Linear least squares

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Let U\mathcal{U} represent a subspace of the linear space Rm\mathbb {R}^{m} of all mm-dimensional column vectors. Take M\mathcal{M} to be the subspace of Rm×n\mathbb {R}^{m \times n} defined by W∈M\mathbf{W} \in \mathcal{M} if and only if W=(w1,…,wn)\mathbf{W} = (\mathbf{w}_{1}, \ldots , \mathbf{w}_{n}) for some vectors w1,…,wn\mathbf{w}_{1}, \ldots , \mathbf{w}_{n} in U\mathcal{U}. Let Z\mathbf{Z} represent the projection (with respect to the usual inner product) of an m×nm \times n matrix Y\mathbf{Y} on M\mathcal{M}, and let X\mathbf{X} represent any m×pm \times p matrix whose columns span U\mathcal{U}. Show that Z=XB∗\mathbf{Z} = \mathbf{X}\mathbf{B}^{*} for any solution B∗\mathbf{B}^{*} to the linear system

X′XB=X′Y(in B). \mathbf{X}'\mathbf{X}\mathbf{B} = \mathbf{X}'\mathbf{Y} \quad \text{(in }\mathbf{B}\text{)}.
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Let X\mathbf{X} represent any n×pn \times p matrix. If a p×np \times n matrix B∗\mathbf{B}^{*} is a solution to the linear system X′XB=X′\mathbf{X}'\mathbf{X}\mathbf{B} = \mathbf{X}' (in B\mathbf{B}), then B∗\mathbf{B}^{*} is a generalized inverse of X\mathbf{X} and XB∗\mathbf{X}\mathbf{B}^{*} is symmetric. Show that, conversely, if a p×np \times n matrix G\mathbf{G} is a generalized inverse of X\mathbf{X} and if XG\mathbf{X}\mathbf{G} is symmetric, then X′XG=X′\mathbf{X}'\mathbf{X}\mathbf{G} = \mathbf{X}' (i.e., G\mathbf{G} is a solution to X′XB=X′\mathbf{X}'\mathbf{X}\mathbf{B} = \mathbf{X}').

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Let the physical quantities xx and yy be related by an unknown linear relationship y=ax+by = a x+b. As a result of nn measurements, the following pairs of values were obtained with some error: (x1;y1)\left(x_{1} ; y_{1}\right), (x2;y2),…,(xn;yn)\left(x_{2} ; y_{2}\right), \ldots ,\left(x_{n} ; y_{n}\right). According to the principle of least squares, the most probable values of the coefficients aa and bb are those for which

∑i=1n(yi−axi−b)2 \sum _{i = 1}^{n}\left(y_{i}-a x_{i}-b\right)^{2}

attains its smallest value. Find the most probable values aˉ,bˉ\bar{a}, \bar{b} for the coefficients aa and bb.

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Let Ax=bA\mathrm{x} = \mathrm{b} be an arbitrary system of linear equations. Prove that the system of equations (ATA)x=ATb\left(A^{T} A\right) \mathbf{x} = A^{T} \mathbf{b} is consistent.

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Using the method of least squares, solve the overdetermined system of linear equations:

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(a)

{x1+x2−3x3=−1,2x1+x2−2x3=1,x1+x2+x3=3,x1+2x2−3x3=1;\begin{cases} x_{1}+x_{2}-3 x_{3}=-1, \\ 2 x_{1}+x_{2}-2 x_{3}=1, \\ x_{1}+x_{2}+x_{3}=3, \\ x_{1}+2 x_{2}-3 x_{3}=1 ; \end{cases}

(b)

{2x1−5x2+3x3+x4=5,3x1−7x2+3x3−x4=−1,5x1−9x2+6x3+2x4=7,4x1−6x2+3x3+x4=8.\begin{cases} 2 x_{1}-5 x_{2}+3 x_{3}+x_{4}=5, \\ 3 x_{1}-7 x_{2}+3 x_{3}-x_{4}=-1, \\ 5 x_{1}-9 x_{2}+6 x_{3}+2 x_{4}=7, \\ 4 x_{1}-6 x_{2}+3 x_{3}+x_{4}=8. \end{cases}

Let f(x)=sin⁡πxf(x) = \sin \pi x on [0,1][0, 1]. The object of this problem is to determine the coefficients αi\alpha_{i} of the cubic polynomial

p(x)=∑i=03αixi p(x) = \sum _{i=0}^{3} \alpha _{i} x^{i}

that is as close to f(x)f(x) as possible in the sense that

r=∫01[f(x)−p(x)]2 dx=∫01[f(x)]2 dx−2∑i=03αi∫01xif(x) dx+∫01(∑i=03αixi)2dx \begin{aligned} r & = \int _{0}^{1} [f(x) - p(x)]^{2} \, dx \\ & = \int _{0}^{1} [f(x)]^{2} \, dx - 2 \sum _{i=0}^{3} \alpha _{i} \int _{0}^{1} x^{i} f(x) \, dx + \int _{0}^{1} \left( \sum _{i=0}^{3} \alpha _{i} x^{i} \right)^{2} dx \end{aligned}

is as small as possible.

