This is the question: let $A$ be a $n\times n$ symmetric matrix and all eigenvalues of $A$ are negatives, and $B$ a $n\times n$ antisymmetric matrix. Show that $x'=(A+B)x$ is stable at the $0$ using $V(x)=<x,x>$ as Lyapunov function.
So far I get that $V'(x)=2(x^tAx+x^tBx$) $, V'(x)\in \mathbb(R)$, then $V'(x)=V'(x)^t$, wich leads to $x^tBx=0$, so the problem resumes to show that $x^tAx\le0$ in an open set that contains the $0$. I think that I have to use some propertie of $A$ having negative eigenvalues, but I can't see how to use that.
If I can get a tip I would appreciate a lot.
Thanks in advance.
Any symmetric matrix can be diagonalized by an orthogonal matrix, that is, there exist an orthogonal matrix $S$ such that $\Lambda=S^tAS$ is diagonal, of course having the eigenvalues in the main diagonal, say $\lambda_1,\ldots,\lambda_n<0$. Therefore, introducing the new coordinate $y=S^tx$ you obtain $$ x^tAx=x^tS\Lambda S^tx=y^t\Lambda y=\sum_{i=1}^n\lambda_i y_i^2\le0 $$ and in fact $x^tAx<0$ for $x\ne0$ (which is the same as $y\ne0$).