Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear all,

As a part of my research, I need to achieve a lower bound to the smallest singular value, $\sigma_{n}\(A+A^{T}\)$ for a stochastic $A$ (as a function of the singular values of $A$), which is generally not Hermitian.

I am aware of the upper bounds (due to Weyl and Fan) and of the fact that for general $\sigma_{i}\(A+B\)$ no lower bound is known. Do you see a way?

Thank you.

Edit: A can be considered a power of a lazy row stochastic matrix. I.e., $A=P^k$ for some strongly diagonally dominant row stochastic $P$.

share|improve this question
1  
See my answer mathoverflow.net/questions/97746 to a similar MO question. However, the assumption that $A$ is stochastic might change the answer. –  Denis Serre Dec 30 '12 at 7:29
    
I came to a relaxation to my problem, wherein A can be a power of a lazy row stochastic matrix. I.e., A=P^k for some strongly diagonally dominant row stochastic P. Hence, P is positive definite, however A is generally not. Do you see a way that it simplifies the problem? –  Daniel86 Jan 3 '13 at 6:22
add comment

1 Answer 1

It does not seem like any obvious bound on $\sigma_n(A+A^T)$ is possible in terms of the singular values of $A$. Indeed, consider $$ A = \left( \matrix{ 1/3 & 1/2 & 1/6 \cr 1/6 & 1/3 & 1/2 \cr 1/2 & 1/6 & 1/3} \right).$$ Clearly, $A$ is stochastic and a computation reveals that it is nonsingular so that all of its singular values are positive. On the other hand, $A+A^T$ is a multiple of the all-ones matrix, so $\sigma_3(A+A^T)=0$.

share|improve this answer
    
Thankyou, it is a nice example, but can this situation occur where $A$ is an integer power of a positive definite matrix? I edited my original post. Thanks. –  Daniel86 Jan 3 '13 at 6:31
    
@Daniel86 - under standard definitions, a real positive definite matrix is automatically symmetric. When you use the words "positive definite" I am guessing you mean that $x^T A x > 0$ for any $x \in R^n$ but $A$ is not necessarily symmetric. Is my understanding correct? –  alex o. Jan 4 '13 at 21:24
    
@alex o. - If $P$ is diagonally dominant but not necessarily symmetric, $P+P^{T}$ is diagonally dominant and symmetric and hence positive definite. However, for $A=P^{k}$ this is not always the case. For such a case, I want to lower bound the smallest singular value of $A+A^{T}$ in terms of the singular values of $P$. –  Daniel86 Jan 4 '13 at 22:30
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.