I am trying to prove that an $n \times n$ matrix is positive definite iff all of its eigenvalues are positive.

I know that if $\lambda$ is an eigenvalue then:

$Ax = \lambda x$ for eigenvalues lambda. From there I can go to:

$x^TAx = x^T\lambda x$

Is there some special property of transpose matrices I can use to just get $\lambda$ on the right side of the equation? Because then the proof would pretty much be done.


$$x^T\lambda x = \lambda x^Tx = \lambda ||x||^2$$

  • $\begingroup$ and $||x||^2$ is always positive, right? $\endgroup$ – Jorge Castillo Apr 19 '15 at 22:13
  • $\begingroup$ @JorgeCastillo assuming $ x\neq 0$, yes ;-) $\endgroup$ – Ant Apr 19 '15 at 22:14

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