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

Cosider the vector space $$V=M_n(\mathbb C)$$ and $$W$$ be a vector subspace of $$V$$ such that every non-zero element of $$W$$ is invertible. Show that dim$$W=1$$.

I know that any matrix over $$\mathbb C$$ is triangulable. I assumed that dim$$W$$ is 2, Can I show that this gives rise to a singular matrix.

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.
• I think we don't need to use triangulable over $\mathbb{C}$, the existence of eigenvalue is enough. Just argue by contradiction as you did. – Hongyi Huang May 15 at 13:37
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!
I'll give you a hint: $$\det\colon W\cong\mathbb{C}^k\to\mathbb{C}$$ is algebraic and obviously nonconstant.