# Do Positive Definite Matrices Must Be Symmetric/Hermitian?

In real spaces, “positive definite” does NOT have to be symmetric. E.g.

$M=\begin{bmatrix}2&0\\2&2\end{bmatrix}$

Because

$\begin{bmatrix} x\\y\end{bmatrix}^T \begin{bmatrix}2&0\\2&2\end{bmatrix} \begin{bmatrix} x\\y\end{bmatrix} = x^2 + y^2 + (x+y)^2$

In complex spaces, positive definite DO implies Hermitian. And actually you only need $$x^{\dagger} A x \in \RR, \forall x \in \CC^n$$.

Proof

If $$x^{\dagger} A x \in \RR, \forall x \in \CC^n$$, to show $$A$$ is Hermitian, we have

$x^{\dagger}Ax = \left( x^{\dagger}Ax \right)^{\dagger} = x^{\dagger}A^{\dagger}x \Longrightarrow x^{\dagger} \left( A-A^{\dagger} \right)x = 0, \forall x \in \CC^n$

So only need to show $$x^{\dagger} B x=0, \forall x \in \CC^n \Longrightarrow B=0$$.

For arbitrary $$x, y \in \CC$$, we have $$\left( x+y \right)^{\dagger} B \left( x+y \right) = 0$$, and $$\left( x+iy \right)^{\dagger} B \left( x+iy \right) = 0$$

Solve the equations, we have $$x^{\dagger} B y=0, \forall y \in \CC$$. Therefore $$x^{\dagger} B =0, \forall x \in \CC$$. Therefore $$B=0$$.