The rank nullity theorem states that for vector spaces $V$ and $W$ with $V$ finite dimensional, and $T: V \to W$ a linear map,
$$\dim V = \dim \ker T + \dim \operatorname{im} T.$$
Does this hold for infinite dimensional $V$? According to this, the statement is false. But according to this, page 4, the statement is still true. I'm thoroughly confused.
Best Answer
The rank formula also holds in infinite dimensions, whether you use cardinal arithmetic for the dimensions, or just say $\infty + n = \infty$, and $\infty + \infty = \infty$ (but one should use cardinal arithmetic). The proof is basically the same as in the finite-dimensional case, you choose a basis $\mathcal{B}_1$ of $\ker T$, a basis $\mathcal{B}_2$ of $\operatorname{im} T$, let $\mathcal{B}_3$ consist of preimages of the elements of $\mathcal{B}_2$ (choose one preimage per element), then $\mathcal{B}_1 \cup \mathcal{B}_3$ is a basis of $V$. In the infinite-dimensional case, some form of the axiom of choice is required, while the finite-dimensional case can be proved without that.