# V is a linear space that has finite dimension $T : V \rightarrow V$ Proof if $Ker T = Ker T^2$ then $Im T = Im T^2$

I have the following question :

V is a linear space that has finite dimension $T : V \rightarrow V$ Proof if $Ker T = Ker T^2$ then $Im T = Im T^2$

I tried to proof it, but I got stuck, This is what I did :

So for all $v \in Ker T \Longleftrightarrow v \in Ker T^2$

We know that (from a theorem) $$dim(ImT)+dim(KerT)=dim(V)=n$$ and also: $$dim(ImT^2)+dim(KerT^2)=dim(V)=n$$

Therefore I can conclude that $dim(ImT^2)=dim(ImT)$

Let $w \in ImT$ so exist $u \in V$ so $T(u)=w \neq 0$

• If $T^2(u)=T(T(u))=T(w) \neq 0$ Therefore we showed that for any $w \in ImT \implies w \in ImT^2$ meaning that $ImT \subseteq ImT^2$ and we also know that $dim(ImT^2)=dim(ImT)$ Therefore $ImT=ImT^2$.

• If $T^2(u)=T(T(u))=T(w) = 0$ Therefore $w \in KerT \Longleftrightarrow w \in KerT^2$

And here I'm stuck I don't think I realize what does it mean to be in $KerT^2$ if $w \in KerT$ then is clear that $T(w)=0$ and also its trivial that $T^2(w)=T(T(w))=T(0)=0$ therefore always if $w \in KerT \implies w \in KerT^2$ isn't that always true? Can someone explain?

Thank you.

• In the first bullet, how does that show that $w\in Im(T^2)$, it only shows that $T(w)\in Im(T^2)$. – Michael Burr Aug 16 '15 at 1:28
• @Michael Burr Oh, you'r right – JaVaPG Aug 16 '15 at 1:30
• I suggest proving that Im$(T^2)\subseteq$ Im$(T)$. Then the equality of dimensions gives what you need. – Andreas Blass Aug 16 '15 at 1:31

All what you did until you get $\dim{\text{Im}\,T}=\dim{\text{Im}\,T^2}$ is perfect.

Now to prove that $\text{Im}\,T=\text{Im}\,T^2$, all you have to do is, as you noticed, to prove that one of them is included in the other. In fact, $\text{Im}\,T^2\subset\text{Im}\,T$ not the opposite and what you did there is fales because to show that $A\subset B$ you have to prove that if you take any element $a\in A$, you can find that it is in $B$, namely: $\forall a\in A, a\in B$ or if $A$ and $B$ are subsets of a certain set say $E$: $\forall x\in E,\,x\in A\Rightarrow x\in B$.

So you took $w\in \text{Im}\,T$. By definition of $\text{Im}\,T$, there exist $u\in V$ such as $T(u)=w$ and then you wrote $T^2(u)=T(w)$ (which is correct) and so $w\in \text{Im}\,T^2$ (which is incorrect) because $w\in \text{Im}\,T^2\Leftrightarrow \exists v\in V\, w=T^2(u)$ but what you did is to prove that $T(w)\in \text{Im}\,T^2$, not $w$ itself.

The true thing is: $\text{Im}\,T^2\subset \text{Im}\,T$ because if $w\in \text{Im}\,T^2$ then $\exists u\in V,\,w=T^2(u)=T(T(u))$. If you put $v=T(u)\in V$ then $w=T(v)$ and so $w\in \text{Im}\,T$.

Edit

About $\ker T^2$ well when $w\in \ker T^2\Leftrightarrow T^2(w)=0$ by definition. Actually, as you proved, we have indeed $\ker T\subset \ker T^2$ but the opposite isn't always true. An easy example is nilpotent linear maps of order $2$, which means when $T:V\to V$ is a linear map such as $T\neq 0_V$ ($0_V$ is the map from $V->V$ such as $\forall v\in V,\,0_V(v)=0$) and $T^2=0_V$. We have $\ker T\neq V$ but $\ker T^2=V$ so there are elements in $\ker T^2$ not in $\ker T$. To view this fact with more intuition for any map (even if it's not nilpotent), well you should use the fact that every linear map $T:V\to W$ where $V$ and $W$ are vector spaces can be entirely defined if all we know is the image by $T$ of the vectors of the basis of $V$ (in the case when $V$ is a finite-dimensional vector space). In particular, if $T:V\to V$ is an operator (a linear map from a vector space to itself is called an operator) where $V$ is a finite-dimensional vector space, then $\ker T$ and $\text{Im}\,T$ are subspaces of $V$. Let $(v_1,\cdots v_n)$ be a basis of $V$. If we know what $T(v_i)$ is for any $i\in\{1,\cdots , n\}$ then $T$ is enterly defined (and you can choose the elements the $T(v_i)$s are equal to freely). Let us take $T(v_1)=O$ and $T(v_2)=v_1$. See that $v_2\notin\ker T$ because $v_1\neq 0$, but $T^2(v2)=T(v_1)=0$. The following picture may help (sorry for the quality I didn't take time to draw it in computer because I want to sleep ^^ ): The green area represents $\ker T$ and it contains $0$. $\forall v\in\ker T,\, T(v)=0$ and so for any element of $\ker T$, in the picture I drow the arrow from it to $0$. But other elements (outside of the green area and so their image isn't $0$) can have an element of $\ker T$ as their image (an arrow coming from white area to green area but not to $0$). If we apply $T$ a second time then we get $0$. This proves why $\ker T^2\neq \ker T$ in general (we can have cases when $\ker T=\ker T^2$ when no vector in the white area reaches the green area, in maths language when the vectors of the basis of the supplementary space of $\ker T$ in $V$ are mapped to vectors in $V\backslash \ker T$, or in other words, this supplementary space is invariant or stable under $T$, i.e, if you call it $S$ then $T(S)\subset S$).

• Thank you for you answer! Just one little thing can you also relate to "what does it mean to be in KerT^2?" That I written above, I really want to understand the concept. – JaVaPG Aug 16 '15 at 1:45
• @JaVaPG You're welcome. I edited my answer. – Scientifica Aug 16 '15 at 2:48
• Great Answer, Clear as the sun :) Thank You! – JaVaPG Aug 16 '15 at 10:00
• @JaVaPG I'm glad to hear it :D You're welcome :D – Scientifica Aug 16 '15 at 11:35

You need to prove first that $$\operatorname{Ker}(T) = \operatorname{Ker}(T^{2})\implies \operatorname{Ker}(T) \cap \operatorname{Im}(T)=\{0\}$$ Since $T$ maps each $x$ to $0$ or nonzero, $\dim(T^{2})\leqslant\dim(T^{})$. If $\dim(T^{2})=\dim(T^{})$, then $T^{}$ is invariant. So there is $$\operatorname{Im}(T)\cap \operatorname{Ker}(T^{})=\{0\}$$ for if not, let $x\in \operatorname{Im}(T^{})\cap \operatorname{Ker}(T^{})$, then $T(x)=0$ and $\dim(T^{2})<\dim(T^{})$. Similarly $$\operatorname{Im}(T^2)\cap \operatorname{Ker}(T^{2})=\{0\}$$ Since $$\dim{\operatorname{Ker}(T)}+\dim{\operatorname{Im}(T)}=n\quad\text{and}\quad\dim{\operatorname{Ker}(T^2)}+\dim{\operatorname{Im}(T^2)}=n$$ There is $\operatorname{Im}(T^2)=\operatorname{Im}(T)$.