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As the title says, am searching for a proof of

If $A,B \in \mathbb{R}^{n\times n}$ and $AB=0$ then $\mathrm{rank}(A)+\mathrm{rank}(B) \leq n$

I am doing this as preparation for an upcoming exam and can't figure a way to start. Please just post small hints as answers. I will try to go from there.

Thank you

ftiaronsem

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4  
If $AB=0$, then the image of $B$ is contained in the kernel of $A$. – Chris Eagle Jan 4 '11 at 17:41
    
More general: math.stackexchange.com/questions/269474/… – Martin Sleziak May 28 at 14:02
up vote 8 down vote accepted

By the Rank-Nullity Theorem, $\mathrm{rank}(A)+\mathrm{nullity}(A)=n$. The problem would be solved if you could show that $\mathrm{rank}(B)\leq\mathrm{nullity}(A)$. Presumably, $AB=0$ will play a role in that, since the result is false otherwise (if $A$ and $B$ were both invertible, for example, then $AB\neq 0$, and $\mathrm{rank}(A)+\mathrm{rank}(B) = 2n\gt n$).

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Yeah, I think I got it: – ftiaronsem Jan 4 '11 at 18:58
    
Lets take a Vector $v \in R^{x \times x}$. $(AB)v = A(Bv)$. Now there are two cases. First $v \in ker(B)$ Than there is nothing to do for $A$. Secondly $v \in Img(B) \Rightarrow (Bv) \in ker(A) \Rightarrow Img(B) \in ker(A)$. So, as Chris sayed, the image of $B$ is contained in the kernel of $A$. However the kernel of $A$ might be larger, which is why we get $dim(Img(B)) \leq dim(ker(A))$. This concludes in $ n = dim(Img(A))+dim(ker(A)) \geq dim(Img(A))+dim(Img(B))$, which is what we wanted to proof. Hopefully this correct ... – ftiaronsem Jan 4 '11 at 19:10
    
Thank you so much. And one more thing ;-). Is $nullity(A)=dim(ker(A))$? – ftiaronsem Jan 4 '11 at 19:12
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@ftiaronsem: Yes, $\mathrm{nullity}(A)=\dim(\ker(A))$. But your argument above is confused. It doesn't matter where $v$ is, $Bv$ lies in the image of $B$ (could be zero, of course). Also, the kernel of $B$ and the image of $B$ are not necessarily mutually exclusive, nor do they have to cover all possibilities (e.g., $B=\left(\begin{array}{cc}0&1\\0&0\end{array}\right)$). The only thing you need to do is show that the image of $B$ is contained in the kernel of $A$, which will automatically give $\dim(\mathrm{Im}(B))\leq\dim(\mathrm{ker}(A))$. Also, "proof" is the noun; "wanted to prove". – Arturo Magidin Jan 4 '11 at 19:16
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@ftiaronsem: Yes, that is much cleaner; you might want to say a word or two about why $v\in\mathrm{Im}(B)$ implies $Av=0$ ("there exists $w$ such that $Bw=v$, so $Av = A(Bw) = (AB)w = 0$ by assumption") as that makes the role of the hypothesis $AB=0$ crystal clear. Otherwise, fine. – Arturo Magidin Jan 4 '11 at 19:44

By the definition of matrix multiplication, every column in the matrix $B$ is a solution to the homogeneous equation $A{\bf x} = {\bf 0}$, since $AB = {\bf 0}$. In other words you can say that the span of columns of $B$ is a subset of the span of the solutions of the equation $A{\bf x} = {\bf 0}$.

To clear it up, $W_B^c \subseteq P(A)$, when $W_B^c$ is the span of the columns of $B$, and $P(A)$ is the span of the solutions of the equation $A{\bf x} = {\bf 0}$.

From here you know that the dimension of $W_B^c$ is smaller or equal to the dimension of $P(A)$, $\dim(W_B^c) \leq \dim(P(A))$. Also, the span of the columns is the rank of columns of $B$ in other words. Combine that with $\dim(P(A)) = n - \rho(A)$ and you get $\rho(B) \leq n - \rho(A)$ which is what you were looking for $\rho(A) +\rho(B) \leq n$.

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