Cayley-Hamilton Theorem I am trying to prove that all strictly upper triangular $n \times n$ matrices $A$, are nilpotent such that $A^n=0$.  
I am having trouble proving: 

$A$'s eigenvalues are all zero implies that $A^n=0$.  

I have the hint to use Cayley-Hamilton, but I'm unsure how to use Cayley-Hamilton to do so.  Is there some part of Cayley-Hamilton that says that $A^n=$ characteristic polynomial of $A$?
 A: You are on the right track. Let's start from the beginning. We wish to prove:
If $A$ is an $n\times n$ strictly upper triangular matrix, $A^n=0$.
We can prove this statement by finding $A$'s characteristic polynomial. To do this, we must look at $A$'s matrix: 
$$A=
\begin{bmatrix} 0        &          & \huge{\star} \\
                         & \ddots   &              \\
                \huge{0} &          &  0           \\ \end{bmatrix}$$
Let's denote $\text{char poly}_x\{A\}$ as $f(x)$.
$$\Rightarrow f(x) = \text{det }\{x\cdot I-A\} = \left|
\begin{bmatrix} x        &          &   \\
                         & \ddots   &   \\
                         &          & x \\ \end{bmatrix}-
\begin{bmatrix} 0        &          & \huge{\star} \\
                         & \ddots   &              \\
                \huge{0} &          &  0           \\ \end{bmatrix}\right|$$
$$\dots= \left|
\begin{bmatrix} x        &          & -\huge{\star} \\
                         & \ddots   &               \\
                \huge{0} &          &  x            \\ \end{bmatrix}\right|$$
The determinant of an upper triangular matrix is the product of it's diagonals (proofs of this can be found elsewhere on this website) so $f(x)=x^n$.
Recall that the Cayley-Hamilton theorem says that $f(A)=0$.
Now $A^n=0.$
A: If $A$ has eigenvalues $\lambda_1,\dots,\lambda_n$, what is the characteristic polynomial of $A$?
If $\lambda_1 = \cdots = \lambda_n = 0$, what is the characteristic polynomial of $A$?
A: I propose a more elementary solution.
Suppose that $A=(C_1|C_2|C_3|\cdots |C_n)$ where $C_1=0$, $C_2=(c_{12},0,0,\ldots)$, $C_3=(c_{13},c_{23},\cdots)$, &c. Consider now $A$ as a map $T:K^n\to K^n$ with $x\mapsto Ax$. Observe that if $e_i$ is the $i$-th canonical vector then $Ae_1=0$ and for $i>1$, $Ae_i=C_i\subseteq\langle e_{1},\ldots,e_{i-1}\rangle$. Prove then by induction that $A^ie_i=0$ for every $i$. In particular, $A^ne_i=0$ for every $i$, so $A^n=0$.
Hint To see what's going on, note that $A^2e_i=AAe_i\subseteq \langle Ae_1,\ldots,Ae_{i-1}\rangle=\langle 0,Ae_2,\ldots,Ae_{i-1}\rangle$. But  we have $Ae_2\in \langle e_1\rangle$, $Ae_3\in \langle e_1,e_2\rangle,\cdots, Ae_{i-1}\in \langle e_1,\ldots,e_{i-2}\rangle $ so that $A^2e_i\in \langle e_1,\ldots,e_{i-2}\rangle$. Since $Ae_2\in \langle e_1\rangle$, we get $A^2e_2=0$, for $Ae_1=0$. For $i>2$ then  $A^2e_i\in \langle e_1,\ldots,e_{i-2}\rangle$. If $i=3$, we get $A^2e_3\in \langle e_1\rangle$, so $A^3e_3=0$, and so on.
