So I think I understand ranks and nullity. I was thinking that the possible values of the rank of $A$ were $\le 10$. Then I thought I could calculate the possible dimensions of the null space by $$\mathrm{rank}A + \mathrm{dim (Null} \ A) = n.$$ And then:
$$10 + \mathrm{dimNul} \ A \le 15,$$
$$\mathrm{dimNul} \ A \le 5, \ \mathrm{nullity} \le 5$$
But maybe I'm completely off. Can someone help me out? Thank you
On
A $10\times 15$ matrix represent a linear transformation from a vector space of dimension $15$ to a vector space of dimension $10$. Now $null(A)=\{v\in domain (A)~|~Av=0\}\subset domain(A)$. Hence nullity of $A$, which is the rank of the null space of $A$, can have dimension $\le15$. While the rank of $A$ is the number of independent columns of $A$. Note that the number of independent columns of a matrix is same as the number of its independent rows. Hence $rank(A)\leq 10$.
You have some good ideas. But your resulting inequality is the wrong way around. Consider if the matrix is only zeroes. Then the nullity is $15$, which just isn't what your inequality says. Rather, instead of $10+\text{nullity} \leq 15$ you should have $\text{rank} + \text{nullity} = 15$. Thus if the rank goes down, the nullity goes up.
You seem to think that because $\text{rank} \leq 10$, we can swap out $\text{rank}$ with $10$, but that is not right. Since we're swapping $\text{rank}$ with something which is potentially larger, but never smaller, we actually get $$ 15 = \text{rank}+\text{nullity} \leq 10 + \text{nullity}\\ 5\leq \text{nullity} $$
As for notation, note that nullity is the dimension of the kernel. So the nullity itself doesn't have a dimension, it is a dimension. Just like rank is the dimension of the column space (which is the same as the range of the corresponding linear map). So rank is a number, and thus doesn't have a dimension.