Let \(V\) be the inner product space consisting of \(\RR^2\) and the inner product whose quadratic form is defined by \[ ||(x_1, x_2)||^2 = (x_1 - x_2)^2 + 3x_2^2. \] Let \(E\) be the orthogonal projection of \(V\) onto the subspace \(W\) spanned by the vector \((3, 4)\). Now answer the four questions of Exercise 6.
Solution
The quadratic form \[ \|(x_1,x_2)\|^2 =(x_1-x_2)^2+3x_2^2 =x_1^2-2x_1x_2+4x_2^2 \] has Gram matrix \[ G=\begin{pmatrix}1&-1\\-1&4\end{pmatrix}, \] so \[ (x\mid y)=x^tGy. \] Let \(v=(3,4)\).
1. The orthogonal projection onto \(W=\operatorname{span}\{v\}\) is\[ \boxed{ E(x_1,x_2) =\frac{-x_1+13x_2}{49}(3,4).} \]
One computes \[ Gv=(-1,13)^t, \] hence \[ (x\mid v)=-x_1+13x_2. \] Also \[ (v\mid v)=v^tGv=49. \] The line-projection formula therefore gives the result [@HK71].
2. In the standard basis,\[ \boxed{ [E]=\frac1{49} \begin{pmatrix}-3&39\\-4&52\end{pmatrix}.} \]
This is obtained by expanding the formula in step 1 coordinatewise.
3. The orthogonal complement is\[ \boxed{W^\perp=\operatorname{span}\{(13,1)\}.} \]
The condition \((x\mid v)=0\) is \[ -x_1+13x_2=0, \] whose solution space is the displayed line.
4. An orthonormal basis adapted to \(W\oplus W^\perp\) is\[ \boxed{ \mathcal B=\left( \frac17(3,4), \frac1{7\sqrt3}(13,1) \right).} \]
The first vector has norm \(1\) because \((v\mid v)=49\). For \(w=(13,1)\), \[ \|w\|^2=(13-1)^2+3=147=49\cdot3, \] so the second displayed vector also has norm one. Step 3 makes them orthogonal. The projection fixes the first and kills the second, hence \[ [E]_{\mathcal B} =\begin{pmatrix}1&0\\0&0\end{pmatrix}. \]