If $A$ is a matrix with complex entries, then there exists a matrix $B$ such that $AB=0$ and $\operatorname{rank} A+ \operatorname{rank}B=n$.

I have some difficulty writing the proof for this one.This problem appeared in a shortlist for a mathematical olympiad (for high-school):

Let $A$ be a $n\times n$ matrix with complex entries.

1. Prove that there exists a $n\times n$ matrix with complex entries $B$ such that $AB=0$ (the null matrix) and $\operatorname{rank} A + \operatorname{rank}B = n$
2. If $1 < \operatorname{rank} A < n$, prove that there exists a $n\times n$ matrix with complex entries $C$ such that $AC=0, CA \neq 0$ and $\operatorname{rank}A + \operatorname{rank} C = n$

Well, since this problem was proposed for the mathematical olympiad, I suppose that the solution involves basic linear algebra concepts i.e. without vector spaces, linear transformations etc.

So far, I have only come up with the fact that if $A$ is invertible, then $B$ is the zero matrix, and using the Sylvester inequality and supposing $A$ singular implies:

$\operatorname{rank} A + \operatorname{rank} B \leq n$

Also, I guess that stating that $\operatorname{rank} B$ exists from the above inequality doesn't imply that exists such $B$ so that $\operatorname{rank} B = n - \operatorname{rank} A$.

• Throw $A$ into Jordan form. It'll have some number of $0$ eigenvectors. Construct your matrix $B$ in the same basis as $A$ by giving it an eigenspace for $1$ where $A$ has an eigenspace for $0$. Then the two matrices have complementary rank and their product is 0. Commented Jan 30, 2014 at 18:29
• Do you still have a link to that shortlist? Or maybe the problems written somewhere? I'm looking for some problems of this kind and it would help me a lot.
– Asix
Commented Feb 19, 2018 at 15:45

For 1, any $$B$$ such that $$\mathrm{Im}(B)=\ker(A)$$ will do. For example, the projection onto $$\ker(A)$$ along some subspace complementary to $$\ker(A)$$.

For 2, modify the above by $$C=BD$$, where $$D$$ is an isomorphism (thus maintaining $$\mathrm{rank}(C)=\mathrm{rank}(B)$$ and $$AC=0$$), such that $$D(\mathrm{Im}(A))$$ intersects $$\ker(A)$$. You can for example take any linear isomorphism $$D$$ that send some non-zero vector in $$\mathrm{Im}(A)$$ to some non-zero vector in $$\ker(A)$$.

If $$\operatorname{rank} A = m < n$$, then there are $$n-m$$ linearly independent vectors $$b_1, \dots, b_{n-m}$$ so that $$Ab_j = 0$$. What would happen if you used these vectors as the columns of $$B$$ (with $$0$$ columns otherwise), and what could you say about the rank of $$B$$?