Assume that $A$ is non-symmetric. "If $A+A^T$ is positive definite, the eigenvalues of $A$ have positive real parts" (originally, it was with negative definite/negative real parts, but I suspect it doesn't matter). This was claimed in a comment of an answer to if eigenvalues are positive, is the matrix positive definite?.
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Could anyone provide a proof for this?
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Is the other direction true? If yes can anyone present a proof, if not a counterexample.
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If $A$ is diagonalizable is the other direction true? If yes can anyone present a proof, if not a counterexample.
Thank you.
(In case anyone wonders what the motivation behind this is, I'm looking into it with the hope of getting some simple geometric intuition for the conditions of asymptotic stability of the origin of a continuous-time LTI system $\dot{x}=Ax$ (i.e. that the eigenvalues have negative real parts). If the other direction of the claim is true, it implies that the angle between the velocity vector $\dot{x}$ and the position vector $x$ is always greater than $\pi/2$, hence the state is always 'being pushed roughly in the direction of the origin'.)
Best Answer
I think the following simple argument works, although I'm surprised not to have heard of it before:
If $\lambda \in \mathbb C$ is an eigenvalue of $A$ with eigenvector $x$, then $\lambda |x|^2 = \langle Ax,x \rangle = \langle x,A^T x \rangle = \overline{\langle A^T x, x\rangle}$, so $\langle A^T x, x\rangle = \bar{\lambda} |x|^2.$
Therefore $2\Re{\lambda} |x|^2 = (\lambda + \bar\lambda)|x|^2 = \langle Ax,x\rangle + \langle A^T x,x \rangle = \langle (A+A^T)x,x \rangle > 0.$
The other direction is false even for diagonalizable matrices: $\left(\begin{array}{cc} 1&100\\0&2\end{array}\right)$ has positive eigenvalues, but not when added to its transpose.
If we make the very strong assumption that $A$ is normal, then the converse holds by looking at $\exp(-At) \exp(-A^T t) = \exp(-(A+A^T)t)$.