Showing that $I - \alpha P$ has no zero eigenvalues when $P$ is a Stochastic matrix

Question

I am trying to to show:

$$ I - \alpha P$$

is non-singular and $\alpha \in [0,1)$. I know how to do it using nullspace arguments (see this) but I wanted to do it by showing the eigenvalues are NOT zero. In particular I am being told that the condition for this being true is when:

$$ | \lambda_i(P)| \leq 1 $$

which I don't believe as of now. Here is why:

Consider

$$B = I - \alpha P$$

let $u$ be an eigenvector of $P$ so:

$$ B u = (I - \alpha P)u = u - \alpha Pu = (1 - \alpha \lambda_u(P))u$$

So any eigenvalue is:

$$ \lambda(B) = \lambda(I - \alpha P) = 1 - \alpha \lambda_u(P) $$

So if we want the above not to be zero we need:

$$ 1 - \alpha \lambda_u(P) \neq 0$$

which to me doesn't translate clearly to $ | \lambda_i(P)| \leq 1 $. In fact, if the eigenvalue is ever less than $1$ and positive then $ 0 \leq \lambda(I - \alpha) < 1$ and if its negative then we have $\lambda(I - \alpha) \leq 1$ which seem fine conditions. The only thing we want it is to NOT be zero...

Am I missing something?

Answer 1

Largest eigenvalue of P must be $1$, this you can prove hmm.. somehow. It should not be too hard if I recall it right.

1. How would it affect largest eigenvalue of $\alpha P$?
2. How would it affect smallest eigenvalue of $I-\alpha P$?