Proving by induction that strictly lower triangular matrix is nilpotent

Question

Let matrix $A \in \mathbb{K}^{n \times n}$ have the property $a_{ij} = 0$ for $1 \leq i \leq j \leq n$. Show that $A^n = 0$.

Proof by induction:

Base Case:

for $n=2: A= \left[ {\begin{array}{cc} 0 & 0 \\ a_{21} & 0 \\ \end{array} } \right]$

So $A^2 = \left[ {\begin{array}{cc} 0 & 0 \\ a_{21} & 0 \\ \end{array} } \right] \cdot \left[ {\begin{array}{cc} 0 & 0 \\ a_{21} & 0 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} } \right]$

Inductive Hypothesis(IH):

Assume $A^n = 0$ holds true for some $n$.

Inductive Step:

$n \rightarrow n+1$, to show: $A^{n+1} = 0 $

$A^{n+1} = A^n \cdot A =^{IH} 0 \cdot A = 0$

It seems to be too simple. Is it correct to prove this by induction?

Answer 1

Yes, induction is certainly a good way to try and prove the result. However, there are some problems with your attempt to do so.

One problem is that the $\ A\ $ in your induction hypothesis is an $\ n\times n\ $ matrix, whereas the one in your induction step has to be an arbitrary $\ (n+1)\times(n+1)\ $ matrix satisfying the stated conditions. That means it cannot be the same $\ A\ $ as the one appearing in the induction hypothesis.

To fix the proof, your induction hypothesis must be something like $"\ A^n = 0\ $ for all strictly lower triangular $\ n\times n\ $ matrices $\ A\ $." Also, in the induction step, an identity you will probably find useful is $$ \pmatrix{B&0 _{n\times1}\\b^\top&0}^k=\pmatrix{B^k&0 _{n\times1}\\b^\top B^{k-1} &0}\ , $$ for any $\ n\times n\ $ matrix $\ B\ $ and $\ n\times1\ $ column vector $\ b\ $.

Answer 2

If you try to deduce the $n=3$ case from $n=2$ that you describe, this happens:

$$ \begin{pmatrix} 0 & 0 & 0 \\ a_{21} & 0 & 0 \\ a_{31} & a_{32} & 0 \end{pmatrix}^{3} = \begin{pmatrix} 0 & 0 & 0 \\ a_{21} & 0 & 0 \\ a_{31} & a_{32} & 0 \end{pmatrix}^{2} \begin{pmatrix} 0 & 0 & 0 \\ a_{21} & 0 & 0 \\ a_{31} & a_{32} & 0 \end{pmatrix} $$ You can't conclude that $A^2 = 0$ because this matrix is $3\times 3$ and not $2\times 2$ like your previous case.