All Eigenvalues of the operator $L(v)= L^2(v).$ Let $L: \Bbb R^n \rightarrow \Bbb R^n $ be a linear operator with the property $L(v)= L^2(v).$ Then find all the eigenvalues of the operator $L$.
My attempt:
$L(v)= L^2(v)=L^3(v)=L^4(v)$ and so on.
So L has to define a matrix such that the result of any exponent on the matrix will result to the matrix itself. So, the matrix could be the identity matrix or any other matrix that satisfies this.
After this, I don't know what to do.
 A: Hint. If $\lambda$ is an eigenvalue of $L$ with corresponding eigenvector $v \in \mathbf R^n$, then 
$$ \lambda v = L(v) = L^2(v) = \lambda^2 v \iff (\lambda^2 - \lambda)v = 0 \iff \lambda^2 - \lambda = 0$$
A: Write down the eigenvalue equation:
$$Lv = \lambda v$$
Then apply what you know. On one hand:
$$
L^2 v = L(Lv) = L(\lambda v) = \lambda Lv = \lambda^2 v
$$
and on the other
$$
L^2 v = L v = \lambda v
$$
so that in particular, $\lambda^2 = \lambda$.
What possible solutions are there to this equation?
A: If $\lambda$ is an eigenvalue then there is $v\neq 0$ such that
$$
\lambda v=L(v)=L^2(v)=L[L(v)]=\lambda L(v)=\lambda^2 v\implies\lambda(1-\lambda)v=0.
$$
Because $v\neq 0$, this means $\lambda$ is either $0$ or $1$. An operator such that $L$ is called idempotent: $L=L^2$. The identity transformation is idempotent but there are non-identity idempotent transformations: let $X$ be $n\times k$ with full column rank, then you can verify that both
$$
P(X)\equiv X(X'X)^{-1}X',\quad M(X)\equiv I_n-P(X)
$$
are idempotent. Moreover, you check that $\text{rank}(P(X))=k$ and $\text{rank}(M(X))=n-k$. In particular, when $0<k<n$, $P(X)$ and $M(X)$ are not $I_n$, which has rank $n$.
