I answered this problem how to raise a matrix to a higher power awhile ago, but I do not know if it is related. I am reading about this problem on the GRE math subject test, and it expects me to know the following given a matrix $P$:
If $P^2=P$, then $P$ is diagonalizable.
I looked up the definition of a diagonizable matrix which is the following:
We say $P$ is diagonizable iff $\exists$ an invertible matrix $A$ ST $A^{-1}PA=R$ where $R$ is some matrix which has all the values zero except its diagonal.
The solution is number $37$ on page $30$ here.
$\textbf{Question:}$ Is there a more concise/efficient way to solve this problem? I have never heard of the word diagonizable anywhere in my undergraduate classes. I am having a hard time going through the solution especially knowing the time restraints are less than $3$ minutes to solve this problem.
If $v\in\mathbb{R}^n$, then$$v=\bigl(v-P(v)\bigr)+P(v).$$Besides, $P\bigl(v-P(v)\bigr)=0$ and $P\bigl(P(v)\bigr)=P(v)$. It follows from this $\mathbb{R}^n$ is the direct sum of the eigenspace that corresponds to the eigenvalue $0$ with the eigenspace that corresponds to the eigenvalue $1$. So, there is a basis $\mathcal{B}=(e_1,\ldots,e_k,e_{k+1},\ldots,e_n)$ of $\mathbb R$ such that$$P(e_j)=\begin{cases}0&\text{ if }j\leqslant k\\1&\text{ otherwise.}\end{cases}$$The matrix of $P$ with respect to $\mathcal{B}$ is clearly diagonal.