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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

Martin Sleziak
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ftiaronsem
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3 Answers3

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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 $\mathrm{rank}(A)+\mathrm{rank}(B) = 2n\gt n$).

Arturo Magidin
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  • Yeah, I think I got it: – ftiaronsem Jan 04 '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 04 '11 at 19:10
  • Thank you so much. And one more thing ;-). Is $nullity(A)=dim(ker(A))$? – ftiaronsem Jan 04 '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 04 '11 at 19:16
  • Ahh, this really helps to clear my head. So hear is the second cleaner atempt: $\left(\forall v \in \mathrm{Im}(B): Av=0 \Rightarrow v \in \mathrm{ker}(A)\right) \Rightarrow \mathrm{Im}(B) \subseteq \mathrm{ker}(A) \Rightarrow \dim(\mathrm{Im}(B))\leq\dim(\mathrm{ker}(A))$. This concludes in $n=\dim(\mathrm{Im}(A))+\dim(ker(A))≥\dim(mathrm{Im}(A))+\dim(\mathrm{Im}(B))$, which is what we wanted to prove ;-). Thanks again for your excellent answer. – ftiaronsem Jan 04 '11 at 19:39
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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 04 '11 at 19:44
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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 $\rho(A) +\rho(B) \leq n$.

Zipzap
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I prefer a different proof than the ones stated above. Let $A$ and $B$ be two $n \times n$ matrices such that $AB = O$. This is equivalent to saying that $[Ab_1 \dots Ab_n] = O$ such that $Ab_i = 0$ for all $i$. First we prove that $\text{coll}(B) \subseteq \text{null}(A)$. Take some vector from the columns space from B. Then this vector, $w$, is a linear combination of the columns of B, i.e., $w = c_1b_1 + \dots c_nb_n$. Then $Aw = A(c_1b_1 + \dots c_nb_n) = c_1(Ab_1) + \dots + c_n(Ab_n) = 0 + 0 + \dots + 0 = 0$. We conclude that $w \in \text{null}(A)$ such that $\text{coll}(B) \subseteq \text{null}(A)$. Therefore, the rank of B must be smaller than the nullity of $A$. Using the rank theorem we obtain $$\text{rank}(A) + \text{rank}(B) \leq \text{rank}(A) + \text{null}(A) = n.$$