I am aware that nullity is the dimension of the null space. So I assume that the null space is empty? But why?
[Math] Why is the nullity of an invertible matrix 0
matricesmatrix equationsmatrix-rank
Related Solutions
I assume for simplicity that we are regarding these infinite matrices as maps from $\ell^2$ to $\ell^2$.
With credit to user1551, here are some examples in which the statement fails:
Failure of 1: The shift operators on $\ell^2(\mathbb R)$ (equipped with the inner product $\langle x,y\rangle=\sum_{k=1}^\infty x_ky_k$) would give you an easy counterexample: $$ A=\pmatrix{0\\ 1&0\\ &1&\ddots\\ &&\ddots&\ddots}, \ A^T=\pmatrix{0&1\\ &0&1\\ &&\ddots&\ddots\\ &&&\ddots}. $$ $Ax=0$ has only the trivial solution, but $A^Te_1=0$.
Failure of 2: On $\ell^2(\mathbb R)$, consider $$ A=\pmatrix{1&1\\ 1&1&1&1\\ &&1&1&1&1\\ &&&&1&1\\&&&&&&\ddots}, \ A^T=\pmatrix{1&1\\ 1&1\\ &1&1\\ &1&1\\ &&1&1\\ &&1&1\\ &&&&\ddots}. $$ We have $Ax=0$ for any $x\in\ell^2$ in the form of $(x_1,-x_1,x_2,-x_2,\ldots)^T$. Therefore the nullity of $A$ is infinite. However, $A^Tx=0$ has only the trivial solution.
Failure of Both: Take $$ A=\pmatrix{ 0 \\ 0&\ddots\\ 1&0\\ 0&0&\ddots\\ &1&0\\ &0&0&\ddots } $$
Now, why do these statements fail in infinite dimensions? Recall that the reason this property holds in the finite dimensional case is due to the rank-nullity theorem. In particular, for $A: \Bbb R^n \to \Bbb R^m$ (that is, $A$ is $m \times n$) $$ \dim im(A) + \dim \ker(A) = n $$ this result no longer applies in infinite dimensions, and we lose some consequences of this theorem.
For example, a fact that holds for a square matrix of finite size is that $A$ is injective (has a trivial null space) if and only if it is surjective. This no longer applies in infinite dimensional spaces. What we can still say, however, is that $A$ is injective if and only if $A^T$ is surjective (whether or not $A$ is square). In our first example, then, we have an example of a map that is injective (has a trivial kernel) but is not surjective. Consequenctly, $A^T$ turns out to be surjective but not injective.
$A$ has smaller left-nullity than $AB$ but not smaller (right)-nullity. For example, if
$$ A = \begin{pmatrix} 1 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 \\ 1 \end{pmatrix}, $$
then $\operatorname{nullity}(A) = 1$ but $\operatorname{nullity}(AB) = 0$.
You can see what the rank-nullity theorem says: if the dimension of $A$ is $m \times n$ and $B$ is $n \times p$ then
$$\operatorname{rank}(A) + \operatorname{nullity}(A) = n,$$
$$\operatorname{rank}(B) + \operatorname{nullity}(B) = p,$$
$$\operatorname{rank}(AB) + \operatorname{nullity}(AB) = p.$$
So we are able to relate the nullity of $B$ to that of $AB$ because they are both related to $p$ but the nullity of $A$ is only related to $n$. If $\operatorname{nullity}(A) > p$ then $\operatorname{nullity}(A) > \operatorname{nullity}(AB)$ since $\operatorname{nullity}(AB) \le p$.
If $A$ and $B$ are square matrices then the inequality is always true: $\operatorname{nullity}(A) \le \operatorname{nullity}(AB)$. One way to see this is to observe that the rank decreases: $\operatorname{rank}(AB) \le \operatorname{rank}(A)$.
Best Answer
The unique vector space of dimension zero is the trivial space $\{0\}$ which includes the zero vector. This space is nonempty, and in fact the empty set is not a vector space because any vector space must have the zero vector.
By the invertible matrix theorem, one of the equivalent conditions to a matrix being invertible is that its kernel is trivial, i.e. its nullity is zero. I will prove one direction of this equivalence and leave the other direction for you to prove.
$(\Rightarrow)$ Suppose $A$ is an invertible $n\times n$ matrix. Let $v\in\ker A$ so that $Av=0$. Then $A^{-1}Av=A^{-1}0\iff Iv=0\iff v=0$. Thus, $\ker A=\{0\}$ so $A$ has nullity zero.