Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am not good with vector spaces so I would be grateful for any help.
As I've been told I need to take $v \in \mathrm{Im}(T)$, $v\neq 0$, and show that if $T(v) = \mathbf{0}$ then $T^2 = 0$. But if $T(v) \neq \mathbf{0}$, then need to show that if $\{e_1,\ldots,e_{n-1}\}$ is a basis of $\mathrm{Ker}(T)$, then $\{e_1,\dots,e_{n-1},v\}$ is a basis of $V$. Then deduce that $T$ is diagonalisable. I don't really know how to do that though...

share|cite|improve this question

Suppose that $0\neq v\in Im(T)$, then $T(v)\in Im(T)$. $\dim(Im(T))=1$ so $Im(T)=\text{span}(v)=F\cdot v$ and you get that there is a $\lambda$ such that $T(v)=\lambda v$.

If $T(v)=0$ then $T^2(V) \subseteq T(Im(T)) = T(F\cdot v)=0$, so $T^2 = 0$. Assume now that $T(v)\neq 0$, then $Im(T) \cap \ker(T) = {0}$ so a basis for the kernel (which is of dimension $n-1$) is a set of eigenvectors of eigenvalue $0$ and $v$ is the n-th eigenvector of eigenvalue $\lambda\neq 0$

share|cite|improve this answer
"v is the n-th eigenvector of eigenvalue \lambda \neq 0" This is false. Nilpotent operators have only zero eigenvalues. If there were an eigenvector of non-zero eigenvalue, it would surely stabilize under action of T (as in the first paragraph) and T^n wouldn't be zero for any n. – Marek Nov 12 '10 at 14:42
In this case it is true, since the operator isn't nilpotent. the assumption is that $T(v)\neq 0$ so $\lambda \neq 0$ and then $T^n(v) = \lambda^n v \neq 0$. – Prometheus Nov 12 '10 at 15:10
Ah, I got confused because you talk about non-nilpotent case in the first paragraph, then you talk about nilpotent case at the beginning of the second paragraph and immediately you switch to non-nilpotent case again. A strange order indeed. – Marek Nov 12 '10 at 15:20
Actually the first paragraph is just the general case for 1 dimensional image - $\lambda$ there can be any number. – Prometheus Nov 12 '10 at 15:38
@Prometheus: Why is it that if {e1,…,en−1,v} is a basis of V then T is diagonalisable? – Maths student Nov 12 '10 at 20:29

Prometheus' answer is essentialy correct. I'd just like to point something out to you so that you can gain a little intuition.

We know that the rank of $T$ is one-dimensional, which means that there exists ($n-1$)-dimensional subspace $W \subset V$ such that $T(W) = 0$ (imagine this in three dimensions as a plane that is being sent to zero). Now pick any $u \in V$ not lying in $W$. We have $T(u) = v \neq 0$ (because otherwise the rank of $T$ would be zero). So what can we say about $v$? Well, it either lies in $W$ and therefore for any $u' \in V$, with decomposition $u' = u + w'$ for some $w' \in W$, we have $T^2(u') = T^2(u + w') = T^2(u) + T^2(w') = T(v) + 0 = 0$ (because both $v$ and $w'$ are in $W$).

So assume that $v \notin W$. But then we know $T(v) = \lambda v$ because the span of $v$ is one-dimensional. Therefore we can write $T = \lambda E_v \oplus 0_W$ which is the desired diagonalization.

Try thinking about it in two dimensions and matrices and convince yourself that this problem is all about this: $$\pmatrix{0&1\cr0&0} \quad \pmatrix{\lambda & 0\cr 0 &0}$$

share|cite|improve this answer
Thank you for the explanation! – Maths student Nov 12 '10 at 20:17

First, if the rank of T is 1, then the dimension of its kernel is n-1. That means 0 is an eigenvalue of order n-1.

Now, as Prometheus said, you just have to find out whether you will find that last eigenvalue or not. To find it out, just take a vector $v \notin Ker(T)$, and look at $T(v)$. Either you'll again fall on $0$. What does it mean ? You have taken a vector from the only dimension not belonging to the kernel of $T$. It means it belongs to $Im(T)$. Any other vector there can be obtained by multiplying $v$ by some coefficient $t \in \mathbb{R}$ (or $\mathbb{C}$, ...). Since T is linear, the same will go for the image of $v$.

So if $T(v) = 0$, all the vectors from that dimension will fall on $0$ when going through $T$. So for any vector $u \in Im(T)$, $T(u) = 0$. That means that for any vector $v \in V$, $T(T(v)) = 0$, that is, $T^2(v) = 0$.

Now if $T(v)$ isn't zero, there will be some vector $u \in V$ (but different from $0$) such that $T(v) = u$. Thus, for any $t \in \mathbb{R}$, $T(t \cdot v) = t \cdot T(v) = t \cdot u$. And $u$ is in $Im(T)$ thus colinear to $v$. Now you have v being an eigenvector of $T$. Taking a basis of $Ker(T)$ and adding $v$ to it provides you a basis of $V$ on which $T$ is diagonalisable.

share|cite|improve this answer
Just FYI. Vector spaces don't have to be over $\mathbb{R}$ or $\mathbb{C}$. Any field will do just fine. – kahen Nov 12 '10 at 16:02
I know but given the level of the question, I decided to stick to them in my answer. – Alp Mestanogullari Nov 12 '10 at 16:33
This is helpful, thank you. – Maths student Nov 12 '10 at 20:24
$v\notin \ker(T)$ doesn't mean that $v \in Im(T)$, though you can find a vector satisfying this, if $T^2 \neq 0$ – Prometheus Nov 12 '10 at 20:50

EDITED after Prometheus's comment to use the characterization using sum of dimension of eigenspaces instead of counting eigenvalues. This argument should be correct.

Obviously $\dim V = n \geq 1$. Consider $T$'s restriction to its image. This is a linear map on a 1-dimensional space, so we can write $T\vert_{T(V)}(v) = cv$ for some $c \in k$. There are now two possibilities. Either $c = 0$ or $c \neq 0$.

  • $c=0$: Then $T^2 = 0$ (why?) and all of $T$'s eigenvalues are $0$ since we must have $0 = T(T(v)) =T(\lambda v) = \lambda T(v) = \lambda^2 v$ for any eigenvalue $\lambda$ and any eigenvector $v$. We conclude that $T$ can't be diagonalizable since all of its eigenvalues are $0$, but the eigenspace corresponding to $0$ is $\ker T$ which has dimension $n-1$.
  • $c \neq 0$: First note that clearly $T^2 \neq 0$. We show that the sum of the dimensions of $T$'s eigenspaces is $n$, so $T$ must be diagonalizable [Wikipedia]. And this is very easy: $T(V)$ is a 1-dimensional eigenspace with eigenvalue $c$, and $\ker T$ is an n-1-dimensional eigenspace with eigenvalue $0$. Thus the sum of the dimensions of the eigenspaces is at least $n$, but it can't be greater than $n$, so it must be exactly $n$ (not 100% sure on this... LinAlg is a long time ago...)
share|cite|improve this answer
Diagonalizable doesn't mean n distinct eigenvalues. It means that (from the link you gave) sum of the dimensions of its eigenspaces is equal to n. – Prometheus Nov 12 '10 at 13:49
D'oh. You're absolutely right... Man, my linear algebra is rusty :( – kahen Nov 12 '10 at 13:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.