# 3D Hermitian Matrices

Consider two operators represented in a three-dimensional space by matrices

$H=\epsilon\begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}$  and $A=\alpha\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}$

a) Are these operators Hermitian?

b) Demonstrate that they commute.

c) Find a basis of eigenvectors common to the two of them. (Consider which of these operators to use to construct such a basis).

$A$ matrix is Hermitian if $A=(A^T )^*$ in our case since values are real $A=A^T$

$H^T=ϵ\begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}$  and  $A^T=α\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}$  so that they are hermitian

b)

$[H,A]=HA-AH=$

$=ϵα\begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}-αϵ\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&-1\end{pmatrix}=$

$=ϵα\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}-αϵ\begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}=0$

c)

Find eigenvalues and eigen vectors of $A$ (because they commute they share the same eigenvectors). These eigen vectors are a common basis for each $H$ and $A$.

Eigenvalues of $A$

$det⁡(A-λI)=0$  or $\begin{vmatrix}|1-λ&0&0\\0&-λ&1\\0&1&-λ\end{vmatrix}|=λ^2 (1-λ)-(1-λ)=(1-λ)(λ^2-1)=0$

$λ_1=1$ and $λ_2=1$ and $λ_3=-1$

Eigen vectors $X$ are :

$(A-λ*I)X=0$ for each $λ$ found above

$X_1=(0,-1,1)$,  $X_2=(0,1,1)$  and $X_3=(1,0,0)$

Reference Wolfram Alpha