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Given a real matrix $A_{n\times n}$ with at least one eigenvalue having positive real part, how can I find $x\in\mathbb{R}^n$ such that $\langle\, Ax, x\rangle > 0$?

I tried to use the real part of the eigenvector, but it didn't quite work. With $x=u+vi$ eigenvector of the eigenvalue $\lambda = a+bi$, $a>0$, we have $Au = au-bv$. Then,

$$\langle\, Au, u\rangle = a|u|^2-b \langle\, u, v\rangle.$$

But I don't know if the RHS is always greater than 0.

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    $\begingroup$ Look at the complex conjugate of your eigenvalue. Add them together. See what happens. $\endgroup$ – uniquesolution Nov 18 '20 at 16:28
  • $\begingroup$ Do you mean adding the eigenvalues or the eigenvectors? I tried these, but I couldn't solve it. $\endgroup$ – JPMarciano Nov 18 '20 at 16:34
  • $\begingroup$ If $\lambda$ is an eigenvalue of $A$, is the complex conjugate $\overline{\lambda}$ also an eigenvalue? Answer yourself this, first. $\endgroup$ – uniquesolution Nov 18 '20 at 16:37
  • $\begingroup$ Yes, because A is a real matrix, so its characteristic polynomial $p$ has real coefficients and this implies that if $\Im (\lambda) \not=0$ then $p(\bar{\lambda})=0$. And in the case of $\Im (\lambda) =0$ the problem can be solved by taking the eigenvector of $\lambda$ as $x$. $\endgroup$ – JPMarciano Nov 18 '20 at 16:43
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We have $Ax = \lambda x$. This means that $$ Au + i Av = A(u + vi) = (a + bi)(u + vi) = (au - bv) + (bu + av)i. $$ Now, note that $$ \langle Au,u \rangle + \langle Av,v \rangle = \langle au - bv,u\rangle + \langle bu + av,v \rangle\\ = a \|u\|^2 + a \|v\|^2 - b \langle v,u \rangle + b \langle u,v \rangle\\ = a (\|u\|^2 + \|v\|^2) > 0. $$ It follows that we must have $\langle Au,u \rangle > 0$ or $\langle Av,v \rangle > 0$.

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  • $\begingroup$ Awesome! Thank you! $\endgroup$ – JPMarciano Nov 18 '20 at 17:23

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