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.
$R(T) = F(T)$ 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.