Interesting subspace of $M_n(\mathbb{C})$ [CMI 2019]

Let $A,B\in W$ be two linearly independent matrix, then $A^{-1}B$ is invertible (not necessarily to be in $W$). Now $A^{-1}B$ has an eigenvector $v$, corresponding to eigenvalue $\lambda\in \mathbb{C}$ (the field $\mathbb{C}$ implies the existence of eigenvalue). We have $A^{-1}Bv = \lambda v$. This implies $(B-\lambda A)v = 0$ and so $B-\lambda A$ is not invertible.

Hint: Suppose its dimension is not one. Then it has a basis $\mathcal{B}$ which contains at least two independent invertible matrices. Call this two matrices as $A$ and $B$. Then $$(\forall k \in \Bbb C):\;A \neq kB$$ Since $W$ is a vector space, so $$(\forall k):\;A-kB \in W $$ Now consider $\det (A-kB)$ to arrive a contradiction!