1.16 Computing the best approximation

Recall that if $P$ is an orthogonal projection onto a subspace $A$ , we can write any $x$ as

where $Px\in A$ and $(I-P)x\perp A$ . We now turn to how to actually find $P$ .

We begin with the finite-dimensional case, assuming that $\{v_1,...,v_N\}$ is a basis for $A$ . If $(I-P)x\perp A$ then we have that for any $x$

We also note that since $Px\in A$ , we can write $Px={\sum }_{k=1}^N{c}_k{v}_k$ . Thus we obtain

from which we obtain

We know $x$ and $v_1,...,v_N$ . Our goal is to find $c_1,...,c_N$ . Note that a procedure for calculating $c_1,...,c_k$ for any given $x$ is equivalent to one that computes $Px$ .

To find $c_1,...,c_N$ , observe that [link] represents a set of $N$ equations with $N$ unknowns.

More compactly, we want to find a vector $c\in {\mathbb{C}}^N$ such that $Gc=b$ where

Thus, since $G^{-1}$ exists, we can write $c=G^{-1}b$ to calculate $c$ .

As a special case, suppose now that $\left\{v_j\right\}$ is an orthobasis for $A$ ? What is $G$ ? It is just the identity matrix $I$ ! Computing $c$ just got much easier, since now $c=b$ . Plugging this $c$ back into out formula for $Px$ we obtain

Just to verify, note that $P$ is indeed a projection matrix:

Example Suppose $f\in L_2\left([0,4]\right)$ is given by

Suppose $f\in L_2\left([0,4]\right)$ is given by

$$f\left(t\right)=\left\{\begin{array}{cc}t\hfill & \mathrm{if}t\in [0,\frac{1}{2}]\hfill \\ 1-t\hfill & \mathrm{if}t\in [\frac{1}{2},1].\hfill \end{array}\right)$$

Let $A=\{\text{piecewise constant functions on}[0,\frac{1}{4}),[\frac{1}{4},\frac{1}{2}),[\frac{1}{2},\frac{3}{4}),[\frac{3}{4},1]\}$ . Our goal is to find the closest (in $L_2$ ) function in $A$ to $f\left(t\right)$ . Using $v_1,...,v_4$ from before, we can calculate $c_1=\frac{1}{4}$ , $c_2=0$ , $c_3=\frac{-\sqrt{2}}{16}$ , $c_4=\frac{\sqrt{2}}{16}$ . Thus, we have that

$$\widehat{f}\left(t\right)=\frac{1}{4}{v}_1-\frac{\sqrt{2}}{16}{v}_3+\frac{\sqrt{2}}{16}{v}_4.$$