Show that matrix of $T$ has at least dim range $T$ nonzero entries. 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 range $T$ nonzero entries.
I have no idea on how to solve this.
 A: Take a basis $v_1, ...,  v_n$ of $V$ and a basis $w_1, ..., w_n$ of $W$. The rank of $\text{Mat}(v_i, w_i, T)$ is at most the amount of nonzero collumns, as zero vectors in a list do not contribute to the span of the list of collumns. But the number of nonzero collumns is bounded by the number of nonzero entries. Therefore the rank of $\text{Mat}(v_i, w_i, T)$ can be no more than the number of nonzero entries it contains.
A: This is an exercise from Linear Algebra Done Right, so here's an answer in line with Axler's style: Choose bases $v_1,v_2,...,v_n$ of $V$ and $w_1,w_2,...,w_m$ of $W$ and let $\mathcal{M}(T)$ be the matrix of $T$ with respect to these bases. Column $j$ of $\mathcal{M}(T)$ will consist entirely of zeros iff $Tv_j=0$; equivalently column $j$ has at least one nonzero entry iff $Tv_j \not=0$. So out of the $n$ basis vectors of $V$ say $k \leq n$ of them are mapped to zero; assume without loss that $Tv_1,Tv_2,...,Tv_k=0$. Then $\mathrm{range} \: T = \mathrm{span}(Tv_{k+1},Tv_{k+2},...,Tv_n)$, which has dimension $\leq n-k$ (since these vectors need not be linearly independent.) Now for each $j=k+1,k+2,...,n$, $Tv_j \not=0$ so at least one of the scalars in its basis expansion with respect to the $w_i$'s is nonzero; hence the corresponding column of the matrix has at least one nonzero entry. So each of these $n-k$ vectors contributes at least one nonzero entry to the matrix, so we have at least $n-k$ nonzero entries (which is at least $\mathrm{dim} \: \mathrm{range} \: T$.)
A: Suppose we have a basis {$v_{1},...,v_{n}$} of $V$ and {$w_{1},...,w_{m}$} of $W$, and so $M(T)$ is an $m$ by $n$ matrix. Let $p$ denote the number of zero-columns in the matrix. Thus, we know the dimension of the null space is at least $p$. And so by the fundamental theorem of linear maps, $n = \dim \operatorname{ran}(T)+\dim \operatorname{null}(T) \geq \dim \operatorname{ran}(T)+p$. Therefore, $n-p \geq \dim \operatorname{ran}(T)$, and we are done.
