Example
Here is another view of an orthogonal reflection. It displays its effect on a box in 3-space.
Find the matrix representing the orthogonal reflection \(M\) of \(\RNrSpc{3}\) about the hyperspace perpendicular to \(\Vect{c}=(1,2,1)\).
We know that the matrix \(B\) representing \(M\) is of size \((3,3)\), and that its \(j\)-th column consists of the coordinates of \(M(\StdBssVec{j})\), \(1\leq j\leq 3\). Therefore we compute
| \(M(\StdBssVec{1})\) | \(=\) | \((1,0,0) - 2\cdot \dfrac{\DotPr{(1,0,0)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((1,0,0) - \tfrac{2}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(4,-4,-2)\) |
| \(M(\StdBssVec{2})\) | \(=\) | \((0,1,0) - 2\cdot \dfrac{\DotPr{(0,1,0)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((0,1,0) - \tfrac{4}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(-4,-2,-4)\) |
| \(M(\StdBssVec{3})\) | \(=\) | \((0,0,1) - 2\cdot \dfrac{\DotPr{(0,0,1)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((0,0,1) - \tfrac{2}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(-2,-4,4)\) |
The vectors \(M(\StdBssVec{1})\), \(M(\StdBssVec{2})\), \(M(\StdBssVec{3})\) form the columns of the matrix \(\Mtrx{B}\)representing \(M\):
\[ B = \dfrac{1}{6} \left[ \begin{array}{rrr} 4 & -4 & -2 \\ -4 & -2 & -4 \\ -2 & -4 & 4 \end{array} \right]\ =\ \dfrac{1}{3} \left[ \begin{array}{rrr} 2 & -2 & -1 \\ -2 & -1 & -2 \\ -1 & -2 & 2 \end{array} \right] \]Here is another view of an orthogonal reflection. It displays its effect on a box in 3-space.