{Let X be a finite dimensional vector space over F, and $\displaystyle f:X\to X$ a linear transformation. If the rank of $\displaystyle f^2=f\circ f$ is equal to the rank of f, prove that $\displaystyle \text{ker} f \cap \text{Im}f=\{0\}$

Since $\displaystyle \text{rank}f\circ f=\text{rank} f\Rightarrow \text{Im}f\circ f=\text{Im} f$.

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