How to prove $A^n = 0$ if $A$ is an $n\times n$ strictly upper triangular matrix I would like to prove the statement on the title. I tried to prove it using the definition of multiply of matrices, however it seems like weak to me. Can't decide how to make a correct proof of the statement above.
Edit: Those ns are the same ns. So that, the dimension of the Matrix have to match the power to which it is raised.
 A: Write $a^{(k)}_{ij}$ for the $(i,j)^{\text{th}}$ component of $A^k$. We'll prove by induction on $k \ge 1$ that $a_{ij}^{(k)}=0$ whenever $i \le j + k - 1$. Note that when $k=1$, this says precisely that $A$ is strictly upper-triangular, and so the base case is trivial.
Now for the induction step, fix $k \ge 1$ and suppose that $a^{(k)}_{ij} = 0$ when $i \le j + k - 1$. Fix $i,j$ and suppose that $i \le j + k \ (=j+(k+1)-1)$. Then
$$a^{(k+1)}_{ij} = \sum_{\ell=0}^n a^{(k)}_{i\ell} a_{\ell j}$$
Now we know that $a_{\ell j} = 0$ whenever $\ell \le j$ so that
$$a^{(k+1)}_{ij} = \sum_{\ell=j+1}^n a^{(k)}_{i\ell} a_{\ell j}$$
Moreover, if $j+1 \le \ell \le n$ then
$$\ell+k-1 \ge (j+1)+k-1 = j+k \ge i$$
so that $a^{(k)}_{i\ell} = 0$ by the induction hypothesis. Hence $a^{(k+1)}_{ij} = 0$, as required.
Applying this when $k=n$ gives $a^{(n)}_{ij} = 0$ whenever $i \le j+n-1$. But this is always true, since $i \le n$ and $j+n-1 \ge n$ for all $1 \le i,j \le n$. So indeed $A^n=0$.
A: A proof using Cayley-Hamilton's theorem if you are interested.
Looking at the form of $A$ (an upper triangular matrix), you can see that:
$$\chi_A(X)=X^n$$
where $\chi_A$ is its characteristic polynomial.
By Cayley-Hamilton's theorem:
$$A^n=\chi_A(A)=0.$$
A: It suffices to show $A^n x = 0$ for any $x \in F^n$.  To see this, define $W_i := \{ ( x_1, x_2, \ldots, x_{n-i}, \underbrace{0, \ldots, 0}_{i~\mathrm{zeros}}) : x_1, \ldots, x_{n-i} \in F \}$ which are each linear subspaces of $F^n$.  (Note that this only requires at least $i$ trailing zeros and allows more since it allows for $x_{n-i} = 0$.)  Then the condition on $A$ implies that $A \cdot W_i \subseteq W_{i+1}$ for each $i$.  Thus, for any $x \in F^n$:
\begin{align*}
x & \in W_0 = F^n \\
A x & \in W_1 \\
A^2 x & \in W_2 \\
& \vdots \\
A^n x & \in W_n = \{ 0 \}.
\end{align*}
A: Similar to the question at: How to show that a $4 \times 4$ strictly upper triangular matrix is nilpotent?. Read through the answers and comments posted and you should be able to prove for all $n$.
