# If all the entries of a matrix on and below the diagonal are zero, then it is nilpotent

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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• 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? – NazimJ Mar 14 at 22:15
• @NazimJ you mean column? don't know how to write a formal proof – Eden Hazard Mar 14 at 22:16

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 $$j - i < k$$.