any non-zero vector in V can be expressed as a linear combination of eigenvectors for the eigenvalues 1 and −1. Let $V$ be a ﬁnite-dimensional vector space and let $T, P$ be linear
operators on $V$ such that that $T^2 = I$ and $P^2 = P$.
(b) Show that any non-zero vector in $V$ is either an eigenvector for $T$
with eigenvalue −1 or can be expressed as a linear combination
of eigenvectors for $T$ with the eigenvalues 1 and −1.
In part (a),I have already shown, $v−T(v)$ is either an eigenvector
for $T$ with eigenvalue −1 or the zero vector. And $v+T(v)$ is either an eigenvector
for $T$ with eigenvalue 1 or the zero vector.
What's the point of the question? any non-zero vector $v$ can be expressed as a linear combination of 
$v-T(v)$ and $v+T(v)$:
$v= (v-T(v))/2+(v+T(v))/2$ ,why we have the "either" here? Do I need to show uniqueness?
Also, what can we say about eigenvalues of $P$?
Thx in advance~
 A: $P$ has $1$ as an eigenvalue if $P \neq 0$. Notice that $P^{2}=P$ so $P^{2}(v)=P(v)$, therefore $P(v)$ is an eigenvector with eigenvalue $1$. Now let $v$ be any other eigenvector of $P$ then $P(v)=\lambda v$ for some $\lambda$. Thus $P^{2}(v)=\lambda P(v)$ which implies $\lambda =1$. I can also have $0$ as an eigenvalue if it is not injective. 
A: If v-T(v)=0 or if v+T(v)=0, then v itself is an eigenvector of T.  For example, if v-T(v)=0, then T(v)=v, which means that v is an eigenvector of T with eigenvalue 1.  Analogously, if v+T(v)=0, then T(v)=-v in which case v is an eigenvector of T with eigenvalue -1. This means you've now treated all possible cases.  
As for P, we see that it satisfies x^2-x=0.  This means that the minimum polynomial of P must divide x^2-x.  Thus the only possibilities for the min pol are x, x-1, or x^2-x.  At this point we can say that 0 and 1 are the only possible eigenvalues for P.  (Note that if P is neither I nor O, then 0 and 1 both occur as eigenvalues of P.)
A: You are right, the question should have been to show that any (non necessarily non-zero) vector is a linear combination of eigenvectors for $-1$ and $1$, under the assumption that such eigenvectors exist at all. And it would have been better to ask to show that the space is the direct sum of the eigenspaces for $-1$ and $1$ (eigenspaces do contain $0$, and can be taken to contain only $0$ in for the "eigenspace" asssociated to a non-eigenvalue). Or equivalently that $T$ is diagonalisable with eigenvalues contained in $\{-1,1\}$. And the argument you gave indeed already shows that.
