# Rank and nullity of matrices

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 $$\text{rank} (C)=n$$ ?

• Have you sen the rank-nullity theorem? Commented Feb 9, 2020 at 0:05
• Yes, but we have that the nullity of $D$ is $0$ and we want to show that the rank of $c$ is $n$. How can we use the theorem here? @Bernard Commented Feb 9, 2020 at 8:35

It follows by the rank-nullity theorem: $$n = \dim(\Bbb R^n) = \operatorname{rank}(C) + \operatorname{nullity}(C)$$
• Or is there a typo at the statement of the exercise and it should be the nullity of $C$ instead of $D$ ? Commented Feb 9, 2020 at 20:53