Gram-Schmidt Process

Given n linearly independent vectors that form the subspace $W$, you can use the Gram-Schmidt Process to find the orthogonal basis for $W$.

$2$ Vectors

We take one of the 2 vectors and find its ‘z’ using the other vector. Now this new vector and ‘other’ vector are the new orthogonal basis. As math: Given the linearly independent vectors: $\vec{v},\vec{u}$ the orthogonal basis is \{\vec{v},\vec{u}-\text{Proj}_\vec{v}{\vec{u}}\}.

More formally, given the linearly independent vectors: $\overrightarrow{x_1},\overrightarrow{x_2}$,

$$\begin{array}{rl}\overrightarrow{v_1}& =\overrightarrow{x_1}\\ \overrightarrow{v_2}& =\overrightarrow{x_2}-\frac{\overrightarrow{x_2}\cdot \overrightarrow{x_1}}{\overrightarrow{x_1}\cdot \overrightarrow{x_1}}\overrightarrow{x_1}\end{array}$$

Then the set $\{\overrightarrow{v_1},\overrightarrow{v_2}\}$ is the orthogonal basis.

n-Vectors

Given the linearly independent vectors: $\overrightarrow{x_1},\overrightarrow{x_2},\dots ,\overrightarrow{x_n}$

$$\begin{array}{rlr}\overrightarrow{v_1}& =\overrightarrow{x_1}& W_1=\text{Span}\{v_1\}\\ \overrightarrow{v_2}& =\overrightarrow{x_2}-{\text{Proj}}_{W_1}\overrightarrow{x_2}& W_2=\text{Span}\{v_1,v_2\}\\ \overrightarrow{v_3}& =\overrightarrow{x_3}-{\text{Proj}}_{W_2}\overrightarrow{x_3}& W_3=\text{Span}\{v_1,v_2,v_3\}\\ ⋮& & ⋮\\ \overrightarrow{v_n}& =\overrightarrow{x_n}-{\text{Proj}}_{W_{n-1}}\overrightarrow{x_n}& W_n=\text{Span}\{v_1,\dots ,v_n\}\end{array}$$

The orthogonal basis will be $\{\overrightarrow{v_1},\overrightarrow{v_2},\dots \overrightarrow{v_n}\}$ We will use the Orthogonal Decomposition Theorem that can be found in Formulas