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The whole question: 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.

My answer: Let $v_1, \ldots, v_n$ be a basis of $V$, $Tv_1, \ldots, Tv_p$ be a basis of range $T$, $w_1, \ldots, w_m$ a basis of $W$. Then because each $Tv_j$ is nonzero (because the list is lin. ind.) for each $Tv_j = A_{1,j}w_1 + \cdots + A_{m,j}w_m$, at least one of $A_{i,j}w_i$ is nonzero, and so $\mathcal M (T)$ has at least dim range $T$ nonzero entries.

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Disregarding that $Tv_1,\dots Tv_p$ does not necessarily form a basis (e.g. never if $Tv_1=0$) , your proof is correct.
This problem can be resolved by either using indices $i_1,\dots i_p$ or by rearranging the basis of $V$.

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  • $\begingroup$ There was a previous problem where I showed that there is a subspace $U$ of $V$ s.t. null $T \cap U = {0}$ and range $T = \{Tu : u \in U\}$. Then if I set $v_1, \ldots , v_p$ a basis of $U$ and $v_{p+1}, \ldots , v_n$ a basis of null $T$ that would also work? $\endgroup$
    – BMSmudge
    Jun 20, 2018 at 1:18

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