What are the possible eigenvalues of a linear transformation $T$ satifying $T = T^2$ Let $T$ be a linear transformation $T$ such that $T\colon V \to V$.
Also, let $T = T^2$. What are the possible eigenvalues of $T$?
I am not sure if the answer is only $1$, or $0$ and $1$.
It holds that $T = T^2$, thus $T(T(x)) = T(x)$. Let's call $T(x) = v$, so $T(v) = v$. which means that $\lambda=1$. But I am not sure about this, while I have seen a solution that says that $0$ is possible as well.
Thanks in advance !
 A: Let $v\neq 0$ be an eigenvector of $T$ with eigenvalue $\lambda$, so $Tv=\lambda v$. Using $T=T^2$ we have
$$
Tv = T^2 v = T(Tv) = T(\lambda v) = \lambda(Tv) = \lambda^2 v.
$$
Hence, $\lambda v = \lambda^2 v$. Since $v\neq 0$ we conclude $\lambda = \lambda^2$. The only solutions to this equation are $0$ and $1$.
A: Think of this as follows:
$$T^2=T\implies T(T-I)=0$$
Thus, $\;T\;$ is a root of $\;x(x-1)\;$ and thus the characteristic polynomial of $\;T\;$ can only have $\;0\;$ or $\;1\;$ as its roots, and thus these precisely are the only possible eigenvalues of $\;T\;$ .
So you were half right...:) 
A: Another side remark: You say that you are not sure if 1, or both 0 and 1 can be eigenvalues.
In some cases, it is worthwhile to think of specific examples and see what they can tell us. So what are some examples of matrices $T$ that satisfy $T^2 = T$?
Well, the identity is certainly one, and its eigenvalues are all 1.
However, another such matrix is the zero matrix! It also trivially satisfies $\mathbf{0}^2 = \mathbf{0}$. Its eigenvalues are all zero, so zero can certainly be an eigenvalue as well.
Anyhow, this of course just tells you that both 0 and 1 are possible eigenvalues of such a matrix, but not that they are the only possible eigenvalues. For that, the other answers provide a full solution.
