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Given

  • $A \in R^{n \times n}$,
  • $v \in R^n$,
  • $\sigma \in R$, $\sigma=\frac{1}{v^T v}>0$,

is $$B=A A^T - \sigma A v v^T A^T$$ positive definite?

What I currently know is that

  • $A A^T$ being positive definite and
  • $-\sigma A v v^T A^T$ is a negative semidefinite rank-1 matrix with eigenvalues 0 and $- \sigma v^T A^T A v$.

Additionally, I have the following inequality (forgot whose law that was...) $$\lambda_{min} (B)\ge \lambda_{min}(AA^T) - \sigma v^T A^T A v$$ with $\lambda_{min}$ being the smallest eigenvalue of the matrix.

Can I say anything about $\lambda_{min} (B)\ge \lambda_{min}(AA^T) - \sigma v^T A^T A v >0$ ?

Thanks.

user115779
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  • not if $A,$ hence $A^T,$ is singular. – Will Jagy Jan 19 '14 at 20:21
  • A is regular in my case. – user115779 Jan 19 '14 at 20:26
  • In that case, you can just find $w^T B w$ for $w$ a column vector, see what happens. Three cases, $w= v,$ next $w \perp v$, if that is not conclusive then $w = v \cos \theta + u \sin \theta,$ where $u \perp v.$ – Will Jagy Jan 19 '14 at 20:36
  • Hmm, I think I don't understand. I would like to show that B is pos. def. in general. Finding one particular vector w, for which it holds, would not help in this case, would it? – user115779 Jan 19 '14 at 20:44
  • Same idea, $w^T B w,$ first $w = Av,$ next $w \perp Av,$ third a mixeture. – Will Jagy Jan 19 '14 at 20:44
  • If $w^T B w > 0$ for all nonzero $w,$ that is positive definiteness. – Will Jagy Jan 19 '14 at 20:45
  • If dimension is at least $3$ and $v$ is not an eigenvector of $A,$ I guess you need to work in three directions, two are $v$ and $Av,$ the third is anything orthogonal to both. – Will Jagy Jan 19 '14 at 20:52
  • With Av do you mean the v from my B above? Or do you mean some arbitrary vector? If I understand correctly, I have to show w^T B w >0 for arbitrary w, right? – user115779 Jan 19 '14 at 20:56

1 Answers1

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It's false. Take two by two $A=I,$ and take the column vector $v = (0,1)^T.$ Calculate $B.$ I was trying to get you to write down some simple 2 by 2 examples to see what actually happens, rather than fiddle with random theorems when you don't actually know the outcome yet. I'll need to look this up, I think it was Olga Taussky-Todd who said that if a matrix statement is false, there is a two by two counterexample.

Will Jagy
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