Information about $T$ if $T^2=0$ What can we say about $T$ if $T^2=0$? How can we prove that it has zero as an eigenvalue?
 A: If  $T \ne 0$, then there is a non-zero vector $\mathbf v$ such that $T\mathbf v \ne 0$, and then $T(T\mathbf v) = T^2\mathbf v = 0$, showing that $0$ is an eigenvalue of $T$ with eigenvector $T\mathbf v$.  In fact every eigenvalue of $T$ is $0$, since if $T \mathbf v = \lambda \mathbf v$ for nonzero $\mathbf v$, then $0 = T^2\mathbf v = \lambda^2 \mathbf v$, so since $\mathbf v \ne 0$, $\lambda^2 = 0$, whence $\lambda =0$ as well.
More can be said about $T$:  for example, $I \pm T$ is invertible, since $(I + T)(I -T) = I^2 - T^2 = I$.    The list of "what we can say about T" can be on the long side, so having shown $0$ is an eigenvalue, per specific request, I'll leave further facts for those who seek them to discover.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: If $T^2=0$, then $det(T^2)=det(0)=0$.  But as $det(AB)=det(A)det(B)$, we now see that $det(T)^2=0$, so $det(T)=0$.  As $det(T)$ is the product of all eigenvalues for $T$, we see that $0$ is an eigenvalue of $T$. 
A: Doc's solution is great if your vector space is finite-dimensional.  If you have something bigger like a Hilbert space, you can do this:
The kernel of $T^2$ is the whole vector space, so the image of $T$ is contained entirely within the kernel of $T$.  If the kernel of $T$ is just $0$, then the image of $T$ is just $0$, a contradiction.  Hence the kernel of $T$ is nontrivial, and there is therefore a nonzero vector in the eigenspace corresponding to eigenvalue zero (and thus zero is an eigenvalue).
A: Any linear transformation of $V$ to $V$ with nonzero kernel has $0$ as an eigenvalue. If there exists $x\neq 0$ such that $T(x)=0$, then certainly $T(x)=0x$.
There are two cases: 
Case 1: $T(x)=0$ for all $x$, whence any nonzero $x$ is an eigenvector for $0$ or
Case 2: $T(x)\neq 0$ for some $x\neq 0$, whence $T(T(x))=0$, and $T(x)$ is an eigenvector for $0$.
