I hope you can help me, I'm very new to linear algebra, I am given the linear transformation $T:V\rightarrow V$ that fullfills $T^2 = 0$, what can be said of the relationship that exists between Im$(V)$ and $\ker(V)$? What I did so far is the following:

$T(Tx) = 0 \implies Tx \in \ker(T) \wedge Tx \in $Im$(T) \implies Tx \in \ker(T)\bigcap $ Im$(T)$

but $Tx$ should be equal to $0$ according to transformation's definition because $Tx \in $ Im$(T)$ also belongs to $V$ and given the transformation's definition $T(Tx) = 0$ and picking $w=Tx$ whe could say $T(w) = 0$ $ \forall$ $w \in V$ so, can we conclude that $\ker(T)\bigcap$ Im$(T) = 0?$, so $\ker(T)$ and Im$(T)$ are disjoint sets? can we conclude anything else? am I right?

thanks for any replies, cheers.

  • $\begingroup$ You have some good work. But you seem to be concluding that $T=0$, which does not follow. The trouble is in this $\forall w\in V$ bit, I think — you're really assuming that $T$ is surjective. $\endgroup$
    – Hoot
    Jul 3, 2016 at 23:08
  • $\begingroup$ @Hoot yes, I should have said $\forall w \in img(T)$, can I conclude anything else? $\endgroup$
    – dennisbot
    Jul 3, 2016 at 23:22

3 Answers 3


This one's a little difficult to answer without giving away the solution, but I'll try.

Your logic is good until we get to the sentence that starts "but $Tx$ should be equal to $0$..." This does not follow from the fact that $TTx = 0$. Yes, it's true for particular values of $x$, but not for $x$ arbitrary.

For example, consider the map $\mathbb{R}^3 \to \mathbb{R}^3$ which sends $(x, y, z)$ to $(y, 0, 0)$. Clearly, the square of this map is $0$. The image of this map is the $x$-axis. The kernel of this map is the $(x, z)$-plane. What is the relation between them? (It falls right out of the implication you have already proven.)


We have the inclusion: $\text{Im } T \subseteq \ker T$.

Proof: let $w \in \text{Im } T$. There exists $v \in V$, such that $w = Tv$. Hence, $Tw = T^2 v = 0$.


quite easy

Look, $T^{2}\left(X\right)=0$ $\Longrightarrow $T $\left(X\right)\in {Ker}\left(T\right)$

We know that $T\left(X\right)$ is range of $T$

it implies $R\left(T\right)$ $\subseteq$ $N\left(T\right)$

That means Range of T $\subseteq$ Nullity Of T

  • $\begingroup$ Hi, you have relatively quickly deleted your recent meta question: How to search a question on mathstackexchange. I have mentioned some advice on searching here in chat, maybe some of that might be useful for you. If you have more questions about searching, you can ask there or - probably even better - in the chatroom dedicated to searching. $\endgroup$ Nov 26, 2017 at 13:28
  • $\begingroup$ How do you conclude equality rather than simply inclusion? (Also, instead of $T(X)\in Ker(T)$, it should be $T(X)\subseteq Ker(T)$.) $\endgroup$ Nov 30, 2017 at 15:46
  • $\begingroup$ @AndrésE.Caicedo Yes brother you right !.I have edited my post $\endgroup$ Dec 1, 2017 at 3:23

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