0

Incorrect theorem: Suppose $A$ is a real matrix, if $\forall x \neq 0,\ x \in \Bbb R^n,\ x^TAx > 0$, then all eigenvalues $> 0$.

False proof: If $Ax = \lambda x$, then $$ x^TAx = \lambda x^Tx = \lambda ||x||^2> 0$$

Since $||x||^2> 0$, so $\lambda > 0$ q.e.d.

So I'm pretty sure this isn't true, since we can take a rotation matrix as a counter-example. However, I can't find the problem with this false proof.

Sean Lake
  • 1,757
  • 10
  • 23
Tony Tarng
  • 491
  • 6
    This proof is perfectly fine; it proves that if an eigenvalue of $A$ is real, then it is greater than $0$. – Redundant Aunt Sep 16 '16 at 11:28
  • 2
    Rotation matrices fail as a counter-example. They don't satisfy $x^T A x > 0\ \forall, x$ if $x$ is defined over the complex field, $\mathbb{C}$ (nb: then you need to use conjugate transpose in place of transpose). If you limit yourself to the real number field, $\mathbb{R}$, then rotation matrices don't have eigenvalues. – Sean Lake Sep 16 '16 at 11:35
  • @SeanLake I'm supposing x is $\in \Bbb R^n$ here. Also, this answer here constructs such a rotation matrix: http://math.stackexchange.com/a/653161/308438 – Tony Tarng Sep 16 '16 at 11:41
  • 3
    Right, and the problem is that general rotation matrices only have eigenvalues for eigenvectors in $\mathbb{C}^n$. For vectors in $\mathbb{R}^n$ the only possible eigenvalues are $\pm 1$, and, if the rotation matrix has eigenvalues, and obeys $x^T Rx > 0 \ \forall x\in \mathbb{R}^n$ then all of it's eigenvalues will be $1$. – Sean Lake Sep 16 '16 at 11:46

1 Answers1

1

Your condition is equivalent to $A+A^T$ is symmetric $>0$. That implies that every eigenvalue $\lambda$ of $A$ satisfies $Re(\lambda)>0$.