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

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 "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.


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