If rankT = rank $T^2$，How could I prove $\ker(T) \cap \text{Im}(T) = \{0\}$? Let $T: V \rightarrow V$ be a linear transformation. Assume that
$$
\operatorname{rank} T=\operatorname{rank} T^{2}
$$
Show that $\operatorname{Ker} T \cap \operatorname{Im} T=\{0\}$.
Actually, I have an idea.
I remember a corollary: if dim V = dim W <=> V and W is isomorphism.Therefore ,T is an isomorphism. And $T(v) = 0$ if and only if $v = 0$, so $\ker(T) = \{0\}$.
And ImT is a subspace of V, so ImT have an zero vector .Then we got $\ker(T) \cap ImT = \{0\}$
But if I prove in this way ,$rank T = rank T^2$ is useless.
And another thing is ,I learn that course linear algebra here.
And I do homework, but I can not find the solution, Does anyone else who find the solution of homework?

Well, you nearly got it right. The condition $\mathrm{rank}\;T = \mathrm{rank}\;T^2$ implies that if you restrict $T$ onto the subspace $\mathrm{Im}\;T$ , which of course is invariant, you will get an isomorphism.
Thus, any vector in $\mathrm{Im}\;T$ except $\mathbf{0}$ is non-zero under $T$. This is equivalent to say that $\mathrm{Im}\;T\;\cap\;\mathrm{Ker}\;T=\{\mathbf{0}\}$
