Suppose $V,W$ are both two dimensional vector spaces. Let $f_1,f_2 \dots f_5$ be linear transformations from $V \rightarrow W$. Show there exists choices for real numbers $a_1, \dots, a_5$, which are not all $0$, so that $a_1f_1 + a_2f_2 + a_3f_3 + a_4f_4 + a_5f_5$ is the zero transformation. It makes sense intuitively but I'm not so sure of a rigorous proof.
Essentially we want to show $\exists a_1 \dots a_5$ that the Kernel is all of $V$. So can I essentially just say this can be represented as a matrix so that
$A\begin{bmatrix} a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$
Where $A$ is a $2\times 5$ matrix $a_{i,j}$ represents the image of $f_i$ in the $j^{th}$ dimension of $V$, and since this matrix has rank at most $2$, it must have a nontrivial nullspace. Is this an okay way to show it? I'm not as comfortable with linear independence of linear transformations as I am with independence of vectors themselves.