Jtdylktuy

How can I show that for an $ntimes n$ strictly lower triangular matrix $A$, $A^n = [0]$, but $A^{n-1} neq 0$? I can see it from some quick examples but I'm having trouble formalizing those observations.

The $3 times 3$ matrix

$$left[begin{array}{cc} 0 & 0 & 0 \ 0 & 0 & 0 \ 1 & 0 & 0end{array}right]$$

gives a counterexample to the claim.

See Corollary 3.78 in my Notes on linear algebra for the proof of $A^n = 0_{n times n}$. (The proof is not difficult, but is significantly lengthened by the fact that I could assume nothing for granted in that class.)

As @T.Bongers pointed out in his answer, $A^{n-1} neq 0_{n times n}$ is not always satisfied. However, for each $n > 0$, there exists a lower-triangular $n times n$-matrix $A$ that satisfies $A^{n-1} neq 0_{n times n}$. To prove this, set
begin{equation}
A = begin{pmatrix}
0 & 0 & 0 & cdots & 0 & 0 \
1 & 0 & 0 & cdots & 0 & 0 \
0 & 1 & 0 & cdots & 0 & 0 \
vdots & vdots & vdots & ddots & vdots & vdots \
0 & 0 & 0 & cdots & 0 & 0 \
0 & 0 & 0 & cdots & 1 & 0
end{pmatrix} ;
end{equation}
this is the $n times n$-matrix whose $left(i,jright)$-th entry is $delta_{i, j+1}$ for all $i$ and $j$ (where we are using the Kronecker delta notation; see Definition 2.37 in op. cit.). Then, the $left(i,jright)$-th entry of $A^k$ is $delta_{i, j+k}$ for all $i$ and $j$ and $k geq 0$ (indeed, this can be proven by induction on $k$, where the induction step relies on the definition of the product of two matrices). Thus, in particular, the $left(n,1right)$-th entry of $A^{n-1}$ is $delta_{n, 1+left(n-1right)} = delta_{n, n} = 1 neq 0$; therefore, $A^{n-1} neq 0_{ntimes n}$. (See also Show that $A^n = 0$ but $A^{n-1} neq 0$ for an $ntimes n$ strictly lower triangular matrix for variants of this proof.)