# What does the inner product of two complex general vectors have to do with complex conjugation?


I must not have understood this. I tried the following example. What's the inner product of $\v{2+3i}{7+3i} \cdot \v{1+2i}{2+2i}$? It's $4 + 27i$. If I commute $\v{1+2i}{2+2i} \cdot \v{2+3i}{7+3i}$, I get the same $4 + 27i$ because inner products are commutative. Taking the complex conjugate $4 + 27i$, I get $4 - 27i$ which is not the same as its complex conjugate, $4 + 27i$.

So I'm pretty lost here. What does he mean by what he said?

Inner product on comlex vetorspace have conjugate symmetry (so not commutative).

An inner product is a function that assigns to every ordered pair of vectors $x$ and $y$ in V a scalar in $\mathbb{C}$, denoted $\langle x,y\rangle$, such that $\forall\ x, y, z \in V$ and $c \in \mathbb{C}$: \begin{equation*} V \rightarrow F: \quad x,y \rightarrow \langle x,y\rangle \end{equation*}

\begin{array} ((a)\quad \langle x+z,y\rangle= \langle x,y\rangle+ \langle z,y\rangle \\[5pt] (b)\quad \langle cx,y\rangle= c \langle x,y\rangle \\[5pt] (c)\quad \overline{\langle x,y\rangle} =\langle y,x\rangle \\[5pt] (d)\quad \langle x,x\rangle >0 \ \text{if} \ x\ne 0 \\[5pt] \end{array}

Here $\overline{x}$ denotes the complex conjugate of $x$.

In complex n dimensional euclidean space the inner product of two vectors $(a_1,\ldots,a_n)$ and $(b_1,\ldots,b_n)$ defined by $\sum_{j=1}^n \overline{a_j}b_j$

So in your example the correct way of calculation is

$$\langle{2+3i},{7+3i}\rangle \cdot \langle{1+2i},{2+2i}\rangle=28+9i$$

and

$$\langle{1+2i},{2+2i} \rangle \cdot \langle{2+3i},{7+3i}\rangle =28-9i$$

which is consistent with $(c)\ \overline{\langle x,y\rangle} =\langle y,x\rangle$.

To a physicist, the inner product of vectors $(a_1,\ldots,a_n)$ and $(b_1,\ldots,b_n)$ is $\sum_{j=1}^n \overline{a_j}b_j$. Also a physicist would write $z^*$ for the complex conjugate of $z$ rather than $\overline z$.