Do Positive Definite Matrices Must Be Symmetric/Hermitian?

\( \renewcommand{\vec}[1]{\boldsymbol{#1}} \DeclareMathOperator*{\E}{\mathbb{E}} \DeclareMathOperator*{\Var}{\mathrm{Var}} \DeclareMathOperator*{\Cov}{\mathrm{Cov}} \DeclareMathOperator*{\argmin}{\mathrm{arg\,min\;}} \DeclareMathOperator*{\argmax}{\mathrm{arg\,max\;}} \def\ZZ{{\mathbb Z}} \def\NN{{\mathbb N}} \def\RR{{\mathbb R}} \def\CC{{\mathbb C}} \def\QQ{{\mathbb Q}} \def\FF{{\mathbb FF}} \def\EE{{\mathbb E}} \newcommand{\tr}{{\rm tr}} \newcommand{\sign}{{\rm sign}} \newcommand{\1}{𝟙} \newcommand{\inprod}[2]{\left\langle #1, #2 \right\rangle} \newcommand{\set}[1]{\left\{#1\right\}} \require{physics} \)

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



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

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


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 \(B=0\).

No notes link to this note