Finding $x$ with $\langle\, Ax, x\rangle > 0$ when matrix $A$ has at least one eigenvalue with positive real part.

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.

• Look at the complex conjugate of your eigenvalue. Add them together. See what happens. – uniquesolution Nov 18 '20 at 16:28
• Do you mean adding the eigenvalues or the eigenvectors? I tried these, but I couldn't solve it. – JPMarciano Nov 18 '20 at 16:34
• If $\lambda$ is an eigenvalue of $A$, is the complex conjugate $\overline{\lambda}$ also an eigenvalue? Answer yourself this, first. – uniquesolution Nov 18 '20 at 16:37
• 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 $x$. – JPMarciano Nov 18 '20 at 16:43

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 $$\langle Av,v \rangle > 0$$.