# 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. – Gyu Eun Lee May 5 '13 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? – symplectomorphic May 5 '13 at 1:04
• @proximal Perhaps OP hasn't shown any effort that's why some one might have downvoted. – srijan May 5 '13 at 12:17

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? – James44 May 6 '13 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). – Ruslan May 8 '13 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 $x=\pm1$. – Andreas Blass May 8 '13 at 21:07
There are $2^{3}$ elements are there. Hint:-$A$=diag($\pm1$,$\pm1$,$\pm1$), counting all, we get 8 elements.