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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 $2011$.

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

-- edit2 -- dimension instead of rank

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    $\begingroup$ I'd have choosen $2013$ instead of $2011$ :). $\endgroup$ – user26857 Jan 7 '13 at 11:21
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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.

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

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