Find the matrix representing the orthogonal projection \(P\) of \(\RNrSpc{3}\) onto the hyperspace perpendicular to \(\Vect{a}=(1,2,1)\).
We know that the matrix \(A\) representing \(P\) is of size \((3,3)\), and that its \(j\)-th column consists of the coordinates of \(P(\StdBssVec{j})\), \(1\leq j\leq 3\). Therefore we compute
| \(P(\StdBssVec{1})\) | \(=\) | \((1,0,0) - \dfrac{\DotPr{(1,0,0)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((1,0,0) - \tfrac{1}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(5,-2,-1)\) |
| \(P(\StdBssVec{2})\) | \(=\) | \((0,1,0) - \dfrac{\DotPr{(0,1,0)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((0,1,0) - \tfrac{2}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(-2,2,-2)\) |
| \(P(\StdBssVec{3})\) | \(=\) | \((0,0,1) - \dfrac{\DotPr{(0,0,1)}{(1,2,1)}}{\DotPr{(1,2,1)}{(1,2,1)}} \cdot (1,2,1)\) |
| \(\) | \(=\) | \((0,0,1) - \tfrac{1}{6}(1,2,1)\) |
| \(\) | \(=\) | \(\tfrac{1}{6}(-1,-2,5)\) |
\(P(\StdBssVec{1})\), \(P(\StdBssVec{2})\), and \(P(\StdBssVec{3})\) form the columns of the matrix \(\Mtrx{A}\) representing \(P\):
\[ A = \dfrac{1}{6} \left[ \begin{array}{rrr} 5 & -2 & -1 \\ -2 & 2 & -2 \\ -1 & -2 & 5 \end{array} \right] \]