Question

If y is an eigenvalue of an orthogonal matrix ,show that 1/y is also it's eigenvalue

Answer 1

Let $A$ be an orthogonal matrix of order $n$ Then $AA^t=I_n$ ....(1) and $A$ is non-singular.

Since $A$ is non-singular , $|A|\neq0$ and $y\neq0.$

Since $y$ is an eigen value of $A$ , $|A-yI_n|=0$

$\implies |A-yAA^t|=0$ , from (1)

$\implies |A|.|I_n-yA^t|=0$ $[\because|AB|=|A||B|]$

$\implies |I_n-yA^t|=0$ , since $|A|\neq0$

$\implies (-1)^n.y^n.|A^t-\frac{1}{y}I_n|=0$

$\implies |A^t-\frac{1}{y}I_n|=0$ , as $(-1)^n.y^n\neq0$ ........(2)

Again we know that , $|A^t-\frac{1}{y}I_n|= |A-\frac{1}{y}I_n|^t= |A-\frac{1}{y}I_n|$ $[\because|A^t|=|A|^t and |A^t|=|A|]$

Therefor from (2) we have, $|A-\frac{1}{y}I_n|=0$

This proves that $\frac{1}{y}$ is an eigen value of orthogonal matrix $A.$