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Let $A\in \mathcal{M_{n\times n}}(\mathbb{C})$ be a matrix.

Does $x^*Ax>0$ for all $x\in \mathbb{C}^{n\times 1}$ such that $x\neq 0_{\mathbb{C}^{n\times 1}}$ imply that $A$ is hermitian?

Please provide a proof or a counter-example accordingly.

Are positive-definite matrices defined for all matrices or just for hermitian ones?

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Set $x=\left[ \begin{array}{c} i \\ 1 \end{array} \right]$. We get $x^*Ax=2-i$ which isn't even real, let alone greater than zero :( – Linna Dec 30 '12 at 23:15
Good point. I was too hasty to respond. Thanks. – Daryl Dec 30 '12 at 23:25
up vote 1 down vote accepted

Yes. Actually, $x^*Ax\in\mathbb R$ for all $x\in\mathbb C^n$ is enough to force $A$ to be hermitian.

Given $\alpha\in\mathbb C$, we have $$ (x+\alpha y)^*A(x+\alpha y)=x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay. $$ As the term on the left is real, we have that the right hand side equals its conjugate: $$ x^*Ax+|\alpha|^2y^*Ay+\overline\alpha y^*Ax+\alpha x^*Ay=x^*A^*x+|\alpha|^2y^*A^*y+\alpha x^*A^*y+\overline\alpha y^*A^*x. $$ As $x^*Ax, y^*Ay\in\mathbb R$, we have $x^*A^*x=x^*Ax$, $y^*A^*y=y^*Ay$. Then $$ \overline\alpha y^*Ax+\alpha x^*Ay=\alpha x^*A^*y+\overline\alpha y^*A^*x. $$ Taking $\alpha=1$, we get $$\tag{1} y^*Ax+x^*Ay=x^*A^*y+y^*A^*x. $$ Taking $\alpha=i$, we get $$\tag{2} -y^*Ax+x^*Ay=x^*A^*y-y^*A^*x. $$ Subtracting $(2)$ from $(1)$, we get $$ y^*Ax=y^*A^*x. $$ As $x,y$ we arbitrary, we conclude that $A=A^*$.

Regarding your second question, there is no agreement. Some require positive-definite matrices to be hermitian, while others simply ask for $x^*Ax>0$. The two things the same in the complex case, but not in the real one.

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If $x^* A x$ is real, that says $x^* A x = x^* A^* x$.

Now the polarization identity

$$ x^* A y = \frac{1}{4} \left( (x + y)^* A (x + y) - (x - y)^* A (x - y) - i (x + i y)^* A (x + i y) + i (x - i y)^* A (x - i y)\right)$$

determines $x^* A y$ in terms of $z^* A z$ for some vectors $z$. Thus if $x^* A x = x^* A^* x$ for all $x$, we also have $x^* A y = x^* A^* y$ for all $x$ and $y$, and then $A = A^*$, i.e. $A$ is hermitian.

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Thanks a lot, again!! – Linna Dec 30 '12 at 23:40

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