91Ó°ÊÓ

If Vis a subspace of ${\mathbf{R}}^{\mathbf{n}}$with an orthonormal basis ${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}$thenlocalid="1659451887578" ${\mathbf{proj}}_{\mathbf{v}}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{=}{\stackrel{\mathbf{→}}{\mathbf{x}}}^{\mathbf{ll}}\mathbf{=}\mathbf{(}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{)}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{(}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{.}\stackrel{\mathbf{→}}{\mathbf{x}}\mathbf{)}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}$

For all $\stackrel{→}{x}$in localid="1659436140715" $R^n$.

Let us write the matrix in the $\stackrel{→}{V_i}.\stackrel{→}{V_i}$notation.

Consider the terms below.

localid="1659451940364" $A=\left[\begin{array}{ccc}3& 5& 11\\ 5& 9& 20\\ 11& 20& 49\end{array}\right]=\left[\begin{array}{ccc}{\stackrel{→}{v}}_1.{\stackrel{→}{v}}_1& {\stackrel{→}{v}}_1.{\stackrel{→}{v}}_2& {\stackrel{→}{v}}_1.{\stackrel{→}{v}}_3\\ {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_1& {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_2& {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_3\\ {\stackrel{→}{v}}_3.{\stackrel{→}{v}}_1& {\stackrel{→}{v}}_3.{\stackrel{→}{v}}_2& {\stackrel{→}{v}}_3.{\stackrel{→}{v}}_3\end{array}\right]=\left[\begin{array}{ccc}{||{\stackrel{→}{v}}_1||}^2& {\stackrel{→}{v}}_1.{\stackrel{→}{v}}_2& {\stackrel{→}{v}}_1.{\stackrel{→}{v}}_3\\ {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_1& {||{\stackrel{→}{v}}_2||}^2& {\stackrel{→}{v}}_2.{\stackrel{→}{v}}_3\\ {\stackrel{→}{v}}_3.{\stackrel{→}{v}}_1& {\stackrel{→}{v}}_3.{\stackrel{→}{v}}_2& {||{\stackrel{→}{v}}_3||}^2\end{array}\right]$

To obtain the span. Since it is a singleton vector. So by normalizing it. Therefore, an orthonormal basis will be.

$\frac{{\stackrel{→}{v}}_2}{||{\stackrel{→}{v}}_2||}=\frac{{\stackrel{→}{v}}_2}{3}=\stackrel{→}{u}$.

So, according to the above theorem

$$\begin{gathered} pro{j}_{{\stackrel{→}{v}}_2}\left({\stackrel{→}{v}}_1\right)=\left(\stackrel{→}{u}.{\stackrel{→}{v}}_1\right)\stackrel{→}{u} \\ =\frac{{\stackrel{→}{v}}_2}{3}.{\stackrel{→}{v}}_1\frac{{\stackrel{→}{v}}_2}{3} \\ =\frac{{\stackrel{→}{v}}_2.{\stackrel{→}{v}}_1}{3}\frac{{\stackrel{→}{v}}_2}{3} \\ =\frac{5}{9}{\stackrel{→}{v}}_2 \end{gathered}$$

Hence, the role="math" localid="1659438206334" $pro{j}_{{\stackrel{→}{v}}_2}\left({\stackrel{→}{v}}_1\right)=\frac{5}{9}{\stackrel{→}{v}}_2$.