# 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 $P$?

• eigenvector of whom $T$ or $P$? state your problem clearly. – Abishanka Saha Sep 14 '13 at 4:30
• For T, thx Abishanka~ – Bob Sep 14 '13 at 4:46
• Then why did you mention $P$ at all? Does it play any role in this problem – Abishanka Saha Sep 14 '13 at 4:52
• oh, sorry for that, P is about the other parts,just add that. – Bob Sep 14 '13 at 5:02
• The question has a problem for $T=I$: there are no eigenvectors for $-1$, and (even though I can conceive linear combinations of an empty set of vectors) I cannot conceive what a linear combination involving a non-existent vector would be. – Marc van Leeuwen Sep 14 '13 at 7:35

$P$ 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.
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.