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:

$$A = A* = A^{-1} \text{ where } a_{i,j}=0 \rightarrow i\ne j \ \forall i,j \in \{1,2,3\}$$

• Whoever downvoted should at least offer some constructive reason. If you aren't familiar with this sort of classification problem, then it might seem confusing at first. You'll have to explore a few examples and come up with your own conjecture, then try to prove it. For now, forget the "unitary" requirement, and consider: if a matrix must be Hermitian and diagonal, then what can we say about its coefficients? Once you have that, adding the unitary condition should be simple enough. Commented May 5, 2013 at 1:02
• do you know what the terms mean? start with the condition of being simultaneously diagonal and hermitian (which means equal to its conjugate transpose). but the transpose of a diagonal matrix is itself, so the matrix must be equal to its conjugate... what does this tell you about the entries on the diagonal? Commented May 5, 2013 at 1:04
• @proximal Perhaps OP hasn't shown any effort that's why some one might have downvoted. Commented May 5, 2013 at 12:17
• Why is no answer accepted? Commented Jan 5, 2020 at 3:22

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 $A_1 = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_2 = \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right)$, $A_3 = \left( \begin{array}{cc} -1 & 0 \\ 0 & 1 \\ \end{array} \right)$, $A_4 = \left( \begin{array}{cc} -1 & 0 \\ 0 & -1 \\ \end{array} \right)$

• Thanks for the help. I have two questions though. Why must they be real if $A$ is Hermitian? Also, why must they be of absolute value 1? Commented May 6, 2013 at 11:06
• @James44 they must be real if $A$ is Hermitian because they are the eigenvalues of $A$. Eigenvalues of Hermitian matrix are always real. As for why they must have absolute value of 1 is that unitary matrix has columns (and rows) with norm of 1 (unit, thus the name). Commented May 8, 2013 at 16:43
• It may be easier to ignore eigenvalues and just work directly with the definitions. "Diagonal" tells you that all entries are $0$ except those on the main diagonal; so it remains to figure out what numbers $x$ can be on the diagonal. The definition of "Hermitian" say that each such $x$ equals its complex conjugate $x^*$, and the definition of "unitary", together with the information that the off-diagonal entries are $0$, gives you $x^*x=1$. Together, these force $x=\pm1$. Commented May 8, 2013 at 21:07
There are $2^{3}$ elements are there. Hint:-$A$=diag($\pm1$,$\pm1$,$\pm1$), counting all, we get 8 elements.