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Let $A$, $B \in Mat_{n,n}(\mathbb R)$. Suppose $AB =0 $. Show $\operatorname{rank}(A) + \operatorname{rank}(B) = \operatorname{rank}([A^T|B])$.

I've already prove without the assumption $AB=0$ that $\operatorname{rank}(AB) \le \operatorname{rank}(A)$ and $\operatorname{rank}(A+B) \le \operatorname{rank}(A)+ \operatorname{rank}(B)$.

A hint to prove my question is provided: $\operatorname{Col}(A^T) \cap \operatorname{Col}(B) = \{0\}$, but I don't really see how I can benefit from this or how this is proved.

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$\DeclareMathOperator\Col{Col} % \DeclareMathOperator\rank{rank} % \DeclareMathOperator\span{span} $You benefit from $\Col(A^T)\cap\Col(B)=\{0\}$ by using it to show that $$ \Col([A^T|B]) = \Col(A^T) + \Col(B) $$ is a direct sum, so $\dim\Col([A^T|B]) = \dim\Col(A^T)+\dim\Col(B)$. This gives you what you want, since $\rank(A)= \dim\Col(A) = \dim\Col(A^T)$.

To prove $\Col(A^T)\cap\Col(B)=\{0\}$ consider any $y\in\Col(B)$. By definition of $\Col(B)$ there is some $x\in\mathbb R^n$ such that $y=Bx$. Therefore $Ay=ABx=0$. Now let $A_1, \dots, A_n\in\mathbb R^{1\times n}$ be the rows of $A$. Since $Ay=0$, we have $A_i y=0$ for all $i$, so $$ y \in \span(A_1, \dots, A_n)^\perp = \Col(A^T)^\perp, $$ where $^\perp$ denotes the orthogonal complement.

Since $\Col(A^T)\cap\Col(A^T)^\perp=\{0\}$ and $\Col(B)\subseteq\Col(A^T)^\perp$, we get $$\Col(A^T)\cap\Col(B)=\{0\}.$$

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Could you explain a bit more, please ? Why is it true that the intersection is $\{0\}$ ? Also how can I use this ? –  Nicolas Lykke Iversen Feb 19 at 15:22
    
@NicolasLykkeIversen, I just added the argument for that in the answer. –  Christoph Feb 19 at 15:30
    
Thank you, Christoph Pegel. I guess they just added the hint to use, because we have not yet studied orthogonal complement in our class, but I understand it well. –  Nicolas Lykke Iversen Feb 19 at 15:33

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