5

whether every positive definite matrix has to be symmetric? If not, what will be the example?

thomus
  • 577
  • 3
    Usually, "positive definite" means "symmetric positive definite" (or Hermitian positive definite), but I think you can find examples of authors for whom "positive definite" does not necessarily imply "symmetric" (or Hermitian). – littleO Sep 28 '16 at 06:02
  • 2
    The matrix $A:=\begin{bmatrix}1 & 1 \ -1 & 1\end{bmatrix}$ satisfies $v^T A v > 0$ for all $v$, which may or may not be a counterexample depending on which definitions you are using. You can see this by computing $\begin{bmatrix}x & y\end{bmatrix}A\begin{bmatrix}x \ y\end{bmatrix} = x^2 + y^2$ – Nick Alger Sep 28 '16 at 06:03
  • Isn't this by definition? If $f: V \to V$ is positive operator, then $(fv,v) = (v,fv)$. – IAmNoOne Sep 28 '16 at 06:43

2 Answers2

11

If the definition is simply $x^TAx\geq0$ for all $x $, this can fail over the reals: $$ A=\begin{bmatrix}1&1\\0&1\end {bmatrix} $$ satisfies $x^TAx\geq0$ for all $x\in\mathbb R^2$.

For complex matrices, the condition $x^*Ax\geq0$ for all $x\in\mathbb C^n $ implies that $A $ is selfadjoint.

Martin Argerami
  • 205,756
1

By definition it is assumed to be symmetric.

angryavian
  • 89,882