The inner product for vectors in $\mathbb{R}^n$ is just the sum of elementwise multiplications:

\[\langle x, y \rangle = x_1 y_1 + \cdots + x_n y_n = \sum_{k=1}^n x_k y_k\]

But the inner product is not that simple for vectors in $\mathbb{C}^n$:

\[\langle x, y \rangle = x_1 \bar{y_1} + \cdots + x_n \bar{y_n} = \sum_{k=1}^n x_k \bar{y_k}\]

where $\bar{y_k}$ is the complex conjugate of $y_k$ (i.e., if $y_k = a + bi$, $\bar{y_k} = a - bi$).

Why is the inner product for vectors in $\mathbb{C}^n$ defined this way? You could easily imagine the inner product formula being the same for vectors in both $\mathbb{R}^n$ and $\mathbb{C}^n$.

It turns out this this definition of the inner product is important for preserving the idea that the Euclidean norm of a vector is a measure of its magnitude.

To parse a set of numbers as “magnitudes”, we need two things of them:

  • We should be able to order them.
  • They should be $\geq 0$.

This is why complex numbers with nonzero $i$ components cannot be magnitudes. We cannot order them (and therefore, the question of whether or not they are greater than zero does not make sense). The only set of numbers which satisfies both these desiderata are nonnegative reals.

Now, recall that for $x \in \mathbb{R}^n$ the Euclidean norm can be calculated as

\[\|x\| = \sqrt{\langle x, x \rangle}.\]

This is very neat! We’d like it to be true for vectors $x \in \mathbb{C}^n$ as well.

For concreteness, say $x = [a_1 + b_1 i \; \ldots \; a_n + b_n i]^t$. With the definition of the inner product imported directly from $\mathbb{R}^n$-land,

\[\begin{aligned} \| x \| &= \sqrt{\langle x, x \rangle} = \sqrt{x_1^2 + \cdots + x_n^2} \\ &= \sqrt{(a_1 + b_1 i)^2 + \cdots + (a_n + b_n i)^2} \\ &= \sqrt{(a_1^2 - b_1^2 + 2a_1b_1i) + \cdots + (a_n^2 - b_n^2 + 2a_nb_ni)} \\ &= \sqrt{(a_1^2 + \cdots + a_n^2) - (b_1^2 + \cdots + b_n^2) + 2(a_1 b_1 + \cdots + a_n b_n)}. \end{aligned}\]

We cannot guarantee that this root will be real. Even for the simple case where $a_j = 0$, i.e., every entry in $x$ is some multiple of $i$, we get

\[\| x \| = \sqrt{- (b_1^2 + \cdots + b_n^2)} = i \sqrt{(b_1^2 + \cdots + b_n^2)}.\]

If, instead, we define the inner product for vectors in $\mathbb{C}^n$ as shown at the top of the post, we get something much nicer:

\[\begin{aligned} \| x \| &= \sqrt{\langle x, x \rangle} = \sqrt{x_1 \overline{x_1} + \cdots + x_n \overline{x_n}} \\ &= \sqrt{(a_1 + b_1 i)(a_1 - b_1 i) + \cdots + (a_n + b_n i)(a_n - b_n i)} \\ &= \sqrt{(a_1^2 + b_1^2) + \cdots + (a_n^2 + b_n^2)} \\ &= \sqrt{(a_1^2 + \cdots + a_n^2) + (b_1^2 + \cdots + b_n^2)}. \end{aligned}\]

This root is now guaranteed to be a nonnegative real. It is also more obviously the magnitude of a vector than the root we got with the naive definition of the inner product. Each entry $a_k + b_k i$ of the vector $x$ is a complex number with the magnitude $\sqrt{a_k^2 + b_k^2}$. It makes a lot of sense that the magnitude of the vector made up of these entries is $\sqrt{\sum_{k=1}^n (a_k^2+b_k^2)}$.