Definition 6.1.1.
A set of vectors \(\{v_1,\ldots,
v_k\}\) is called orthogonal if
\begin{equation*}
v_i\cdot v_j = \delta_{ij}=\begin{cases}1 \amp \text{ if \(i=j\) } \\ 0 \amp \text{ otherwise } \end{cases}
\end{equation*}
Section 6.1 Orthogonality
In this chapter we deal with orthogonality of vectors and various properties.
Recall, that if \(S=\{v_1,\ldots,
v_k\}\) is linearly independent subset of \(\R^n\) and \(v_{k+1}\notin span(S)\text{,}\) then \(S\cup \{v_{k+1}\}=\{v_1,\ldots,
v_k,v_{k+1}\}\) is linearly independent subset of \(\R^n\text{.}\)
If \(x=(x_1,\ldots, x_n) \in \R^n\text{,}\) then \(\norm{x}^2: =x_1^2+\cdots+x_n^2\text{.}\)
Checkpoint 6.1.2.
Let \(\{u_1,\ldots,
u_k\}\) be an orthogonal set of vectors in \(\R^n\text{.}\) Let \(u\in \R^n\) and define
\begin{equation*}
u_{k+1}:=u-\frac{u_1\cdot u_{k+1}}{\norm{u_1}^2}u_1-\frac{u_2\cdot u_{k+1}}{\norm{u_2}^2}u_2-\cdots -\frac{u_k\cdot u_{k+1}}{\norm{u_k}^2}u_k\text{,}
\end{equation*}
Then
(i) \(u_i\cdot u_{k+1}=0\) for all \(i=1,\ldots, k\)
(ii) If \(u\notin span(\{u_1,\ldots, u_k\})\text{,}\) then \(u_{k+1}\neq 0\) and \(\{u_1,\ldots,
u_k,u_{k+1}\}\) is an orthogonal set.
Checkpoint 6.1.3.
If \(\{u_1,\ldots,
u_n\}\) is orthogonal set then it is linearly independent.
Definition 6.1.4.
A basis \(\beta =\{u_1,\ldots,
u_n\}\) is called an orthogonal basis if \(\beta\) is an orthogonal set in \(\R^n\text{.}\) In addition if \(\norm{u_1}=1\) for all \(i\text{,}\) then \(\beta\) is called an orthonormal basis.
Checkpoint 6.1.5.
- The standard basis \({\cal E} = \{e_1,\cdots, e_n\}\) is an orthonormal basis of \(\R^n\text{.}\)
- \(\beta =\left\{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\right\}\) is an orthonormal basis of \(\R^2\text{.}\)
- \(\beta=\{(2,1,1), (-1,1,1),(0,-1,1)\}\) is an orthogonal basis of \(\R^3\text{.}\) However, it is not an orthonormal basis.
- \(\beta'=\left\{\left(\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right), \left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right),\left(0,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right\}\) is an orthonormal basis of \(\R^n\text{.}\)
What is an advantage of having an orthonormal basis?
Let \(\beta =\{u_1,\ldots, u_n\}\) be an orthonormal basis of \(\R^n\) and \(x\in \R^n\text{.}\) Then there exist scalars \(\alpha_1,\ldots, \alpha_n\) such that \(x=\alpha_1u_1+\cdots +\alpha_nu_n\text{.}\) Then is is easy to check that \(\alpha_i = x\cdot u_i\) for all \(i\text{.}\) In particular, the scalars \(\alpha_i\) can be explicitly written in terms of \(u\) and \(u_i\text{.}\) This is advantage of having an orthonormal basis.
Checkpoint 6.1.6.
(i) Find the coordinates of a vector \((1,2)\) with respect to an orthonormal basis \(\beta =\left\{\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right), \left(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\right\}\) of \(\R^2\text{.}\)
(ii) Find the coordinates of the vector \((2,5,7)\) with respect to an orthonormal basis \(\beta'=\left\{\left(\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}\right), \left(\frac{-1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right),\left(0,\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right\}\) of \(\R^3\text{.}\)
Next we deal with to find an orthonormal basis of \(\R^n\) or any subspace of \(\R^n\text{.}\)
