If treating linear algebra in a more geometric fashion, the length of a vector occurs naturally, and is what you would expect from its name. With complex numbers, we will define a similar function. Recall that if $c$ is a complex number, then $\modulus{c}$ denotes its modulus (Definition MCN).
The norm of the vector $\vect{u}$ is the scalar quantity in $\complex{\null}$ \begin{equation*} \norm{\vect{u}}= \sqrt{ \modulus{\vectorentry{\vect{u}}{1}}^2+ \modulus{\vectorentry{\vect{u}}{2}}^2+ \modulus{\vectorentry{\vect{u}}{3}}^2+ \cdots+ \modulus{\vectorentry{\vect{u}}{m}}^2 } = \sqrt{\sum_{i=1}^{m}\modulus{\vectorentry{\vect{u}}{i}}^2} \end{equation*}
Computing a norm is also easy to do.
Notice how the norm of a vector with real number entries is just the length of the vector. Inner products and norms are related by the following theorem.
Suppose that $\vect{u}$ is a vector in $\complex{m}$. Then $\norm{\vect{u}}^2=\innerproduct{\vect{u}}{\vect{u}}$.
When our vectors have entries only from the real numbers Theorem IPN says that the dot product of a vector with itself is equal to the length of the vector squared.
Suppose that $\vect{u}$ is a vector in $\complex{m}$. Then $\innerproduct{\vect{u}}{\vect{u}}\geq 0$ with equality if and only if $\vect{u}=\zerovector$.
Notice that Theorem PIP contains three implications: \begin{align*} \vect{u}\in\complex{m}&\Rightarrow\innerproduct{\vect{u}}{\vect{u}}\geq 0\\ \vect{u}=\zerovector&\Rightarrow\innerproduct{\vect{u}}{\vect{u}}=0\\ \innerproduct{\vect{u}}{\vect{u}}=0&\Rightarrow\vect{u}=\zerovector \end{align*}
The results contained in Theorem PIP are summarized by saying “the inner product is positive definite.”