Editing Gram–Schmidt process (section)

===Euclidean space===
Consider the following set of vectors in <math>\mathbb{R}^2</math> (with the conventional [[Inner product space#Euclidean vector space|inner product]])
<math display="block">S = \left\{\mathbf{v}_1=\begin{bmatrix} 3 \\ 1\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}2 \\2\end{bmatrix}\right\}.</math>

Now, perform Gram–Schmidt, to obtain an orthogonal set of vectors:
<math display="block">\mathbf{u}_1=\mathbf{v}_1=\begin{bmatrix}3\\1\end{bmatrix}</math>
<math display="block"> \mathbf{u}_2 = \mathbf{v}_2 - \operatorname{proj}_{\mathbf{u}_1} (\mathbf{v}_2)
= \begin{bmatrix}2\\2\end{bmatrix} - \operatorname{proj}_{\left[\begin{smallmatrix}3 \\ 1\end{smallmatrix}\right]} {\begin{bmatrix}2\\2\end{bmatrix}}
= \begin{bmatrix}2\\2\end{bmatrix} - \frac{8}{10} \begin{bmatrix} 3 \\1 \end{bmatrix}
= \begin{bmatrix} -2/5 \\6/5 \end{bmatrix}. </math>

We check that the vectors <math>\mathbf{u}_1</math> and <math>\mathbf{u}_2</math> are indeed orthogonal:
<math display="block">\langle\mathbf{u}_1,\mathbf{u}_2\rangle = \left\langle \begin{bmatrix}3\\1\end{bmatrix}, \begin{bmatrix} -2/5 \\ 6/5 \end{bmatrix} \right\rangle = -\frac{6}{5} + \frac{6}{5} = 0,</math>
noting that if the [[dot product]] of two vectors is 0 then they are orthogonal.

For non-zero vectors, we can then normalize the vectors by dividing out their sizes as shown above:
<math display="block">\mathbf{e}_1 = \frac{1}{\sqrt {10}}\begin{bmatrix}3\\1\end{bmatrix}</math>
<math display="block">\mathbf{e}_2 = \frac{1}{\sqrt{40 \over 25}} \begin{bmatrix}-2/5\\6/5\end{bmatrix}
= \frac{1}{\sqrt{10}} \begin{bmatrix}-1\\3\end{bmatrix}. </math>