I was given the following problem in class, and I'm not really sure how to begin this proof.
Describe all $3 \times 3$ matrices that are simultaneously Hermitian, unitary, and diagonal. How many such matrices are there?
Here's what I have so far. A Hermitian matrix is a complex matrix that is equal to its conjugate transpose:
$$A \text{ is Hermitian} \Leftrightarrow A=A^*$$
A unitary Matrix is a complex matrix whose conjugate transpose equals its inverse:
$$A \text{ is Unitary} \Leftrightarrow A^*=A^{-1}$$
A diagonal matrix is a matrix where the entries outside the main diagonal are all zero.
$$A \text{ is Diagonal} \Leftrightarrow a_{i,j}=0 \rightarrow i\ne j \ \forall i,j \in \{1,2,..,n\} $$
To satisfy all conditions we can say that a $3 \times 3$ matrix $A$ is simultaneously Hermitian, unitary, and diagonal when:
$$A = A* = A^{-1} \text{ where } a_{i,j}=0 \rightarrow i\ne j \ \forall i,j \in \{1,2,3\}$$
Best Answer
Hint : Here I have done for $2 \times 2$ matrix.
Let $A = \left( \begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} \right)$
be a diagonal matrix with complex entries. Its eigenvalues are precisely $a$, $b$. Because $A$ is Hermitian, they must be real. Also $A$ is unitary, they must each be of absolute value $1$. There are exactly four matrices satisfying these conditions:
Let $A_1 = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$, $A_3 = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_4 = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} \right)$
I hope this may help you.