Question

1. a) Show that the eigenvalues of a hermitian matrix are real.

Answer 1

Answer on Question #85214 – Math – Linear Algebra

Question

1. a) Show that the eigenvalues of a Hermitian matrix are real.

Solution

Let $\lambda$ be eigenvalue of a hermitian matrix $A$. We have:

$$Ax = \lambda x$$

for some $x \neq 0$.

Since $A$ is hermitian,

$$\langle Au, v \rangle = \langle u, Av \rangle \quad (1)$$

for all vectors $u, v$.

Then

$$\begin{array}{l} \langle Ax, x \rangle = \langle \lambda x, x \rangle = \lambda \langle x, x \rangle, \\ \langle x, Ax \rangle = \langle x, \lambda x \rangle = \bar{\lambda} \langle x, x \rangle. \end{array}$$

From (1) we have then:

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

Since $x \neq 0$, $\langle x, x \rangle \neq 0$, from which

$$\lambda = \bar{\lambda},$$

$\operatorname{Re}\lambda + i\operatorname{Im}\lambda = \operatorname{Re}\lambda - i\operatorname{Im}\lambda$,

$\operatorname{Im}\lambda = 0$,

which means $\lambda$ is real.

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