Let $V$ be a subspace of ${\mathbb{R}}^n$ of dimension $k$.

We say that a basis $\{u_1,\dots ,u_k\}$ for $V$ is an orthogonal basis if for all distinct $i,j\in \{1,\dots ,k\}$, $u_i$ and $u_j$ are orthogonal (i.e. $u_i\cdot u_j=0$). Furthermore, it is an orthonormal basis if, in addition, $u_i$ is a unit vector (i.e. $\parallel u_i\parallel =1$ or $u_i\cdot u_i=1$) for each $i=1,\dots ,k$.

Examples

$\{\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]\}$ is an orthonormal basis for ${\mathbb{R}}^3$.

$\{\left[\begin{array}{c}\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right]\}$ is an orthonormal basis for the nullspace of $A=\left[\begin{array}{cccc}1& -2& 0& 0\\ 0& 0& 1& 1\end{array}\right]$

One benefit of having an orthonormal basis $\{u_1,\dots ,u_k\}$ is that if $\mathrm{\Gamma }$ denotes the ordered basis $(u_1,\dots ,u_k)$, then for $v\in V$, $[v{]}_{\mathrm{\Gamma }}$ is given by $\left[\begin{array}{c}{u}_1\cdot v\\ u_2\cdot v\\ ⋮\\ u_k\cdot v\end{array}\right]$.

For example, $v=\left[\begin{array}{c}2\\ 1\\ -1\\ 1\end{array}\right]$ is in the nullspace of $A$ above. Letting $\mathrm{\Gamma }=(u_1,u_2)$ where $u_1=\left[\begin{array}{c}\frac{2}{\sqrt{5}}\\ \frac{1}{\sqrt{5}}\\ 0\\ 0\end{array}\right]$ and $u_2=\left[\begin{array}{c}0\\ 0\\ \frac{1}{\sqrt{2}}\\ -\frac{1}{\sqrt{2}}\end{array}\right]$, we obtain $$[v{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}\frac{2}{\sqrt{5}}(2)+\frac{1}{\sqrt{5}}(1)+0(-1)+0(1)\\ 0(2)+0(1)+\frac{1}{\sqrt{2}}(-1)+(-\frac{1}{\sqrt{2}})(1)\end{array}\right]=\left[\begin{array}{c}\sqrt{5}\\ -\sqrt{2}\end{array}\right].$$ One can easily check that $\sqrt{5}{u}_1-\sqrt{2}{u}_2=v$.

To construct an orthonormal basis for a subspace of ${\mathbb{R}}^n$, one can use the Gram-Schmidt orthonormalization process. The details of this process can be found here.

To see why $[v{]}_{\mathrm{\Gamma }}=\left[\begin{array}{c}{u}_1\cdot v\\ u_2\cdot v\\ ⋮\\ u_k\cdot v\end{array}\right]$, let ${\lambda }_1,\dots ,{\lambda }_k$ be scalars such that $v={\lambda }_1{u}_1+\cdots +{\lambda }_k{u}_k.$

Then for each $i=1,\dots ,k$, $$\begin{array}{rcl}{u}_i\cdot v& =& u_i\cdot ({\lambda }_1{u}_1+\cdots +{\lambda }_k{u}_k)\\ & =& {\lambda }_1{u}_i\cdot u_1+\cdots {\lambda }_k{u}_i\cdot u_k\\ & =& {\lambda }_1{u}_i\cdot u_i\\ & =& {\lambda }_1\parallel u_i{\parallel }^2\\ & =& {\lambda }_1\end{array}$$

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