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Let $V$ be a vector space. Determine all linear transformations $T: V → V$ such that $T = T^2.$

My solution goes like this:

Let $T:V\to V$ be such that $T^2=T.$

Let $F(T)=\{x\in V: T(x)=x\}.$

Claim: $V=F(T)+N(T)$

We note that,

$$x=T(x)+(x-T(x)),$$ for all $x\in V.$

Now, $T(x-T(x))=T(x)-T^2(x)=0\implies x-T(x)\in N(T).$

Again, $T(T(x))=T^2(x)=T(x)\implies T(x)\in F(T).$

Thus, $x\in F(T)+N(T)\implies V\subseteq F(T)+N(T).$

Hence, $F(T)+N(T)=V.$ So, our claim is established.

Let $y\in R(T).$ This means that $\exists x\in V$ such that $T(x)=y.$ We note that, $T^2(x)=y=T(x)=T(T(x))=T(y).$ So, $y\in F(T).$ Hence, $R(T)\subseteq F(T).$

Again, $x\in F(T)\implies \exists x\in V$ such that $T(x)=x\implies x\in R(T).$ Thus, $F(T)\subseteq R(T).$

Finally, we have, $R(T)=F(T).$

But this means, $T=I_V,$ where $I_V:V\to V$ such that $I_V(x)=x,\forall x\in V.$

If $T=I_V$ then $T^2=T$ is trivially satisfied.

So, $T=I_V$ is the only possible case such that $T^2=T$ holds, for a linear operator $T$ on $V.$

However, the answer given states that,

$T$ must be the projection on $W_1$ along $W_2$ for some $W_1$ and $W_2$ such that $W_1 ⊕ W_2 = V .$

Is the above solution valid? I couldn't find any mistake with my approach.

If something is wrong, with my solution, then I want to know which step causes the problem explicitly ? In this case, is there any way to fix the problem.

Any clarifications regarding this issue will be greatly appreciated.

Thomas Finley
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  • Yes, this is called projection. See other duplicates here, e.g., this one. And for the direct sum decomposition $\ker(T)\oplus {\rm im}(T)$ see for example this duplicate. You can compare with it. – Dietrich Burde Dec 12 '23 at 15:53
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    Since the user is asking about possible errors in their proof, I would not consider this a duplicate. – DominikS Dec 12 '23 at 15:56
  • @DietrichBurde Thanks for your comment! But the thing is, I want to know whether my solution is correct and valid? Taking a look at those links you gave, it seems both my solution and the answer in the book are valid way of answering this problem, if I am getting what you tried you to imply ,correctly. – Thomas Finley Dec 12 '23 at 15:56
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    @DominikS I didn't vote for closing as duplicate. But certainly it makes sense to compare the solution with the standard solutions. One remark - where is $N(T)$ defined? Of course $N(T)=\ker(T)$. – Dietrich Burde Dec 12 '23 at 15:57
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    @Thomas Finley: How do you reach the conclusion that $T=I_V$? In the previous line you conclude $R(T) = F(T)$, which is correct, but I don't see how $T=I_V$ follows from this? – DominikS Dec 12 '23 at 15:58
  • It may help to see examples where this holds that are not the trivial case of the identity. For instance, the map $T~:~\Bbb R^3\to \Bbb R^3$ given by $T((a,b,c))= (a,0,0)$ is one such map which is not the identity while clearly $T(T((a,b,c))) = T((a,0,0)) = (a,0,0) = T((a,b,c))$ – JMoravitz Dec 12 '23 at 16:02
  • Yur proofs that $V=F(T)+N(T)$ and $F(T)=R(T)$ are fine, but this does not imply $T=I_V.$ Any projection $T$ satisfies $V=F(T)+N(T)$ and $F(T)=R(T)$. – Anne Bauval Dec 12 '23 at 16:12

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$R(T) = F(T)$ doesn't necessarily imply that $T = I_V$, unless $F(T)$ happens to be all of $V$. You've shown that $T$ is identity when restricted to the subspace $F(T)$, which is quite distinct from being the identity map on $V$.

Think about the intersection $F(T) \cap N(T)$, and what this means geometrically. You should be able to convince yourself of the solution provided.

ocamo
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