Question #109981
Show that eigen values of hermitian matrice are real numbers. Expalin.
1
Expert's answer
2020-04-21T15:45:23-0400

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

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

Let λ\lambda be a eigen value of AA and XX be corresponding eigen vector of λ\lambda .

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

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

XAX=λXX.................(2)X^*AX=\lambda X^*X .................(2)

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

(XAX)=(λXX){(X^*AX)}^*={(\lambda X^* X)}^*

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

    XAX=λˉXX\implies X^* A X=\bar{\lambda} X^* X

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

From equation (2) and (3) we have

λXX=λˉXX\lambda X^*X=\bar{\lambda}X^*X

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

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


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