Answer to Question #109981 in Matrix | Tensor Analysis for MS

Question

Show that eigen values of hermitian matrice are real numbers. Expalin.

Answer 1

A complex square matrix $A$ is called hermitian matrix iff $A=A^*$ .

Where $A^*=$ Transpose conjugate of the square matrix $A$ .

Let $\lambda$ be a eigen value of $A$ and $X$ be corresponding eigen vector of $\lambda$ .

Then $AX=\lambda X$ ...............(1).

Premultiplying both side of (1) by $X^*$ ,we get

$X^*AX=\lambda X^*X .................(2)$

Taking conjugate transpose of both side of (2) ,we get

${(X^*AX)}^*={(\lambda X^* X)}^*$

$\implies X^* A^*{(X^*)}^*=\bar{\lambda} X^* {(X^*)}^*$

$\implies X^* A X=\bar{\lambda} X^* X$

$(\because {(X^*)}^*=X \ \text{and} \ A^*=A)$

From equation (2) and (3) we have

$\lambda X^*X=\bar{\lambda}X^*X$

$\implies (\lambda -\bar{\lambda})X^*X=O$

But $X$ is not a zero vector as it is eigen vector ,therefore $X^*X\neq O$$\implies \lambda -\bar{\lambda}=0 \ \text{so that } ,\lambda=\bar{\lambda}$ and consequently $\lambda$ is real.