Question

A 3x3 matrix P is such that, $P^3=P$. Then the eigen values of P are

• A. 1, 1, -1
• B. 1, 0.5 + j0.866, 0.5 - j0.866
• C. 1, -0.5 + j0.866, -0.5 -j 0.866
• D. 0, 1, -1

Answer 1

The correct option is D 0, 1, -1

$P^3=P$

According to the Cayley-Hamilton Theorem, every square matrix satisfies its own characteristic equation.

$\therefore$ $P^3-P$ = 0

$\Rightarrow$ ${\lambda }^3-\lambda$ = 0

$\Rightarrow$ $\lambda ({\lambda }^2-1)$ = 0

$\Rightarrow$ $\lambda (\lambda -1)(\lambda +1)$ = 0

$\Rightarrow$ $\lambda$ = 0, 1, -1
So eigen values are $0,1,-1$