Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define $A$ and $B$ as being square matrices of dimension $2011$. Prove that if $AB=0$, then at least one of matrices $A+A^{T}$ or $B+B^{T}$ have rank below $2011$.

-- edit --

Rank of a matrix is a number of linear independent rows.

-- edit2 -- dimension instead of rank

share|improve this question
2  
I'd have choosen $2013$ instead of $2011$ :). –  user26857 Jan 7 '13 at 11:21
add comment

1 Answer

up vote 5 down vote accepted

If $A,B\in M_n(K)$, $K$ a field, $n$ odd, and $AB=0$, then $A+A^T$ or $B+B^T$ is singular.

If $A+A^T$ and $B+B^T$ are invertible, then their rank is $n$. But $\operatorname{rank}(A+A^T)\le 2\operatorname{rank}(A)$ and similarly $\operatorname{rank}(B+B^T)\le 2\operatorname{rank}(B)$. Set $n=2k+1$. Then $\operatorname{rank}(A)\ge k+1$ and $\operatorname{rank}(B)\ge k+1$, so $\operatorname{rank}(A)+\operatorname{rank}(B)\ge n+1$. From Sylvester Rank Inequality we have $\operatorname{rank}(A)+\operatorname{rank}(B)-n\le \operatorname{rank}(AB)=0$, a contradiction.

share|improve this answer
    
Please describre, why if $A+A^T$ is invertible, then it's rank is n. –  Steve Jan 7 '13 at 17:16
    
Probably we are assuming that $rank(A+A^T)=rank(B+B^T)=2011$, and showing that this is a contradiction. –  Jonny Jan 7 '13 at 17:28
    
A square $X$ matrix of size $n$ is invertible iff $\det X\neq 0$ iff rank$(X)=n$. –  user26857 Jan 7 '13 at 18:45
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.