91Ó°ÊÓ

If Vis a subspace of ${\mathit{R}}^{\mathbf{n}}$with an orthonormal basis$\mathit{u}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{u}{\mathbf{⃗}}_{\mathbf{m}}$then

$\mathit{p}\mathit{r}\mathit{o}{\mathit{j}}_{\mathbf{v}}\mathit{x}\mathbf{⃗}\mathbf{=}{\mathit{x}}^{\mathbf{I}\mathbf{I}}\mathbf{⃗}\mathbf{}\mathbf{=}\mathbf{(}\mathit{u}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{.}\mathit{x}\mathbf{⃗}\mathbf{)}\mathit{u}{\mathbf{⃗}}_{\mathbf{1}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathbf{(}\mathit{u}{\mathbf{⃗}}_{\mathbf{m}}\mathbf{.}\mathit{x}\mathbf{⃗}\mathbf{)}\mathit{u}{\mathbf{⃗}}_{\mathbf{m}}$

For $x⃗$all in $R^n.$

Let us write the matrix in the$v{⃗}_i.v{⃗}_j$notation.

Consider the terms below.

$$\begin{gathered} A=\left|3511 \\ 5920 \\ 112049\right|=\left[{\stackrel{â‡\P Ä}{V}}_1.{\stackrel{â‡\P Ä}{V}}_1{\stackrel{â‡\P Ä}{V}}_1.{\stackrel{â‡\P Ä}{V}}_2{\stackrel{â‡\P Ä}{V}}_1.{\stackrel{â‡\P Ä}{V}}_3 \\ {\stackrel{â‡\P Ä}{V}}_2.{\stackrel{â‡\P Ä}{V}}_1{\stackrel{â‡\P Ä}{V}}_2.{\stackrel{â‡\P Ä}{V}}_2{\stackrel{â‡\P Ä}{V}}_2.{\stackrel{â‡\P Ä}{V}}_3 \\ {\stackrel{â‡\P Ä}{V}}_3.{\stackrel{â‡\P Ä}{V}}_1{\stackrel{â‡\P Ä}{V}}_3.{\stackrel{â‡\P Ä}{V}}_2{\stackrel{â‡\P Ä}{V}}_3.{\stackrel{â‡\P Ä}{V}}_3\right]=\left[{\left|\left|{\stackrel{â‡\P Ä}{V}}_1\right|\right|}^2{\stackrel{â‡\P Ä}{V}}_1.{\stackrel{â‡\P Ä}{V}}_2{\stackrel{â‡\P Ä}{V}}_1.{\stackrel{â‡\P Ä}{V}}_3 \\ {\stackrel{â‡\P Ä}{V}}_2.{\stackrel{â‡\P Ä}{V}}_1{\left|\left|{\stackrel{â‡\P Ä}{V}}_2\right|\right|}^2{\stackrel{â‡\P Ä}{V}}_2.{\stackrel{â‡\P Ä}{V}}_3 \\ {\stackrel{â‡\P Ä}{V}}_3.{\stackrel{â‡\P Ä}{V}}_1{\stackrel{â‡\P Ä}{V}}_3.{\stackrel{â‡\P Ä}{V}}_2{\left|\left|{\stackrel{â‡\P Ä}{V}}_3\right|\right|}^2\right] \end{gathered}$$

Consider the vector $v⃗=v{⃗}_2-pro{j}_{}\left(v{⃗}_3\right)v{⃗}_2$

Now, according to the above theorem an orthonormal basis of span$v{⃗}_3$, ,which is

$u⃗=\frac{v{⃗}_3}{\left|\right|v{⃗}_3\left|\right|}=\frac{v{⃗}_3}{7}$

To find out that the above $\stackrel{→}{V}=\stackrel{→}{V_z}-\frac{20}{{49}_{}}{\stackrel{â‡\P Ä}{v}}_3$term is orthogonal to or not consider the equations below.

$$\begin{gathered} \stackrel{â‡\P Ä}{v}.{\stackrel{â‡\P Ä}{v}}_3={\stackrel{â‡\P Ä}{v}}_2-\frac{20}{49}{\stackrel{â‡\P Ä}{v}}_3 \\ ={\stackrel{â‡\P Ä}{v}}_2.{\stackrel{â‡\P Ä}{v}}_3-\frac{20}{49}{\stackrel{â‡\P Ä}{.v}}_3.{\stackrel{â‡\P Ä}{v}}_3 \\ =20-\frac{20}{49}×49 \\ =0 \end{gathered}$$

Thus,${\stackrel{â‡\P Ä}{v}}_{}$ is orthogonal to ${\stackrel{â‡\P Ä}{v}}_3$.

Hence,$\stackrel{â‡\P Ä}{v}={\stackrel{â‡\P Ä}{v}}_2-pro{j}_{v_3}{\stackrel{â‡\P Ä}{v}}_2$ is orthogonal to ${\stackrel{â‡\P Ä}{v}}_3$.