# 3.3: Least Squares Solution

Let $$x$$ be a particular solution of (1a). Denote by $$x_{A}$$ its unique projection onto the range of $$A$$ (i.e. onto the space spanned by the vectors $$a_{i}$$) and let $$x_{A^{\perp}}$$ denote the projection onto the space orthogonal to this. Following the same development as in the proof of the orthogonality principle in Lecture 2, we find

$x_{A}=A \prec A, A \succ^{-1} \prec A, x \succ \ \tag{3a}$

with $$x_{A^{\perp}}= x- x_{A}$$. Now (1a) allows us to make the substitution $$y= \prec A, x \succ$$ in (3a), so

$x_{A}=A \prec A, A \succ^{-1} y\ \tag{3b}$

which is exactly the expression we had for the solution $$\check{x}$$ that we determined earlier by inspection, see (2b).

Now note from (3b) that $$x_{A}$$ is the same for all solutions $$x$$, because it is determined entirely by $$A$$ and $$y$$. Hence it is only $$x_{A^{\perp}}$$ that is varied by varying $$x$$. The orthogonality of $$x_{A}$$ and $$x_{A^{\perp}}$$ allows us to write

$<x, x>=<x_{A}, x_{A}>+<x_{A^{\perp}}, x_{A^{\perp}}>\nonumber$

so the best we can do as far as minimizing $$<x,x>$$ is concerned is to make $$x_{A^{\perp}}=0$$. In other words, the optimum solution is $$x = x_{A} = \check{x}$$.

Example 3.3

For the FIR filter mentioned in Example 3.1, and considering all input sequences $$x[k]$$ that result in $$y[0] = 7$$, find the sequence for which $$\sum_{i=-N}^{N} x^{2}[i]$$ is minimized. (Work out this example for yourself!)

Example 3.4

Consider a unit mass moving in a straight line under the action of a force $$x(t)$$, with position at time $$t$$ given by $$p(t)$$. Assume $$p(0)=0, \dot{p}(0)=0$$, and suppose we wish to have $$p(T ) = y$$ (with no constraint on $$\dot{p}(T )$$). Then

$y=p(T)=\int_{0}^{T}(T-t) x(t) d t=<a(t), x(t)> \ \tag{4}$

This is a typical underconstrained problem, with many choices of $$x(t)$$ for $$0 \leq t \leq T$$ that will result in $$p(T ) = y$$. Let us find the solution $$x(t)$$ for which

$\int_{0}^{T} x^{2}(t) d t=<x(t), x(t)> \ \tag{5}$

is minimized. Evaluating the expression in (2a), we find

$\check{x}(t)=(T-t) y /\left(T^{3} / 3\right) \ \tag{6}$

How does your solution change if there is the additional constraint that the mass should be brought to rest at time $$T$$, so that $$\dot{p}(T ) = 0$$?

We leave you to consider how weighted norms can be minimized.