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Show that, if all the entries of $A$ on and below the diagonal are zero, then $A$ is nilpotent.

I know this has been asked before , but I want to solve this question without the use of Cayley-Hamilton theorem, which I am able to do.

All I know so far is it has eigenvalue $0$ with algebraic multiplicity $n$. Any hints on how to proceed would be appreciated, thanks.

I also notice that a matrix with the above form size $2\times2$ and $3\times3$ has order $2$ and $3$, respectively.

Is there a way to show for an $m \times m$ matrix that is $A^m=(0)$?

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  • $\begingroup$ Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. $\endgroup$ – Shaun Mar 14 at 21:24
  • $\begingroup$ Well the row at the far right is the one that lives the longest, since it has the most non-zero entries. So can you show that after m iterations, the rightmost row becomes all 0? $\endgroup$ – NazimJ Mar 14 at 22:15
  • $\begingroup$ @NazimJ you mean column? don't know how to write a formal proof $\endgroup$ – Eden Hazard Mar 14 at 22:16
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This is easy. Just examine what happens to a vector when you multiply it by such a matrix and apply induction.

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Hint If you try some small cases, you'll notice that for all positive integer $k$ the first $k - 1$ superdiagonals of $A^k$ have all zero entries. In particular, this shows that if $A$ is $m \times m$ then $A^m = 0$.

So, formulate the property "the first $k - 1$ superdiagonals of $A^k$ have all zero entries" in terms of the entries $(A^k)_{ij}$ and prove that the property holds using induction on $k$.

Additional hint Since the diagonal and subdiagonal entries are all zero, the condition that the first $k - 1$ superdiagonals entries are (also) zero is exactly that $(A^k)_{ij} = 0$ for all $j - i < k$.

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