# 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. – Ian Coley Jan 30 '14 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 Feb 19 '18 at 15:45

If $\textrm{rank} A = m < n$, 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 $B$?
For 1, any $B$ such that $Im(B)= Ker(A)$ 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 $rank(C)= rank(B)$ and $AC=0$), such that $D(Im(A))$ intersects $Ker(A)$. You can for example take any linear isomorphism $D$ that send some non-zero vector in $Im(A)$ to some non-zero vector in $Ker(A)$.