3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ That's a nice solution. Thank you. That means my logic was not sufficient? $\endgroup$ – MathCosmo May 15 at 13:34
  • 1
    $\begingroup$ 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. $\endgroup$ – Hongyi Huang May 15 at 13:37
2
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

I'll give you a hint: $\det\colon W\cong\mathbb{C}^k\to\mathbb{C}$ is algebraic and obviously nonconstant.

$\endgroup$
  • 1
    $\begingroup$ I'm not getting anywhere with this hint please explain $\endgroup$ – MathCosmo May 15 at 13:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.