Vector Magnitude as Dot Product of Self

Magnitudes and dot products

Magnitude of a vector in $\mathbb{R}^n$

Consider a vector $\vec{v}=(v_1) \in \mathbb{R}^1$. We define its magnitude to be $\left|v_1\right|$. Thus $\left|\left|\vec{v}\right| \right|^2=v_1^2$. Now consider $\vec{v} \in \mathbb{R}^2$. We have that $\vec{v}$ is the hypotenuse of a triangle formed by $v_1$ and $v_2$. Thus its length is $\sqrt{v_1^2+v_2^2}$. Now, we can show inductively that $||\vec{v}||$ for a vector in $\mathbb{R}^n$ is $\sqrt{v_1^2+v_2^2 + \cdots + v_n^2}$, it is the hypotenuse of the triangle with legs a vector in $\mathbb{R}^{n-1}$ and $|v_n|$. We assume that $||(v_1, \cdots, v_{n-1})|| = \sqrt{v_1^2+v_2^2 + \cdots + v_{n-1}^2}$. Now using the Pythagorean theorem we get $||\vec{v}||^2 = \left(\sqrt{v_1^2+v_2^2 + \cdots + v_{n-1}^2}\right)^2 + v_n^2$ $\implies ||\vec{v}|| = \sqrt{v_1^2+ \cdots + v_n^2}$.

Magnitude in terms of the dot product

Recall that $\vec{u} \cdot \vec{v}=\sum{u_iv_i}$, where $i$ goes from $1$ to $n$, and $n$ is the dimension of the space containing $\vec{u}$ and $\vec{v}$. Now, if we have $\vec{v}=\vec{u}$, then the dot product is $\sum{v_i^2}$. But this is also the square of the magnitude of $\vec{v}$. Thus we have a new formula for the magnitude of a vector $\vec{v}$: $\left|\left|\vec{v}\right| \right| = \sqrt{\vec{v}\cdot\vec{v}}$.

So, given a vector $\vec{v}$ with coordinates $(a, b)$, its $magnitude$, or length, is $\sqrt{a^2+b^2}$, just like in the Pythagorean Theorem.