Injective Transformations are Not Necessarily Isomorphisms for Infinite Dimensional Spaces

Question

In this StackExchange question, there is a comment that says the statement

Let V,W be the vector spaces over a field K, and assume that $\dim V=\dim W$. Let $F:V\to W$ be a linear map. If $F$ is surjective, prove that $F$ is an isomorphism.

is not true for infinite dimensional spaces. Why is this the case? It seems to me that the proof in the answer does not ever assume that the dimensions are finite. Where does that assumption come into play?

Answer 1

The simplest example is the space $V=K^{\mathbb{N}}$ of sequences of elements of $K$. The function $T\colon V\to V$ that maps the sequence $(a_0,a_1,\dotsc)$ into the sequence $(0,a_0,a_1,\dots)$ is obviously linear, injective and not surjective.

Conversely, the function $S\colon V\to V$ that maps the sequence $(a_0,a_1,\dotsc)$ into the sequence $(a_1,a_2,\dotsc)$ is linear, surjective and not injective.

Thus the finite dimension assumption in the theorem you cite is necessary. The fact that $W$ is distinct or not from $V$ is irrelevant: two spaces having the same dimension are isomorphic. Note that, using also infinite cardinals for dimensions, the rank nullity theorem still holds: if $T\colon V\to W$ is a linear map, then $$ \dim\ker T+\dim\operatorname{im}T=\dim V $$ However, when $\dim V$ is infinite, this just says that $\dim\ker T=\dim V$ or $\dim\operatorname{im}T=\dim V$, because for infinite cardinals the sum is the same as the maximum between the two. There's no “subtraction of cardinals”.

Answer 2

It is used here:$$\dim(\ker(T)) +\dim(W) =\dim(V) \Longrightarrow\dim(\ker(T)) = 0.$$How would you deduce this if $\dim(W)=\dim(V)=+\infty$?