3.3: Least Squares Solution
- Page ID
- 24243
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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.