Finding a basis that allows the matrix of a linear transformation to have [1,0,0,...,0] as its first row I'd like some help with problem 4 from section 3C of Linear Algebra Done Right:

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

I've found a solution here, but it only provides a formulation of the basis vectors and doesn't explain why they lead to the desired first row of $M(T)$.
I am finding it hard to believe such a basis exists.

Take for instance $T:\Bbb R^2\to \Bbb R^2$ such that $T(v_1)=v_2;T(v_2)=v_1$

The matrix will be given by \begin{bmatrix} 0 &amp;1\\1&amp;0\end{bmatrix}

Now suppose that (as claimed in the problem) there exists  a  basis $\{v_3,v_4\}$ such that the matrix becomes \begin{bmatrix} 1 &amp;0\\0&amp;0\end{bmatrix}.

Then $T(v_3)=v_4;T(v_4)=v_3$.

By the matrix given $T(v_3)= v_3;T(v_4)=0\implies v_3=0$ which is false as $\{v_3,v_4\}$ is a basis
I am finding it hard to believe such a basis exists.

Take for instance $T:\Bbb R^2\to \Bbb R^2$ such that $T(v_1)=v_2;T(v_2)=v_1$

The matrix will be given by \begin{bmatrix} 0 &amp;1\\1&amp;0\end{bmatrix}

Now suppose that (as claimed in the problem) there exists  a  basis $\{v_3,v_4\}$ such that the matrix becomes \begin{bmatrix} 1 &amp;0\\0&amp;0\end{bmatrix}.

Then $T(v_3)=v_4;T(v_4)=v_3$.

By the matrix given $T(v_3)= v_3;T(v_4)=0\implies v_3=0$ which is false as $\{v_3,v_4\}$ is a basis