Suppose $w_1,\dots,w_n$ is a basis of $W$ and $V$ is finite-dimensional. Suppose $T \in \mathcal L(V,W)$. Prove that there exists a basis $v_1,\dots,v_m$ of $V$ such that all the entries in the first row of $\mathcal M(T)$ (with repsect 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 $W$ instead of being able to choose a basis of $W$.)

If $w_1 \in \text{range }T_1$ set $v_1$ so $Tv_1 = w_1$, otherwise (TODO)
then extend to a basis $v_1,\dots,v_n$ of $V$. For each $k > 1$ we have

Meaning $T(v_k - c_kv_1)$ has no $w_1$ component, so consider the list

It's linearly independent because $a_1v_1 + \dots + a_n(v_n - c_nv_1) = 0$ implies

Since $v_1,\dots,v_n$ are independent we have $a_2 = \dots = a_n = 0$ meaning the original equation is $a_1v_1 = 0$ thus $a_1 = 0$ as well. So our new list is a basis by 2.39, and we clearly have

As desired, meaning $\mathcal M(T)$ with respect to the $v_1,\dots,(v_n-c_nv_1)$ basis will have the first row be $(1,0,\dots,0)$.

This proof is effectively a matrix algorithm style proof, but written in terms of linear maps.