# Proving something is an inner product

I am faced with the following question:

Prove that this is an inner product on V.

I understand to show something is an inner product I must verify the following conditions:

1. $$\langle\,f,g\rangle = \overline{\langle\,g,f\rangle}$$ (Conjugate symmetry)

2. $$\langle\,\overline\alpha (f + g),h\rangle = \overline\alpha(\langle\,f,h\rangle + \langle\,g,h\rangle)$$ (Linearity in the first argument)

3. $$\langle\,f,f\rangle \ge 0$$ with equality iff $$f=0$$

(Note: conditions (1) and (2) are slightly different to usual as we are dealing with complex numbers)

I have seen other examples of this being done, however, I am not sure how to do it in this particular case. Any help with this would be greatly appreciated.

• Could you be more specific about where you're running into trouble? If you tried copying what you've seen for similar examples, then where exactly did you get stuck? Jan 26, 2021 at 20:38
• @James, this is one of those grind-em-out problems, but straightforward. Write out the products and compute the traces. If it's any comfort for you, all of this holds for $n \times n$ matrices too. Jan 26, 2021 at 20:40
• Shouldn't 2. be $\langle\bar{\alpha}f+\bar{\beta}g,\,h\rangle=\bar{\alpha}\langle f,\,h\rangle+\bar{\beta}\langle g,\,h\rangle$?
– J.G.
Jan 26, 2021 at 20:51
• @Aruralreader sorry I don't fully understand what you mean by this? Jan 26, 2021 at 20:57
• @BenGrossmann I got stuck at knowing where to begin. The other examples didn't involve complex numbers and were more straightforward. Jan 26, 2021 at 21:00

If you have $$A$$ and $$B$$, compute $$C_{ij} := \sum_k A^*_{ki}B_{kj}$$. That's what $$A^\dagger B$$ is all about, and with it you can compute its trace, giving \begin{align} \langle A, B\rangle &= \sum_{i} A^*_{ki}B_{ki}. \end{align}

Straightforward this way to work through each of the conditions you list for an inner product.