Finding a Projection Along a Subspace

Example

Find a projection $E$ that projects ${\mathbb{R}}^3$ onto $R=span\{(1,2)\}$ along $N=span\{3,-1\}$

Solns

Naive Solns

1. As $\beta =\{{\alpha }_1,{\alpha }_2\}$ forms a basis for ${\mathbb{R}}^n$
2. Given $\alpha =(x_1,x_2)\in {\mathbb{R}}^2$, we can solve $a,b\in \mathbb{R}$ (in terms of $x_1,x_2$) such that $a=a{\alpha }_1+b{\alpha }_2$
3. $(x_1,x_2)=a{\alpha }_1+b{\alpha }_2$
4. For our example, we can solve the system of equations to get: $a=\frac{(x_1+3x_2)}{7}$, $b=\frac{(2x_1+6x_2)}{7}$
5. We can then define $E$ by $E(a{\alpha }_1+b{\alpha }_2)={\alpha }_1$ which explicitly gives $E(x_1,x_2)=\frac{(x_1+3x_2)}{7}(1,2)=(\frac{x_1+3x_2}{7},\frac{2x_1+6x_2}{7})$

Matrix Soln

1. As $E({\alpha }_1)={\alpha }_1$
2. As $E({\alpha }_2)=0$
3. Then, this implies $[E{]}_{\beta }=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$ Finally, we show that

$$[E^2{]}_{\beta }=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]=[E{]}_{\beta }$$

Then,

$$\left[\begin{array}{cc}1& 3\\ 2& -1\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]{\left[\begin{array}{cc}1& 3\\ 2& -1\end{array}\right]}^{-1}$$