Geometry :: Linear Algebra Review
1 Linear Algebra Review
A vector is an object with a direction and magnitude. We often denote a vector \(u\) as \(u = [u_1,u_2, \ldots u_n]\), where \(n\) is the dimension of the vector.
Let \(u\) and \(v\) be vectors of dimension \(n\). The algebraic dot product of \(u\) and \(v\) is
\[u \cdot v = \sum_i^n u_iv_i.\]
The length of a vector \(u\) is \[\|u\| = \sqrt{u \cdot u}.\]
The geometric dot product of \(u\) and \(v\) is \(u \cdot v = \|u\| \|v\| \cos \theta\), where \(\theta\) is the angle between \(u\) and \(v\).
The geometric and algebraic dot products are the same.
For points \(P = (x_0,y_0)\) and \(Q = (x_1,y_1)\) we can define a vector \(u\) which starts at \(P\) and ends at \(Q\) by \(u = [(x1-x0), (y1-y0)]\).
For any two vectors \(u\) and \(v\) of dimension \(n\),
\[|u\cdot v| \leq \|u\| \|v\|.\]
Let \(u\) and \(v\) be two vectors, and \(\theta\) the angle between them.
\begin{align*} |\cos \theta| &\leq 1 \\ \|u\| \|v\| |\cos \theta| &\leq \|u\| \|v\| \\ |u \cdot v| &\leq \|u\| \|v\|. \end{align*}For any two vectors \(u\) and \(v\) of dimension \(n\),
\[\|u + v\| \leq \|u\| + \|v\|.\]
For vectors \(u\) and \(v\),
\begin{align*} \|u + v\|^2 &= (u+v)\cdot (u+v) \\ &= u\cdot u + 2u\cdot v + v\cdot v\\ &= \|u\|^2 + 2u\cdot v + \|v\|^2\\ &\leq \|u\|^2 + 2\|u\|\|v\| + \|v\|^2\\ \|u + v\|^2 &\leq (\|u\| + \|v\|)^2. \end{align*}Taking the square root of each side gives the result.
For points \(P,Q\), and \(R\), \(PQ \leq PR + RQ\).
Let \(P=(x_0,y_0), Q=(x_1,y_1)\), and \(R=(x_2,y_2)\) be points. Define \(u\) to be the vector starting at \(P\) and ending at \(R\) and \(v\) to be the vector starting at \(R\) and ending at \(Q\). Notice that
\begin{align*} u &= [(x_2-x_0), (y_2-y_0)] \\ v &= [(x_1-x_2), (y_1-y_2)] \\ u+v &= [(x_1-x_0), (y_1-y_0)]. \end{align*}Then
\begin{align*} \|u+v\| &\leq \|u\| + \|v\| \\ \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2} &\leq \sqrt{(x_2-x_0)^2 + (y_2-y_0)^2} + \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2} \\ PQ &\leq PR + RQ. \end{align*}