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Question

Answer 1

We know that $Px = \lambda x$

So $(P^2 + 2P + I)x = P^2x + 2Px + Ix =\color{Brown}{ \lambda^2 + 2\lambda + 1}$

$\lambda_1 = -1$

$\lambda_2 = \frac{1}{2}$

$\lambda_3 = 3$

let $\lambda^{'} _1,\lambda^{'} _2,\lambda^{'} _3$ be eigen values of $(P^2 + 2P + I)$

$\lambda^{'} _1 = \lambda _1^2 + 2 \lambda _1 + 1 =  (-1)^2 + 2(-1) + 1 = 1 + 1 -2 = 0$

$\lambda^{'} _2 = \lambda _2^2 +2 \lambda _2 + 1 = (\frac{1}{2}) ^2 + 1 + 1 =\large \frac{9}{4}$

$\lambda^{'} _3 = \lambda _3^2 +2 \lambda _3 + 1 =  3^2 + 2(3) + 1 = 16$

Comments

Cayley-Hamilton Theorem:-Every Square Matrix satisfies its own characteristics equations.