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This is from Axler's 1996 "Down with Determinants," I do not see why this basis would give the desired matrix representation. I would appreciate any help in seeing this.

Emilio Minichiello
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  • If it is zero below the main diagonal then it is upper triangular. And nilpotent matrices are just that. –  Jul 03 '19 at 16:23
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    Because $T(\ker T^n) \subseteq \ker T^{n-1}$. – Joppy Jul 03 '19 at 16:25
  • @Shogun There are nilpotent matrices that are not of that form! They are alway similar to a strict upper triangular matrix, but that is what the op wants to understand. – Con Jul 03 '19 at 16:26
  • @Joppy I am sorry. I just realized before your last comment that nilpotent matrices can have rank less than $n$. But Thanks. :) – Kumar Jul 03 '19 at 16:29
  • It's quite easy to see that. Ker $T$ will have alteast one Linearly independent vector(L.I.V.), and the rank of $T$ $\leq$ $n-1$. Now, when you compose $T$ with $T$, then Rank of $T^2$ $\leq$ $n-2$. and Therefore, you can find atleast two L.I.V's such that they will form a basis for Ker $T$. You can proceed like this and you will prove the lemma. – Kumar Jul 03 '19 at 16:31
  • I could have worded it better. Im walking. I mostly was trying to say there is probably a duplicate if you search for nilpotent representation as upper triangular. –  Jul 03 '19 at 16:34
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    Possible duplicate of [Nilpotent operator have a basis respresentation that is strictly upper triangular?, https://math.stackexchange.com/questions/532372/are-strictly-upper-triangular-matrices-nilpotent](https://math.stackexchange.com/questions/1003525/nilpotent-operator-have-a-basis-respresentation-that-is-strictly-upper-triangula, https://math.stackexchange.com/questions/532372/are-strictly-upper-triangular-matrices-nilpotent) – Kumar Jul 03 '19 at 16:36
  • @Shogun Thanks. I have flagged it. – Kumar Jul 03 '19 at 16:37

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