Suppose $v_1,\dots,v_m$ is a basis of $V$ and $W$ is finite-dimensional. Suppose $T \in \mathcal L(V,W)$. Prove that there exists a basis $w_1,\dots,w_n$ of $W$ such that all the entries in the first column of $\mathcal M(T)$ (with respect to the bases $v_1,\dots,v_m$ and $w_1,\dots,w_n$) are $0$ except for possibly a $1$ in the first row, first column.

(In this exercise, unlike 3c/3, you are given the basis of $V$ instead of being able to choose a basis of $V$.)

If $T(v_1) \ne 0$ define $w_1 = T(v_1)$ then extend to a basis $w_1,\dots,w_m$ of $W$. We have

Clearly $w_1 = 1w_1 + 0w_2 + \dots + 0w_m$, and since $w_1,\dots,w_m$ are independent this is the only linear combination that gives $w_1$ meaning $A_{j,1} = 1$ if $j=1$ and $A_{j,1} = 0$ otherwise.

If $T(v_1) = 0$ then the first column of $\mathcal M(T)$ must be zero for any basis $w_1,\dots,w_m$ since

Implies $A_{j,1} = 0$ for all $j$ by the independence of $w_1,\dots,w_m$.

Interestingly we "hardly" had to choose the basis of $W$ as we only had to pick $w_1$. If we handpicked all of $W$ one thing I think we could do is get 3c/3 except without the ordering, ie. the diagonal is littered with ones and zeros as opposed to "all ones, then all zeros".