Prove that if P is idempotent a $I- \lambda P$ is invertible Let $P\in K^{nxn},P^2=P$ and $\lambda \in K,\lambda \ne 1$. I need to prove that $I- \lambda P$ is invertible.
I'm quite confused with this problem, because I know that if $P^2=P$ and $P\ne I$, then the determinant of P should be $0$, but we have that $(I- \lambda P)^2=I- \lambda P$ , so $I-\lambda P$ should also be idempotent.
That would mean that if $\lambda \ne 0$, the determinant of $I- \lambda P$ should be zero, but something does not seem right for me, what am I doing wrong? How can I find the inverse of $I- \lambda P$ ?
The book says that $(I- \lambda P)^{-1}=I+\frac{\lambda}{1-\lambda}P$ 
Thanks, any help will be appreciated.
 A: Let's assume that $\exists\lambda\neq 1,\, I-\lambda P$ non inversible. This means that $\exists x\in\mathbb{K}^n,\, x\neq 0\land x=\lambda Px$. And this leads immediately by multiplying by $P$ to the left to $Px=\lambda Px$ i.e $(1-\lambda)Px=0$ and because $\lambda\neq 1$ we have $Px=0$ and therefore $x=\lambda Px=0$ a contradiction and we have proven $\forall \lambda\neq 1,\, I-\lambda P$ inversible. This answers the question.
Now if we want to find the inverse denote for $\lambda\lt 1$
$$
Q = \sum_{n=1}^{\infty} (\lambda P)^n = (\sum_{n=1}^{\infty} \lambda^n) P \\
=\lambda(1-\lambda)^{-1}P
$$
Now for $\lambda\neq 1$ let's compute keeping in mind $P^2=P$
$$(I-\lambda P)(I+Q)=(I-\lambda P)(I+{\lambda\over 1-\lambda}P)=I$$
And the inverse is the one given in the book
A: Well I think we could proceed like this:
\begin{equation}
(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=(I-\lambda^{n}P^{n})
\end{equation}
Which if $\lambda<1$ gives you
\begin{equation}
(I-\lambda P)(I+\lambda P+\lambda^{2}P^{2}+\lambda^{3}P^{3}+...)=I
\end{equation}
from there you see that since $P^n=...=P^2=P$ then:
\begin{equation}
(I-\lambda P)(I+P(\lambda +\lambda^{2} +\lambda^{3} +...))=I
\end{equation}
which is just the geometric series 
\begin{equation}
(I-\lambda P)(I+P(\frac{\lambda}{ 1-\lambda}))=I
\end{equation}
Or otherwise
\begin{equation}
(I-\lambda P)^{-1}=(I+P(\frac{\lambda}{ 1-\lambda}))
\end{equation}
A: if $|\lambda| \lt 1$ then we may define
$$
Q = \sum_{n=1}^{\infty} (\lambda P)^n = (\sum_{n=1}^{\infty} \lambda^n) P \\
=\lambda(1-\lambda)^{-1}P
$$
and 
$$
(I-\lambda P)(I+Q) = I
$$
it can be seen by calculation that the restriction on $\lambda$ is not required, as long as $\lambda \ne 1$
A: For $\lambda=0$ the assertion is obvious. Else we can write $$I-\lambda P=\lambda(\lambda^{-1} I -P)$$
Note that the only eigenvaules of $P$ are $0$ and $1$, hence the latter is invertible whenver $\lambda \neq 0,1$.
A: First I want to say that $$(I-\lambda P)^2 = (I-\lambda P)(I - \lambda P) = I - \lambda IP - \lambda PI + \lambda^2 P $$ and not what you concluded.
Secondly, here is a different approach to the problem. Since you are working with a finite dimensional vector space you only need to show that $I-\lambda P$ is injective or surjective to get that it is invertible. I like injectivity, so suppose that $$ x-\lambda Px = y - \lambda Py$$ Then $$x-y = \lambda P(x-y) $$ From there you should be able to show that $x-y=0$ with some manipulations.
