If $X$ infinite dimensional and $F\colon X\to\mathbb{F}^{n}$ linear, then $\ker(F)$ infinite dimensional subspace of $X$.

Question

Suppose that $X$ is an infinite dimensional vectorspace over the field $\mathbb{F}$ (real or complex numbers). Let $F\colon X\to\mathbb{F}^{n}$ be a linear map. I want to prove that $\ker(F)$ is an infinite dimensional subspace of $X$.

I tried to use the fact that $X/\ker(F)$ and $F(X)$ are isomorphic, but I can't come up with a neat argument. I really want to avoid cardinal arithmetic, i.e. things like "$\infty-n=\infty$" that may show up when you use the "generalized" rank-nullity theorem.

So my question is: Is there a neat (and perhaps short) argument to prove that $\ker(F)$ can not be finite dimensional? Any suggestions are greatly appreciated!

Answer 1

Let $u_1, \ldots, u_p \in X$ be such that $F (u_1),\ldots,F(u_p)$ is a basis of the image of $F$.

Then, show $X=\bigoplus_{i=1}^p{\mathbb{F}u_i} \oplus \ker{F}$.