# Real inner product from a complex one

Let $V$ be a complex vector space. We may view $V$ as a real vector space by simply ignoring non-real scalars. Now suppose that $\langle \cdot,\cdot,\rangle$ is a complex inner product on $V$, and define $$\langle\langle u,v \rangle\rangle = \operatorname{Re}\langle u,v\rangle$$

The problem says:

(a) Show that this defines a real inner product.
(b) Show that the real inner product $\langle\langle u,v \rangle\rangle$ and the complex inner product $\langle u,v \rangle$ define the same norm.
(c) Show that orthogonality with respect to the complex inner product implies orthogonality with respect to the real inner product, but not conversely. It follows that the real Pythagorean Theorem is stronger than the complex version.

For part (a) I have gotten as far as proving that the inner product is positive and symmetric. I am stuck proving the linearity component from the properties of inner product spaces. For most of these problems my main issues comes from figuring out exactly where to start. I'm not looking for someone to give me the answer but to guide me in where to begin.

• Please edit your previous question before asking another, still hardly readable, one. It looks like you are copying/pasting from some html. Can you see how poorly it renders? – Julien Apr 12 '13 at 1:47
• Please show some effort to solve the problem yourself. It's much better to ask about conceptual difficulties you are having, than to just list problems you can't solve. Why can't you solve them? – Ben Millwood Apr 12 '13 at 1:51
• for part (a) I have gotten as far as proving that the inner product is positive and symmetric. I am stuck proving the linearity component from the properties of inner product spaces. For most of these problems my main issues comes from figuring out exactly where to start. I'm not looking for someone to give me the answer but to guide me in where to begin. – user72195 Apr 12 '13 at 1:55

## 1 Answer

For part (a), you can show linearity like this: $$\langle\langle x + y, z \rangle\rangle = \operatorname{Re}(\langle x + y, z \rangle) = \operatorname{Re}(\langle x, z \rangle + \langle y, z \rangle) = \operatorname{Re}(\langle x, z \rangle) + \operatorname{Re}(\langle y, z \rangle) = \langle\langle x, z \rangle\rangle + \langle\langle y, z \rangle\rangle$$ $$\langle\langle cx, y \rangle\rangle = \operatorname{Re}(\langle cx, y \rangle) = \operatorname{Re}(c\langle x, y \rangle) = c\operatorname{Re}(\langle x, y \rangle) = c\langle\langle x, y \rangle\rangle$$ Note that $c \in \mathbb{R}$, so the third equality in the second line is valid.

For part (b), all you need to say is that $\langle x, x \rangle \in \mathbb{R}$ even in complex inner products since it follows from the conjugate symmetric axiom. So taking the real part doesn't change the value and so $\langle\langle x, x \rangle\rangle = \langle x, x \rangle$. Now take square roots to show norms are equal.

For (c), you can clearly see that if the complex inner product is zero then taking the real part is still zero which proves the first part. But it is possible for the complex inner product to be non-zero but its real part zero (i.e. the value only has non-zero imaginary part). So the converse is false.