# Inner product in Complex field

Given $B\in\Bbb C^{n\times n}$ and let $B$ be invertible. Let $x,y\in\Bbb C^n$. Need to show that $$\langle x,y\rangle =(By)^HBx$$ defines an inner product on $\Bbb C^n$. Need help with showing $\langle x, y\rangle=\overline{\langle y, x\rangle}$ and $\langle x, x\rangle \ge 0$.
PS If someone could edit this to use inner product symbol that would be great.

• Welcome to math.SE. Definitely read how to ask. The addition of brief context and any attempts will usually generate more responses. – J. W. Perry Oct 17 '13 at 8:01
• Hint: $\langle x, y\rangle = \sum_{i=1}^n (Bx)_i \overline{(By)_i}$ – Prahlad Vaidyanathan Oct 17 '13 at 8:26

In the following, for $x \in \mathbb{C}^{n}$, $x^{*}$ denotes ${}^t \overline{x}$.
• $(x,y) \, \longmapsto \, \left\langle x,y \right\rangle = (By)^*(Bx) = y^{*} B^{*} B x$ is linear in $x$ and semi-linear in $y$.
• $\left\langle y,x \right\rangle = x^*B^*By = \Big( y^{*}B^{*}Bx \Big)^{*} = \overline{\left\langle x,y \right\rangle}$ (because $(UV)^{*} = V^{*}U^{*}$) for all $x,y$ in $\mathbb{C}$.
• $\left\langle x,x \right\rangle = x^{*} B^{*}Bx = (Bx)^{*}(Bx) = \Vert Bx \Vert^{2}_{\mathbb{C}^{n}} \geq 0$. So, $\left\langle x,x \right\rangle = 0$ if and only if $x = 0$ (because $B$ is invertible).
The three points above prove that $(x,y) \, \longmapsto \, \left\langle x,y \right\rangle$ is an inner product on $\mathbb{C}^{n}$.