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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 $P$.

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    $\begingroup$ See my answer mathoverflow.net/questions/97746 to a similar MO question. However, the assumption that $A$ is stochastic might change the answer. $\endgroup$
    – Denis Serre
    Commented Dec 30, 2012 at 7:29
  • $\begingroup$ 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? $\endgroup$
    – Daniel86
    Commented Jan 3, 2013 at 6:22

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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 $\sigma_3(A+A^T)=0$.

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  • $\begingroup$ 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. $\endgroup$
    – Daniel86
    Commented Jan 3, 2013 at 6:31
  • $\begingroup$ @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? $\endgroup$
    – alex o.
    Commented Jan 4, 2013 at 21:24
  • $\begingroup$ @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 $P$. $\endgroup$
    – Daniel86
    Commented Jan 4, 2013 at 22:30

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