Let $W$ be the subspace of $R^2$ spanned by the vector $(3, 4)$. Using the standard inner product, let $E$ be the orthogonal projection of $R^2$ onto $W$. Find
(a) a formula for $E(x_1, x_2)$ ;
(b) the matrix of $E$ in the standard ordered basis;
(c) $W^\bot$;
(d) an orthonormal basis in which $E$ is represented by matrix
\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}
Attempt: $$E(x_1,x_2)=\frac{((x_1,x_2)|(3,4))}{25}(3,4)= \left( \frac{9x_1+12x_2}{25},\frac{ 12x_1+16x_2}{25} \right) $$
So in the standard basis the matrix is $$\frac{1}{25}\begin{bmatrix}9 & 12 \\ 12 & 16\end{bmatrix}.$$
The set $W^\bot$ is $\{ (x,y) \in R^2 ; ((x,y)|(x',y'))=0 \;\; \forall (x',y') \in W \} =$ $\{ (x,y) \in R^2 ; x = - \frac{4}{3} y \} $.
now I can not find a basis such that the matrix of E in this is \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} Apparently the system $E(a,b)=(1,0);E(c,d)=(0,0) $
has no solution, if I'm doing it right. Any suggestion?
You need an orthonormal basis $u,v$ such that $E(v)=0$ and $E(u)=\pmatrix{1\\0} $ coordinated in the new basis, i.e. $E(u) =1\cdot u+0\cdot v=u$.
But such $u, v$ vectors you already know.