Sum of complex inner product is also an inner product proof

Question

Can someone help with the following exercise?

Let $\langle\cdot\mid\cdot\rangle_1$ and $\langle\cdot\mid\cdot\rangle_2$ be two complex inner products on the complex vector space V. Show that $\langle\cdot\mid\cdot\rangle_=\langle\cdot\mid\cdot\rangle_1 + \langle\cdot\mid\cdot\rangle_2$ is also a complex inner product.

Answer 1

Recall the properties that define a complex inner product. For example, one of them is $$\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle.$$

Let's verify that your inner product satisfies this property. \begin{align} \langle u + v, w \rangle &= \langle u + v, w \rangle_1 + \langle u + v, w \rangle_2 \\ &= \langle u, w \rangle_1 + \langle v, w \rangle_1 + \langle u, w \rangle_2 + \langle v, w \rangle_2 \\ &= \langle u, w \rangle_1 + \langle u, w \rangle_2 + \langle v, w \rangle_1 + \langle v, w \rangle_2 \\ &= \langle u, w \rangle + \langle v, w \rangle. \end{align}

Can you verify the other properties?

Answer 2

If $\langle\cdot\mid\cdot\rangle_1,\langle\cdot\mid\cdot\rangle_2:V\times V\to\Bbb F$ then clearly $\langle\cdot\mid\cdot\rangle:V\times V\to\Bbb F$.

$\langle \alpha v\mid w\rangle=\langle \alpha v\mid w\rangle_1+\langle \alpha v\mid w\rangle_2=\alpha\langle v\mid w\rangle_1+\alpha\langle v\mid w\rangle_2=\alpha\big(\langle v\mid w\rangle_1+\langle v\mid w\rangle_2\big)=\alpha\langle v\mid w\rangle$.

$\langle u+v\mid w\rangle=\langle u+v\mid w\rangle_1+\langle u+v\mid w\rangle_2=\langle u\mid w\rangle_1+\langle v\mid w\rangle_1+\langle u\mid w\rangle_2+\langle v\mid w\rangle_2=\big(\langle u\mid w\rangle_1+\langle v\mid w\rangle_1\big)+\big(\langle u\mid w\rangle_2+\langle v\mid w\rangle_2\big)=\langle u\mid w\rangle+\langle v\mid w\rangle$.

$\langle v\mid w\rangle=\langle v\mid w\rangle_1+\langle v\mid w\rangle_2=\overline{\langle w\mid v\rangle_1}+\overline{\langle w\mid v\rangle_2}=\overline{\langle w\mid v\rangle+\langle w\mid v\rangle_2}=\overline{\langle w\mid v\rangle}$.

$\langle v\mid v\rangle=\langle v\mid v\rangle_1+\langle v\mid v\rangle_2\ge 0$, since $\langle v\mid v\rangle_1,\langle v\mid v\rangle_2\ge 0$.

If $\langle v\mid v\rangle=0$ then $\langle v\mid v\rangle_1+\langle v\mid v\rangle_2=0$, which implies $\langle v\mid v\rangle_1=\langle v\mid v\rangle_2=0$ by the previous property. This means $v=0$.

We are done.