Let $ S \in \mathbb{R}^n $ and $ A \in \mathbb{R}^{n \times n} $ Is it possible to find $ \alpha \in \mathbb{R} $, as a function of $A$, such that $$ S S^T ( \alpha 1_{n \times n} + A ) + ( \alpha 1_{n\times n} + A^T) S S^T \succeq 0 $$ ?
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For a particular $A$? For all $A$? As a function of $A$? – 5xum Sep 26 '18 at 12:09
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as a function of $A$ – C Marius Sep 26 '18 at 12:26
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It could not for all $A$, expand the left side and take $A=-a.1_{n\times n}$ for some big positive $a$. As a function of $A$ is a tricky but the answer is still no we have to find a counter-example – Toni Mhax Sep 26 '18 at 12:35
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1@ToniMhax but then you can just take $\alpha > a$ right ? – P. Quinton Sep 26 '18 at 12:40
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It happens only if $S$ is an eigenvector of $A^T$ – Exodd Sep 26 '18 at 12:57
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The answer still no – Toni Mhax Sep 26 '18 at 12:58
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P Quinton your answer was usefull thou. If posible please restore it – C Marius Sep 26 '18 at 14:09
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Here's a counterexample: take $n=2$, $S = e_1$ and $$ A = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}. $$ you get $$ S S^T ( \alpha 1_{n \times n} + A ) + ( \alpha 1_{n\times n} + A^T) S S^T = A = \begin{pmatrix} 2\alpha & 1\\ 1 & 0 \end{pmatrix} $$ that has $-1$ as determinant, so exactly one eigenvalue is negative.
If you want to probe deeper, try to prove that there exists such an $\alpha$ if and only if $S$ is an eigenvector of $A^T$
Exodd
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