91Ó°ÊÓ

Consider a basis of a subspace Vof${\mathit{R}}^{\mathbf{n}}$ for$\mathit{j}\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathit{m}$ it resolves the vector${\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}$ into its components parallel and perpendicular to the span of the preceding vectors ${\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{1}}\mathbf{,}\mathbf{…}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{V}}}_{\mathbf{j}\mathbf{-}\mathbf{1}}$,

Then

${\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_1\right|\right|}{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{1}}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{\stackrel{→}{v}}_2\right|\right|}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{2}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{j}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\stackrel{→}{v}}_j}^{âŠ\yen }\right|\right|}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{j}}}^{\mathbf{âŠ\yen }}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\stackrel{\mathbf{→}}{\mathbf{u}}}_{\mathbf{m}}\mathbf{=}\frac{\mathbf{1}}{\left|\left|{{\stackrel{→}{v}}_m}^{âŠ\yen }\right|\right|}{{\stackrel{\mathbf{→}}{\mathbf{v}}}_{\mathbf{m}}}^{\mathbf{âŠ\yen }}$

Let assume that $A=\left[a_i\right]$, for$i=1,9$ is a$3×3$ transformation matrix corresponding to T. we get the following.

$T.\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]=\left[\begin{array}{ccc}{a}_1& a_2& a_3\\ a_4& a_5& a_6\\ a_7& a_8& a_9\end{array}\right].\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

Furthermore, the transformation matrix$A^{-1}$ that corresponds to$T^{-1}$ is equal to $A^T$.

Therefore, it gives the following,

role="math" localid="1660128916789" $T^{-1}.\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{ccc}{a}_1& a_4& a_7\\ a_2& a_5& a_8\\ a3& a_6& a_9\end{array}\right].\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}2/3\\ 2/3\\ 1/3\end{array}\right]$

This implies that $a_7=2/3,a_8=2/3,a_9=1/3$.

And the remaining two factors can be found using the Gram-Schmidt algorithm. The easiest solution would be for the absolute values of the entries in each column to be equal to 2/3, 2/3 and 1/3.

The only combinations that satisfy the equation$\left(2/3\right)x+\left(2/3\right)y+\left(2/3\right)z=0$ are$\left(x,y,z\right)=\left(-2/3,1/3,2/3\right)$ and $\left(x,y,z\right)=\left(1/3,-2/3,2/3\right)$.

By put these entries it gives the solution $A=\left[\begin{array}{ccc}-2/3& 1/3& 2/3\\ 1/3& -2/3& 2/3\\ 2/3& 2/3& 1/3\end{array}\right]$.

Thus, if $\left(x,y,z\right)∈R^3$ an orthonormal transformation T, for which $T\left(2/3,2/3,1/3\right)=\left(0,0,1\right)$is $T\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right]\left[\begin{array}{cc}1/3& 2/3\\ -2/3& 2/3\\ 2/3& 1/3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$.

Hence, Solution for the transformation matrix A to T will be $T\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}-2/3\\ 1/3\\ 2/3\end{array}\right]\left[\begin{array}{cc}1/3& 2/3\\ -2/3& 2/3\\ 2/3& 1/3\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$.