Suppose $V$ and $W$ are finite-dimensional and $T \in \mathcal L(V,W)$. Show that with respect to each choice of bases of $V$ and $W$, the matrix of $T$ has at least $\dim \text{range }T$ nonzero entries.

Let $v_1,\dots,v_n$ and $w_1,\dots,w_m$ be the bases for $V$ and $W$ that will be used for the matrix (we don't control these).

Let $A$ be the matrix of $T$, we can clearly find at least $\dim \text{range }T$ vectors in $v_1,\dots,v_n$ with $Tv_k \ne 0$. This implies there is a nonzero element in each row because

Implies at least one $A_{j,k} \ne 0$ (independence of $W$ basis). doing $r$ times implies the number of nonzero entries is at least $\dim \text{range }T$.

This is the best bound we can get using only the dimension of the range, since the identity matrix has the number of nonzero entries equal to the dimension of the range.

todo (easy): Remove "clearly" and make rigorous, but make sure it's still understandable!