Prove that for $T:V \rightarrow V$ such that $T=T^2$, $T$ is diagonalizable.
My Work -
Let $A_n$ represent the representative matrix of $T$, therefore -
$ \forall \,v\in V$ we can say that $Av = A^2v \iff A^2v-Av = 0 \iff A(Av-Iv)=0$
So either our Matrix A is the zero matrix with is diagonal, or $\forall v\in V$ we can say that $Av - Iv = 0 \iff Av = Iv$.
Now, what we can conclude (?) is that $1$ is an eigenvalue of $T$ for every $v\in V$, so can i say that the geometric and algebraic multiplicity of $1$ is $n$ ?
