The only dense linear subspace of $\mathbb{C}^n$.

I want to prove this question: the only dense linear subspace of $\mathbb{C}^n$ is $\mathbb{C}^n$ itself.

My immediat attempt is think of $\mathbb{C}^n$ as a closed subspace of itself, so it is its own closure. Am I right? Is it that simple or there is something more to it?

• Is that the complex numbers? Finite-dim vector spaces are always closed. – mathematician Mar 26 '15 at 14:05
• Yes: $\mathcal{C}^n$ is the n-power of the complex numbers seen as a vector space. Suhail: could you be more explicit, please? I am not an expert in the subject. – gibarian Mar 26 '15 at 14:07
• If $\mathbb{C}^n$ is considered linear space over $\mathbb{Q},$ then it has dense subspace other than $\mathbb{C}^n.$ – Suhail Mar 26 '15 at 14:13

That $\mathbb{C}^n$ is dense in itself is trivial; a topological space is always dense in itself, since the closure of $A$ by definition contains $A$. What is nontrivial is that proper subspaces of $\mathbb{C}^n$ are not dense. For that, think about geometry: given a line in the plane, you can find a point orthogonal to the line, and so you can use the Pythagorean theorem to position that point far away from the line. You can do the same thing with subspaces, for example with the Gram-Schmidt process.
A shorter proof is to show that a subspace of $\mathbb{C}^n$ is closed, therefore it is its own closure. Hence the closure of a proper subspace of $\mathbb{C}^n$ is not $\mathbb{C}^n$.
Any subspace $U \subset \mathbb{C}^n$ is closed, since $U = q^{-1}(\{0\})$ where $q: \mathbb{C}^n \to \mathbb{C}^n/U$ is the quotient map. Another way of seeing that $U$ is closed is thinking about it like an intersection of solution sets of linear equations, which also must be closed (inverse image of 0). Hence, if $U$ is dense, then $U = \overline{U} = \mathbb{C}^n$.
If $V\subset \mathbb{C}^n$ is a proper subspace, then the projection $\mathbb{C}^n \to V^\perp$ is nonzero but vanishes on $V$.