notice that if $M$ is a linear combination of $\sigma_1,\sigma_2,\sigma_3$ then $M\in W$. Thus, $SMS^{-1}\in W$, because $tr(SMS^{-1})=tr(MS^{-1}S)=tr(M)=0.$ Thus, $SMS^{-1}$ is a linear combination of $\sigma_1,\sigma_2,\sigma_3$.