If a Matrix Has Only Zero as an Eigen-Value Then It Is Nilpotent 
Prove that a matrix with only zero eigenvalues must be nilpotent.

How will I be able to prove this? 
 A: Hint: Look at the characteristic polynomial, then use Cayley-Hamilton.
A: Hint: $A = P^{-1}DP$ where $D$ is upper triangular. What are the diagonal entries of $D$? If $A$ is $n\times n$, what is $A^n$?
A: Cayley-Hamilton theorem says a linear transformation (equivalently, of course, its matrix) satisfies its own characteristic polynomial. What is the characteristic polynomial of a matrix with only zero eigenvalues?
A: We have to assume that we are considering complex matrices. Over the reals the assertion is not true, as the example
$$
\begin{bmatrix}0&0&0\\0&0&1\\0&-1&0 \end{bmatrix}
$$
shows.
Over $\mathbb C$, one can do the Schur decomposition, where $A=VTV^*$, with $V$ unitary and $T$ upper triangular. Since the diagonal of $T$ has to contain the eigenvalues of $A$, it has be zero. And it is an easy exercise that if $T$ is an $n\times n$ upper triangular with diagonal zero, then $T^n=0$. So $A^n=(VTV^*)^n=VT^nV^*=0$, and $A$ is nilpotent. 
A: If you have learned schur triangularization (or decomposition), note that matrices with all eigenvalues as zero are unitarily similar to "strictly" Upper Triangular matrices. Now see that strictly upper triangular matrices are always nilpotent. Now look at the converse.
