How can a matrix be Hermitian, unitary, and diagonal all at once? 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:
$$A = A* = A^{-1}  \text{ where } a_{i,j}=0 \rightarrow i\ne j \  \forall i,j \in \{1,2,3\}$$
 A: 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.
A: Start with the definitions of Hermitian, unitary, and diagonal.  You should start with the definition of diagonal...
A: 
Describe all 3 by 3 matrices that are simultaneously Hermitian,
  unitary, and diagonal. How many are there?

There are $2^{3}$ elements are there. Hint:-$A$=diag($\pm1$,$\pm1$,$\pm1$), counting all, we get 8 elements.
