Let $V$ be a finite dimensional vector space over a field $F$. Let $T$ be a linear transformation on $V$ satisfying $T^2 = T$. Find its eigenvalues. 
Let $T \in \mathcal{L}(V)$, and suppose that $\lambda$ is an eigenvalue of $T$. Then for some $v \in V$ with $v \neq 0$, $Tv = \lambda v$. 

But then, this implies that: $T^2 v = T(Tv) = T(\lambda v) = \lambda Tv = \lambda^2 v$. However, since $T^2 = T$, this must hold for all vectors in $V$. 
In particular, $Tv = T^2 v \implies \lambda v = \lambda^2 v \iff \lambda^2 = \lambda$. Hence, we must have that $\lambda_1 = 1$ and $\lambda_2 = 0$.
Have I done something wrong??  
 A: It is correct that if you have an eigenvalue $\lambda$, then it is either $0$ or $1$. However, this is the most you can say; consider when $T$ is the identity transformation. Then $T^2=T$ but $0$ is not an eigenvalue. There are actually linear transformations that have no eigenvalues, so the first statement I mentioned is really all you can say.
A: $0$ and $1$ need not both arise as eigenvalues (consider the cases when $T$ is the identity and when it is $0$), but $0$ and $1$ are indeed the only possible eigenvalues. Your calculation is fine, just be careful because it only shows that $0$ and $1$ are the only possible eigenvalues, not that both actually occur (and the calculation holds for all eigenvectors in $V $, not all vectors). You could also see the result by observing that the minimal polynomial of $T $ divides $X^2 -X $.
A: hint :
use the fact that  the polynomial $X^2-X$ vanish $T$ and splits with simple roots , so T is diagonalisable , thus the eigenvalues of $T$ are the roots of this polynomial 
conclusion : the eigenvalues are $0$ and $1$
A: Observe that $V= \ker(T) \oplus \ker(I-T)$ and $T(V)= \ker(I-T)$. Hence we have:


*

*$0$ is the only eigenvalue of $T$  iff $T=0$

*$1$ is the only eigenvalue of $T$  iff $T=I$

*$0$ and $1$ are eigenvalues of $T$ iff $0 \ne T \ne I$
