Suppose you have an infinite matrix A with real entries. I know the dimension of the null space of A.
Question 1) if the dimension of the null space is a finite number k, is the dimension of the null space of $A^T$ also k?
Question 2) if the dimension of the null space of A is infinite, is the dimension of the null space of $A^T$ also infinite?
Thanks!
Best Answer
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.