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Let $V$ be a finite dimensional vector space over $\Bbb C$ or $\Bbb R$ and let $T$ be a linear operator on $V.$ Prove that $T^2 = T$ if and only if $$\text {rank}\ (T) + \text {rank}\ (I - T) = \dim V.$$

"$\implies$" part $:$ By rank nullity theorem it follows that $$\text {rank}\ (T) + \text {nullity}\ (T) = \dim V.$$ So If we can show that $\text {nullity}\ (T) = \text {rank}\ (I - T)$ then we are through. Now since $T^2 = T$ it follows that $T(I - T) = 0$ which in turn implies that $\text {Im}\ (I - T) \subseteq \text {Ker}\ (T).$ Hence we have $\text {rank}\ (I - T) \leq \text {nullity}\ (T).$ Also for $v \in \text {Ker}\ (T)$ we have $(I - T) (v) = v$ i.e. $v \in \text {Im} (I - T).$ So $\text {Ker}\ (T) \subseteq \text {Im}\ (I - T)$ and hence we have $\text {nullity}\ (T) \leq \text {rank}\ (I - T).$ Combining this with the previous inequality we find that $\text {rank}\ (I - T) = \text {nullity}\ (T),$ as required.

Now how do I prove "$\impliedby$" part?

Any help will be highly appreciated. Thanks in advance.

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If $\operatorname{rank}(T) + \operatorname{rank}(I - T) = \dim V$, then using the rank-nullity theorem on both $T$ and $I - T$, $$\operatorname{null}(T) + \operatorname{null}(I - T) = \dim V.$$ Note that $v \in \operatorname{ker}(T) \cap \operatorname{ker}(I - T)$ implies that $Tv = 0$ and $0 = v - Tv = v$, hence the two kernels sum directly to $V$. Hence, we can adjoin bases for each kernel together to make a full basis $B$ for $V$.

For the vectors $v \in B$ from $\operatorname{ker}(T)$, we note that $(T - T^2)v = (I - T)Tv = (I - T)0 = 0$.

For the vectors $v \in B$ from $\operatorname{ker}(I - T)$, we note that $(T - T^2)v = T(I - T)v = T0 = 0$.

Therefore, $T - T^2$ maps the basis $B$ to $0$, and thus by linearity, maps every vector to $0$. Therefore, $T = T^2$.

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  • $\begingroup$ Very nice proof. +1 Thank you so much. $\endgroup$ – Anacardium Oct 29 '20 at 17:15

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