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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?.

  1. Could anyone provide a proof for this?

  2. Is the other direction true? If yes can anyone present a proof, if not a counterexample.

  3. 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'.)

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jkn
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  • $A+A^T$ is symmetric, thus diagonalizable, and since it is positive definite it is thus similar to a diagonal matrix with strictly positive eigenvalues. $A$'s eigenvalues are equal to $A^T$'s. – akkkk Aug 29 '12 at 11:33
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    @Auke: and what can be concluded? – Fabian Aug 29 '12 at 11:52
  • Fabian: if I'd have a proof I'd give it. These facts can be concluded straight away. Of course the complicating factor here is that $A$'s eigenspaces are different from $A^T$'s. – akkkk Aug 29 '12 at 11:58

3 Answers3

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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 $\exp(-At) \exp(-A^T t) = \exp(-(A+A^T)t)$.

Erick Wong
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  • Wait, maybe I am confused but isn't in general $<x,x>\neq |x|^2$ if $x$ is complex. – Simon Markett Aug 29 '12 at 12:37
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    @SimonMarkett I'm using the complex inner product $\langle x, y\rangle = y^H x$ (my convention is that it's linear in the first argument). – Erick Wong Aug 29 '12 at 12:42
  • Thanks Erick, however what about a proof for the converse in the case that $A$ is diagonalizable? (I meant to put this in the question originally and I forgot ><, adding it now...). – jkn Aug 29 '12 at 12:55
  • @JuanKuntz My mistake: I have edited the answer to show that diagonalizability (even over $\mathbb R$) makes no difference. However, normality seems to be sufficient. – Erick Wong Aug 29 '12 at 13:06
  • @ErickWong Cheers Erick. – jkn Aug 29 '12 at 13:49
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    @ErickWong, bringing up an old post. Thanks for this proof. After doing some digging I found that this is essentially the calculation done in Horn & Johnson "Topics in Matrix Analysis", Property 1.2.5, where they show that the field of values of a matrix $A$, $F(A)$, satisfies the projection: $F(\frac12(A + A^*) ) = \text{Re}F(A)$. – owen88 Feb 23 '15 at 09:25
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consider the following dynamical system: $$\dot{x}=Ax$$ Lets study the stability of this system knowing that $A+A^T<0$.

Consider the following Lyapunov function $$V=x^Tx$$ Taking the derivative along the system dynamics we have $$\dot{V}=\dot{x}^Tx+x^T\dot{x}=x^T(A^T+A)x<0$$ Therefore, the system is globally (the Lyapunov function is radially unbounded) asymptotically stable (here actually it is globally exponentially stable), meaning that $A$ is Hurwitz (the real part of all eigenvalues is negative).

Eric Brown
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    Thanks, sara sa. I'm very familiar with the above ^^. The point of the above exercise was to give some geometric simple intuition to a group of undergrads who are not going to be taught about Lyapunov functions. – jkn Aug 30 '12 at 19:56
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A nice and simple alternative proof is to use Bendixon's bounds.

If $m$ and $M$ are, respectively, the smallest and largest eigenvalues of $(A + A^T)/2$, then $$m < \text{Re}\{\lambda\} < M,$$ where $\lambda$ is any eigenvalue of $A$.

If $(A + A^T)/2$ is negative definite, we have $M < 0$. All the eigenvalues of $A$ will have negative real part, and $A$ is thus stable.

PS: I have found that a similar post mentioned the same theorem.

Alex M.
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