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Given $A_1, ..., A_n$ ($n\geq 3$), where each $A_i$ is a $d$-by-$d$ symmetric, positive definite matrix, define $S = A_1\cdot A_2\cdot...\cdot A_n$ (product of all the $A_i$'s). Let $\lambda_1(A)$ and $\lambda_d(A)$ be the top and bottom eigenvalues of $A$, respectively.

Can we say anything about $\lambda_{d}( S + S^T )$? I know we can lower bound $\lambda_{d}(S+S^T) \geq -2 \lambda_{1}(A_1)\cdot ... \cdot \lambda_{1}(A_n)$ but can we say anything less conservative involving the $\lambda_{d}(A_i)$'s? Ideally, I'd like to be able to say something like $\lambda_{d}(S+S^T) \geq 2\lambda_{d}(A_1)\cdot ... \cdot \lambda_{d}(A_n)$ but I have a feeling this is not true.

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    $\begingroup$ if d is even then all eigenvalues might be negative : According to the theorem of Ballantine (projecteuclid.org/download/pdf_1/euclid.pjm/1102991595) we can write minus identity as a product of positive definite real matrices, since it has positive determinant. $\endgroup$ – jjcale May 22 '14 at 19:14
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A counter-example to your bound is easy to construct. Try with 2 matrices, $$ A_1=\left(\begin{array}{cc} 1+x & 1\\ 1& 1 \end{array}\right)\quad A_2=\left(\begin{array}{cc} 1 & 1\\ 1& 1+\frac{1}{x} \end{array}\right). $$ with $x>0$. $A_1A_2+A_2A_1$ has ** corrected** a negative eigenvalue for all $x\neq1$. For large $x$,$$\lambda_d(A_1)\lambda_d(A_2)\sim x^{-1}, \quad \lambda_1(A_1)\lambda_1(A_2)\sim x\quad \lambda_d(A_1A_2+A_2A_1)\sim \left(1-\sqrt{2}\right)x,$$ the conservative estimate is really what you should expect.

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  • $\begingroup$ Wurlitzer : to "negative definite" : Not true, e.g. for x = 2, there is 1 positive and 1 negative eigenvalue, so its neither positive nor negative definite. $\endgroup$ – jjcale May 22 '14 at 18:03
  • $\begingroup$ @jjcale: sorry, yes. I corrected my statement, thanks. $\endgroup$ – username May 22 '14 at 19:52

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