Prove that $P^k=P$ for any $k \in \mathbb{N}$ where $1$ is the only eigenvalue of $P$ implies $P=I$.

Question

I'm having trouble proving this, using the fact $P^k=P$.

($P \in L(V)$ where $V$ is a finite-dimensional complex vector space.)

Here's my work (I didn't use $P^k=P$, but it still looks valid to me which is why it's doubtful.)

Suppose $P \neq I$.

$\Rightarrow \exists v \in V$ s.t. $Pv \neq v$.

$\Rightarrow Pv-v \neq 0$.

$\Rightarrow v \notin N(P-I) \subseteq N((P-I)^n)$ where $n =$ dim $V$.

But, since $1$ is the only eigenvalue of $P$, the characteristic polynomial of $P$ is $$q(z)=(z-1)^n$$

So, by Cayley-Hamilton, $q(P)=(P-I)^n=0$

Thus, $q(P)v=0$. And, this is a contradiction.

Hence, $P=I$.

Is there anything I'm missing in this proof?

Answer 1

A proof that uses some of the ideas you tried to:

Let $f(x)$ be the minimal polynomial of $P$. The equation $P^k = P$ implies that $f(x)$ divides $x^k -x$. As you stated, the characteristic polynomial is $(x-1)^n$. Since $f(x)$ also divides the characteristic polynomial, $f(x)$ divides $\text{gcd}(x^k-x\ , \ (x-1)^n)$ which is $(x-1)$. So $P - I = 0$.

Answer 2

More general:

let $V$ be a vector space and $P:V \to V$ linear such that $P^2=P$ and $1$ is the only eigenvalue of $P$.

Let $x \in V$, then

$$x=Px+(I-P)x.$$

It is easy to see that $Px \in ker(I-P), (I-P)x \in ker(P)$ and that $ker(I-P) \cap ker(P)=\{0\}.$ Thus

$$V= ker(I-P) \oplus ker(P).$$

Since $0$ is not an eigenvalue of $P$, we have $ker(P)=\{0\},$ hence

$$V= ker(I-P) .$$

This gives $P=I$