A linear operator of rank $1$ is diagonalizable or nilpotent but not both.

Question

$T$ is a linear operator above field $F$. V is of finite dimension and $dim Im T = 1$; show that $T$ is diagonalizable or nilpotent but not both.

This seems like a rather strong statement. I know that $T$ is diagonalizable if there is a basis $B$ of eigenvectors, also from $dimImT = 1$ I think that there are at most 2 eigenvalues because $ImT$ is a span of a single vector, therefore there is a vector $v\in V\;$ such that $T(v) = \lambda v$.

I am kind of missing the rail of the proof here. I would like to get some insight / help.

Thanks in advance!

David C. Ullrich · Accepted Answer · 2019-12-17T16:23:40.847

3

Say $Im T$ is spanned by $v\ne0$. Choose $w$ so $v=Tw$. And choose a scalar $\lambda$ so $Tv=\lambda v$.

If $\lambda=0$ then $Tv=0$; hence $T^2=0$, so $T$ is nilpotent. Further, $T^2=0$, $T\ne0$ imply that $T$ is not diagonalizable.

Suppose $\lambda\ne0$. Then $T^kv=\lambda^kv\ne0$. So $T^k\ne0$, which is to say $T$ is not nilpotent. And if $n=\dim(V)$ and $Z$ is the nullspace of $T$ then $dim(Z)=n-1$. Say $b_1,\dots,b_{n-1}$ is a basis for $Z$. Let $b_n=v$. Then the set $B=\{b_1,\dots,b_n\}$ is an independent set of $n$ vectors. So $B$ is a basis for $V$; since every vector in $B$ is an eigenvector of $T$ this shows that $T$ is diagonalizable.

edited Dec 17 '19 at 16:23

answered Dec 17 '19 at 16:19

David C. Ullrich

92,839

Technically, does $T^2\ne0$ imply that $T$ is not nilpotent? Could it be the case that $T^k=0$ for some $k>2$? – Math1000 Dec 17 '19 at 16:21
@Math1000 Eewps, thanks for noticing that. Luckily the problem is fixed just by changing "$2$" to "$k$" in a few places.... – David C. Ullrich Dec 17 '19 at 16:25
Not trying to be pedantic, I was curious myself :) – Math1000 Dec 17 '19 at 16:26
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@Math1000 Nothing pedantic at all, no problem. Somehow I had the definition of "nilpotent" confused..... – David C. Ullrich Dec 17 '19 at 16:27
My comment was flagged for moderation attention? That's weird, I didn't get any notification of that. Do I not have enough reputation to be alerted of such things? – Math1000 Dec 17 '19 at 16:38
@Math1000 Sometimes it happens that everything I say on a given day is wrong... – David C. Ullrich Dec 17 '19 at 16:40
Why if $T^{2} = 0 , T \neq 0$ then $T$ is not diagonalizable? – GoodWilly Dec 17 '19 at 17:21
@vpam If $T^2=0$ then $0$ is the only eigenvalue. So if $T$ is also disgonalizable then the diagonalization has all zeroes on the diagonal, hence $T=0$. – David C. Ullrich Dec 17 '19 at 19:00

A linear operator of rank $1$ is diagonalizable or nilpotent but not both.

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