Let $1\leq n,m,\ell\in \mathbb{N}$, $A\in \mathbb{R}^{\ell\times m}$, $B\in \mathbb{R}^{m\times n}$ and $C,D\in M_n(\mathbb{R})$, such that $CD=I_n$.
I have shown that $\text{Im}(AB)\subseteq \text{Im}(A)$ and so $\text{rank} (AB)\leq \text{rank}(A)$.
I have also shown that $\text{ker}(B)\subseteq \text{ker}(AB)$ and so $\text{Nullity} (B)\leq \text{Nullity}(AB)$.
I want to show that $\text{rank} (C)=n$ and $\text{Nullity}(D)=0$.
From the above results we get $\text{Nullity} (D)\leq \text{Nullity}(CD)=\text{Nullity}(I_n)$. The kernel of $I_n$ contains only the zero vector and so the dimension is equal to $0$. So we get $\text{Nullity} (D)\leq 0$ and since it cannot be negative it follows that $\text{Nullity} (D)= 0$.
Is this correct?
Could you give me a hint how to show $\text{rank} (C)=n$ ?
It follows by the rank-nullity theorem: $$n = \dim(\Bbb R^n) = \operatorname{rank}(C) + \operatorname{nullity}(C)$$