Eigenvalues and eigenspace when $T^2=1$ I'm trying to find the eigenvalues and the eigenspaces of the branch of transformations T that satisfies $T^2=1$ (the identity).
For $\lambda$ to be an eigenvalue we must have $Tx=\lambda x$.
So: $T^2x=T(T(x)=T(\lambda x)=\lambda T(x)=\lambda^2x=x$. 
Therefore $\lambda^2=1$ and so either $\lambda=1, \lambda=-1$ or both.
I suppose that if T is itself the identity then $\lambda=1$ is an eigenvector of multiplicity n (n being the dimension of the vectorspace).
I do not know what I can say about $\lambda=-1$ as an eigenvector or how I can say anything about the corresponding eigenspaces.
 A: Indeed, $T$'s eigenvalues are $1$, $-1$, or both.  It turns out that $T$ is necessarily diagonalizable, which is to say that every vector can be written as a linear combination of two eigenvectors.  That is, the eigenspaces are such that if $T:V \to V$, then $V$ is the direct sum of the two eigenspaces.
We can get even more precise: every vector $v$ can be written in the form
$$
v = \frac 12 \left(v  + Tv\right) + \frac 12 \left(v  - Tv\right)
$$
$\frac 12 \left(v  + Tv\right)$ will always be an eigenvector associated with $1$, and $\frac 12 \left(v  - Tv\right)$ will always be an eigenvector associated with $-1$.

Note that these conditions are as much as we can say: $T$ is diagonalizable with eigenvalues $1,-1$ if and only if $T^2 = \mathbf 1$.
A: You can't. Any combination of multiplicity $k$ for $1$ and multiplicity $h$ for $-1$ (with $k+h=n$) can be obtained. In the simplest of cases, by taking $T$ to be the diagonal matrix with diagonal 
$$
\overbrace{1,\ldots,1}^{k\text{ times}},\overbrace{-1,\ldots,-1}^{h\text{ times}}.
$$
A: You might want to read about minimal polynomials for matrices (if you are in finite dimensions). Here $(T-1)(T+1)=0$ and $1$ and $-1$ are simple zeroes of the minimal polynomial $(\lambda-1)(\lambda+1)$. This implies that $T$ is diagonalizable, that there are two complementary subspaces $U$ and $V$ spanning the full space and that $Tx=x$ for $x\in U$ and $Tx=-x$ for $x\in V$ 
