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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.

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  • $\begingroup$ If the number of nonzero entries of the matrix is $k$, then the largest that the dimension of the columnspace could possibly be is $k$ (if you had exactly least one nonzero entry in each of $k$ distinct columns and the resulting vectors were not linearly dependent)... $\endgroup$ – Arturo Magidin Feb 28 at 23:13
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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.

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  • $\begingroup$ I haven't learnt rank yet. I am following Linear Algebra Done Right. But, I understand the rank is number of linear independent vectors in the Matrix. Please let me know if my thought below is correct: (Thank you!) The range is a linear combination of the column vectors in the matrix. The dimension of the linear combination (a vector) is the number of non zero elements. So, the columns of the Matrix must have at least dim range $T$ number of non zero entries. $\endgroup$ – JOHN Mar 1 at 22:45
  • $\begingroup$ "The range is a linear combination of the column vectors" - perfect; writing out $T(v_j)$ in terms of $w_i$ gives you the $j$th collumn. $\endgroup$ – Dean Young Mar 2 at 5:28
  • $\begingroup$ "the dimension of the linear combination is the number of nonzero elements" - that's actually not true. If someone says that, then they're refering to the idmension of their span. The dimension of the span is bounded by the number of nonzero vectors in the list (which is what we need), but it's not always equal. For example, consider the span of $\{ v, v \}$, $v \neq 0$, which is $1$-dimensional (as opposed to $2$). We conclude that "the number of nonzero vectors" does not give us all the information we need to determine the span. However, it certainly gives us the bound we need. $\endgroup$ – Dean Young Mar 2 at 5:36
  • $\begingroup$ "the collumns of the matrix must have at least dim range $T$ number of nonzero elements". This is still true, yes. $\endgroup$ – Dean Young Mar 2 at 5:36
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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 $\mathrm{dim} \: \mathrm{range} \: T$.)

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    $\begingroup$ Thank you! I think I follow the logic. For the last sentence: which is less that dim range $T$, should it be $\geq$? Since, the proof said dim range $T \leq n-k$, and we have at least $n-k$ entries. $\endgroup$ – JOHN Mar 1 at 22:35
  • $\begingroup$ Yes, thanks for catching that! $\endgroup$ – Alex Sanger Mar 1 at 22:54

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