Do Positive Definite Matrices Must Be Symmetric/Hermitian?

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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 \(B=0\).


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