The question and its solution are given below:
My question is:
In the last line, why we are sure that $(Av)^H (Av) \geq 0.$? could anyone explain this for me, please?
The question and its solution are given below:
My question is:
In the last line, why we are sure that $(Av)^H (Av) \geq 0.$? could anyone explain this for me, please?
On
For $u=(u_{1},...,u_{n}),v=(v_{1},...,v_{n})\in\mathbb{C}^{n}$, we define \begin{align*} \left<u,v\right>=\sum_{i=1}^{n}u_{i}\overline{v_{i}}. \end{align*} It is routine to check this is an inner product.
On the other hand, if we treat the elements in $\mathbb{C}^{n}$ as column vector, that is, $n\times 1$ matrix, then the conjugate transpose $u^{\ast}$ is such that $u^{\ast}u=\left<\overline{u},u\right>=\|u\|^{2}$, where $\overline{u}=\left(\overline{u_{1}},...,\overline{u_{n}}\right)$.
$(Av)^*(Av)$ is another way to write $\langle Av,Av \rangle$, the inner product in $\mathbb C^n$.