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

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    $\begingroup$ Have you sen the rank-nullity theorem? $\endgroup$
    – Bernard
    Commented Feb 9, 2020 at 0:05
  • $\begingroup$ 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 $\endgroup$
    – Mary Star
    Commented Feb 9, 2020 at 8:35

1 Answer 1

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It follows by the rank-nullity theorem: $$n = \dim(\Bbb R^n) = \operatorname{rank}(C) + \operatorname{nullity}(C)$$

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  • $\begingroup$ But in this case do we not have the rank of C and the nullity of D? $\endgroup$
    – Mary Star
    Commented Feb 9, 2020 at 3:25
  • $\begingroup$ Or is there a typo at the statement of the exercise and it should be the nullity of $C$ instead of $D$ ? $\endgroup$
    – Mary Star
    Commented Feb 9, 2020 at 20:53

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