Are strictly upper triangular matrices nilpotent? An $n\times n$ matrix $A$ is called nilpotent if $A^m = 0$ for some $m\ge1$.
Show that every triangular matrix with zeros on the main diagonal is nilpotent.
 A: WLOG assume that $A$ is upper-triangular (otherwise, a similar argument works with the basis reversed).
Regard $A$ as a linear transformation on $\mathbb{F}^n$ with basis $e_1, \ldots, e_n$. Let $U_i$ be the span of $e_1, e_2, \ldots, e_i$ for $i = 0, 1, \ldots, n$. Observe that $0 = U_0 \subseteq \ldots  \subseteq U_n = \mathbb{F}^n$ and 
also note that $AU_i \subseteq U_{i - 1}$ for $i = 1, \ldots, n$ since $A$ is strictly upper-triangular. Therefore, $A^n = 0$.
A: I think you can prove this using induction.
Prove that for each $1 <= i <= n$, all elements of the first $i$ lines of $A^i$ are zero and for the $(i+1)th$ line, only the first element is not zero and the other elements in this line are zero.
A: Its characteristic polynomial is $T^n$, so by Cayley-Hamilton, $A^n=0$.
A: As therealak12 mentioned, you can prove this by induction. Let $T$ be an arbitrary matrix that is strictly upper triangular.
Induction basis ($k = 1$):
$T_{ij} = 0$ for $i \geq j$ follows from the fact that $T$ is a strictly upper triangular matrix.
Induction hypothesis: $T^k_{ij}=0$ for $i + k - 1 \geq j$ for $k \in \mathbb{N}^+$
Induction step ($k \rightarrow k + 1$): Let $i, j \in \{1, \cdots, n\}$ be arbitrary such that $i  + k \geq j$.
$$
T^{k+1}_{ij} = (TT^k)_{ij} = \sum_{l=1}^n T_{il} T^k_{lj}
= \sum_{l=1}^i T_{il} T^k_{lj} + \sum_{l=i+1}^n T_{il} T^k_{lj}
$$
For our left sum, $T_{il} = 0$ since $i \geq l$ and $T$ is a strictly upper triangular matrix.
For our right sum, we have $l \geq i + 1 \geq j - k + 1$ (since we assumed $i + k \geq j$), from which it follows that $l + k - 1 \geq j$. If we apply the induction hypothesis to this condition, we get $T^k_{ij} = 0$.
Thus:
$$T^{k+1}_{ij} = \sum_{l=1}^i 0 \cdot T^k_{lj} + \sum_{l=i+1}^n T_{il} \cdot 0 = 0
$$
from which the statement then follows.
