In $\mathbb C^2$, Show that $\langle x,y\rangle=xAy*$ is an inner product.

Question

$A= \left[ \begin{array}{cc} 1&i\\ -i&2 \end{array} \right] $

I've shown that

(a) $\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle$.

(b) $\langle cx,y\rangle=c\langle x,y\rangle$.

(c) $\overline{\langle x,y\rangle} = \ \langle y,x\rangle$.

but, i'm not getting how to show $\langle x,x\rangle>0 $, if $x$ is not zero.

Please help!!

Answer 1

$$\left[ \begin{array}{cc} 1 & i \\ -i & 2 \end{array}\right]=\left[ \begin{array}{cc} 1 & 0 \\ -i & 1 \end{array}\right]\left[ \begin{array}{cc} 1 & i \\ 0 & 1 \end{array}\right]$$

Let $A=\left[ \begin{array}{cc} 1 & i \\ -i & 2 \end{array}\right]$ and let $B=\left[ \begin{array}{cc} 1 & 0 \\ -i & 1 \end{array}\right]$.

$$\langle x,x \rangle=xAx^*=xBB^*x^*=(xB)(xB)^*\geq 0$$

Answer 2

We can see that matrix $A$ is a hermitian matrix with positive eigenvalues, namely $\lambda_1 = 0.382$ and $\lambda_2 = 2.618$. This means that $A$ is positive definite. By definition for every vector $x\neq 0 $ we have that $$x^*A x>0$$