Compute inner product $\langle (1+i,4) , (2-3i, 4+5i) \rangle$ in $\mathbb{C}^2$

Question

Compute inner product $\langle (1+i,4) , (2-3i, 4+5i) \rangle$ in $\mathbb{C}^2$

• My Answer:

$= (1+i)(2-3i) + 4(4+5i)$

• Book's answer:

$=(1+i)(2+3i)+4(4-5i)$

Why does the book's method conjugate the terms in the second argument of the inner product?

I didn't know we conjugated when directly computing the standard inner product over a complex field. I thought we only conjugate when pulling out a scalar from the second argument or when we swap the order of arguments.

Answer 1

There are books that define: $$\langle z,w\rangle_1 = \sum_{j=1}^nz_j\overline{w_j},$$ while others define $$\langle z,w \rangle_2 = \sum_{j=1}^n\overline{z_j}w_j,$$so you have to pay attention to that. If we don't conjugate one of the terms, we won't have the property $\langle z,w\rangle = \overline{\langle w,z\rangle}$ we want. Both are called the standard inner products in $\Bbb C^n$. I numbered them so we can compare: $$\langle (1+i,4),(2-3i,4+5i)\rangle_1 = (1+i)(2+3i)+4(4-5i) \\ \langle (1+i,4),(2-3i,4+5i)\rangle_2 = (1-i)(2-3i)+4(4+5i)$$

You forgot to take conjugates there. Your book is using $\langle\cdot,\cdot\rangle_1$.

Answer 2

You have $\langle a, b\rangle = \sum_i a_i\overline{b_i}$ (or $\langle a, b\rangle = \sum_i \overline{a_i}b_i$ depending on the day of the week) so that $\langle a, a\rangle = \|a\|^2$.