On Hermitian matrices

Let $A \in \mathbb{C}^{n\times n}$ be a Hermitian matrix. Prove the following:

a) All the diagonal elements of the matrix $A$ are real.
b) All eigenvalues of $A$ are real.
c) If $\lambda, \mu$ are two dinscint eigenvalues of $A$ and $X, Y$ are their eigenvectors respectively, then $\langle X, Y \rangle=0$.
Solution



a) A matrix is called a Hermitian if
$$A=A^*$$

where $A^*$ is the transpose conjugate matrix of $A$.

All elements of the Herminitian matrix satisfy the equation $a_{ij} = \bar{a}_{ji}$. Hence all elements of the diagonal are real.

b) Let $\lambda_n$ be an eigenvalue. This means that $AX = \lambda X$.  Hence:

$$ \langle AX, Y \rangle = \langle X, AY \rangle$$

Replacing $AX$ with $\lambda X$ we get that:

$$\lambda \langle X, X \rangle=\langle \lambda X, X \rangle = \langle X, \lambda X \rangle = \bar{\lambda} \langle X, X \rangle$$

but since $X \neq 0$ we get that the wanted result.

c) Let $AX = \lambda X$ and $A Y = \mu Y$. Hence:

$$\langle AX, Y \rangle = \langle X, AY \rangle \Leftrightarrow \lambda \langle X, Y \rangle = \bar{\mu} \langle X, Y \rangle \Leftrightarrow (\lambda - \mu ) \langle X, Y \rangle =0$$

But since $\lambda \neq  \mu$ we get the wanted result.