Let \(W\) be the subspace of \(\RR^2\) spanned by the vector \((3, 4)\). Using the standard inner product, let \(E\) be the orthogonal projection of \(\RR^2\) onto \(W\). Find
a formula for \(E(x_1, x_2)\);
the matrix of \(E\) in the standard ordered basis;
\(W^{\perp}\);
an orthonormal basis in which \(E\) is represented by the matrix \[ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}. \]
Solution
Let \[ v=(3,4). \] Then \(W=\operatorname{span}\{v\}\) and \[ (v\mid v)=25. \]
1. The orthogonal projection is\[ \boxed{ E(x_1,x_2) =\frac{3x_1+4x_2}{25}(3,4).} \]
The orthogonal projection onto a line spanned by a nonzero vector \(v\) is \[ E(x)=\frac{(x\mid v)}{(v\mid v)}v \] [@HK71]. Here \[ (x\mid v)=3x_1+4x_2, \] which gives the displayed formula.
2. Its matrix in the standard basis is\[ \boxed{ [E]=\frac1{25} \begin{pmatrix}9&12\\12&16\end{pmatrix}.} \]
Expanding step 1 gives \[ E(x_1,x_2) =\left( \frac{9x_1+12x_2}{25}, \frac{12x_1+16x_2}{25} \right), \] so the displayed coefficient matrix is immediate.
3. The orthogonal complement is\[ \boxed{W^\perp=\operatorname{span}\{(-4,3)\}.} \]
A vector \((a,b)\) is perpendicular to \((3,4)\) exactly when \[ 3a+4b=0. \] The solution space is the line spanned by \((-4,3)\).
4. An orthonormal basis diagonalizing \(E\) is\[ \boxed{ \mathcal B=\left( \frac15(3,4), \frac15(-4,3) \right).} \]
Both vectors have norm one and are orthogonal. The first spans \(W\), so \(E\) fixes it. The second spans \(W^\perp\), so \(E\) kills it. Therefore \[ [E]_{\mathcal B} =\begin{pmatrix}1&0\\0&0\end{pmatrix}. \]