Just wondering if there is a faster or better way of doing this question.
I have 3 matrices: $A = {1\over2}\hbar\left( \begin{smallmatrix} 0&-i\\ i&0 \end{smallmatrix} \right), B = {1\over2}\hbar\left( \begin{smallmatrix} 0&1\\ 1&0 \end{smallmatrix} \right), C = {1\over2}\hbar\left( \begin{smallmatrix} 1&0\\ 0&-1 \end{smallmatrix} \right), $
Let $D = A^2+B^2+C^2$ and that $v$ is an eigenvector of $C$, (with eigenvalue $-{1\over2}\hbar$).
Question: Show that $v$ is an eigenvector of $D$ with eigenvalue ${1\over2}({1\over2}+1)\hbar^2$.
My take: $D={3\over4}\hbar^2\left( \begin{smallmatrix} 1&0\\ 0&1 \end{smallmatrix} \right)$. $Cv=-{1\over2}\hbar v \implies C^2v={1\over4}\hbar^2 v$.
Since $C^2={1\over4}\hbar^2\left( \begin{smallmatrix} 1&0\\ 0&1 \end{smallmatrix} \right)=B^2=A^2$, Therefore $Dv = 3\times {1\over4}\hbar^2v$. So we are done.
Though I do get the numerical answer, it is not immediately in the form asked for. Could anyone suggest another way of doing this? Thank you.