Show that the rank of f composed with f is strictly less than the rank of f if and only if Ker(f) ∩ Im(f) is not equal to {0}.
We understand the situation and have a few examples of functions where this is true, but do not understand how to thoroughly prove this.
On
Suppose $\text{ker }f \cap \text{im }f \neq \{0\}$.
This means that there is some $w \neq 0 \in \text{ker }f \cap \text{im }f$, and some $v \in V$ (the domain of $f$), with:
$f(v) = w \neq 0$
$f(f(v)) = f(w) = 0$.
Extend $\{w\}$ to a basis for $\text{im }f$, say $\{w,w_1,\dots,w_{k-1}\}$, so that $\text{rank }f = k \geq 1$.
Now $\{f(w),f(w_1),\dots,f(w_{k-1})\}$ span $\text{im}(\text{im }f) = \text{im}(f\circ f)$, but since $f(w) = 0$, $\{f(w_1),\dots,f(w_{k-1})\}$ also spans $\text{im}(f\circ f)$, so $\text{rank}(f\circ f) \leq k-1 < k = \text{rank } f$.
On the other hand, suppose $\text{rank}(f\circ f) < \text{rank }f$.
So if $\{w_1,w_2,\dots,w_k\}$ is a basis for $\text{im }f$, then $\{f(w_1),\dots,f(w_k)\}$ is a linearly-dependent set, because it has $\text{rank }f$ elements which is greater than any basis for $\text{im}(\text{im}(f))$ has (any such basis of $\text{im}(\text{im}(f))$ has $\text{rank}(f\circ f)$ elements).
Thus there exists $c_1,\dots c_k$ not all $0$ such that $c_1f(w_1) +\cdots + c_kf(w_k) = 0$.
Set $w = c_1w_1 +\cdots c_kw_k$. Then $w \in \text{im f}$ and $w \neq 0$ since not all the $c_k$ are $0$ and the $w_j$ form a basis for $\text{im }f$.
But $f(w) = f(c_1w_1 +\cdots c_kw_k) = c_1f(w_1) + \cdots + c_kf(w_k) = 0$.
Hence $0 \neq w \in \text{ker }f \cap \text{im }f$, that is: $\text{ker }f \cap \text{im }f \neq \{0\}$.
Note: the above assume the domain of $f$ is a finite-dimensional vector space.
Let $f'$ be the restriction of $f$ to $\DeclareMathOperator{\Im}{Im}\Im f$. Of course, $\Im f'=\Im(f\circ f)$. Now $\ker f'=\ker f\cap\Im f$, hence the rank-nullity theorem implies that $\Im f'=\Im f$, i.e. $\;\DeclareMathOperator{\rk}{rank}\rk(f\circ f)=\rk f$, if and only if $\ker f'=0$.