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Ok so here's the problem.

$T$ is a nilpotent linear transformation on a finite dimensional vector space. (Let's say $V=\mathbb{R}^{n}$, without loss of generality.)

Fact: $T$ has only $0$ as an eigenvalue and there is a smallest nonzero natural number, $m$, such that $\text{Ker}(T^{m})=V$.

Show that $T$ can be written as an upper triangular matrix, $A$, with $0$'s on the diagonal where $A$ is with respect to the basis derived as follows:

First find a basis for $\text{Ker}(T)$. Then expand that basis to one for $\text{Ker}(T^{2})$. Expand again for a basis of $\text{Ker}(T^{3})$ and so on until you get a basis for $Ker(T^m) = V\ $ [of course m could be smaller than 3].

Just a reminder: the columns of a matrix are the image of the basis vectors. Therefore if our final basis after the process just explained is {$u_1, u_2, ..., u_n$}, then the matrix will be $[T(u_1), T(u_2), ..., T(u_n)]$.

Please let me know if you can think of anything. This should be pretty simple, but I'm not seeing something.

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