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(a)

In order to minimize rr, impose the condition that ∂r/∂αi=0\partial r / \partial \alpha_{i} = 0 for each i=0,1,2,3i = 0, 1, 2, 3, and show this results in a system of linear equations whose augmented matrix is [H4∣b][H_{4} \mid b], where H4H_{4} and bb are given by

H4=[1121314]2mm]1213141513141516]2mm]14151617]andb=[2π1π]2mm]1π−4π31π−6π3]. H_{4} = \begin{bmatrix} 1 & \dfrac {1}{2} & \dfrac {1}{3} & \dfrac {1}{4} \\ ]2mm] \dfrac {1}{2} & \dfrac {1}{3} & \dfrac {1}{4} & \dfrac {1}{5} \\[2mm] \dfrac {1}{3} & \dfrac {1}{4} & \dfrac {1}{5} & \dfrac {1}{6} \\ ]2mm] \dfrac {1}{4} & \dfrac {1}{5} & \dfrac {1}{6} & \dfrac {1}{7} \end{bmatrix} \quad \text{and} \quad b = \begin{bmatrix} \dfrac {2}{\pi } \\[2mm] \dfrac {1}{\pi } \\ ]2mm] \dfrac {1}{\pi } - \dfrac {4}{\pi ^{3}} \\[2mm] \dfrac {1}{\pi } - \dfrac {6}{\pi ^{3}} \end{bmatrix}.

Any matrix HnH_{n} that has the same form as H4H_{4} is called a Hilbert matrix of order nn.

(b)

Systems involving Hilbert matrices are badly ill-conditioned, and the ill-conditioning becomes worse as the size increases. Use exact arithmetic with Gaussian elimination to reduce H4H_{4} to triangular form. Assuming that the case in which n=4n = 4 is typical, explain why a general system [Hn∣b][H_{n} \mid b] will be ill-conditioned. Notice that even complete pivoting is of no help.

Perform the following calculations using the matrices:

A=(122412.01)andb→=(121.01). \mathbf{A} = \begin{pmatrix} 1 & 2 \\ 2 & 4 \\ 1 & 2.01 \end{pmatrix} \quad \text{and} \quad \overrightarrow {b} = \begin{pmatrix} 1 \\ 2 \\ 1.01 \end{pmatrix}.
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(a)

Find rk⁡(A)\operatorname {rk}\left(\mathbf{A}\right), and solve Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b} using exact arithmetic.

(b)

Find rk⁡(ATA)\operatorname {rk}\left(\mathbf{A}^T \mathbf{A}\right), and solve ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b} exactly.

(c)

Find rk⁡(A)\operatorname {rk}\left(\mathbf{A}\right), and solve Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b} with 3-digit arithmetic.

(d)

Find ATA\mathbf{A}^T \mathbf{A}, ATb→\mathbf{A}^T \overrightarrow {b}, and the solution of ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b} with 3-digit arithmetic.

Hooke's law says that the displacement yy of an ideal spring is proportional to the force xx that is applied — i.e., y=kxy = kx for some constant kk. Consider a spring in which kk is unknown. Various masses are attached, and the resulting displacements shown below are observed. Using these observations, determine the least squares estimate for kk.

x (lb)y (in)511.1715.4817.51022.01226.3 \begin{array}{c|c} x \text{ (lb)} & y \text{ (in)} \\ \hline 5 & 11.1 \\ 7 & 15.4 \\ 8 & 17.5 \\ 10 & 22.0 \\ 12 & 26.3 \end{array}

Figure 4.6.6Figure 4.6.6

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Show that the slope of the line that passes through the origin in R2\mathbb {R}^{2} and comes closest in the least squares sense to passing through the points {(x1,y1),(x2,y2),…,(xn,yn)}\left\{ (x_1, y_1), (x_2, y_2), \ldots , (x_n, y_n)\right\} is given by m=∑ixiyi/∑ixi2m = \sum_i x_i y_i / \sum_i x_i^2.

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A small company has been in business for three years and has recorded annual profits (in thousands of dollars) as follows.

Year123Sales743 \begin{array}{c|ccc} \text{Year} & 1 & 2 & 3 \\ \hline \text{Sales} & 7 & 4 & 3 \end{array}

Assuming that there is a linear trend in the declining profits, predict the year and the month in which the company begins to lose money.

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An economist hypothesizes that the change (in dollars) in the price of a loaf of bread is primarily a linear combination of the change in the price of a bushel of wheat and the change in the minimum wage. That is, if BB is the change in bread prices, WW is the change in wheat prices, and MM is the change in the minimum wage, then B=αW+βMB = \alpha W + \beta M. Suppose that for three consecutive years the change in bread prices, wheat prices, and the minimum wage are as shown below.

Year 1Year 2Year 3B+\1+\1+\1W+\1+\20\M+\10\−\1 \begin{array}{c|ccc}& \text{Year 1} & \text{Year 2} & \text{Year 3} \\ \hline B & +\1 & +\1 & +\1 \\ W & +\1 & +\2 & 0\ \\ M & +\1 & 0\ & -\1 \end{array}

Use the theory of least squares to estimate the change in the price of bread in Year 4 if wheat prices and the minimum wage each fall by $1.

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Suppose that a researcher hypothesizes that the weight loss of a pint of ice cream during storage is primarily a linear function of time. That is,

y=α0+α1t+ε, y = \alpha _0 + \alpha _1 t + \varepsilon ,

where y=y = the weight loss in grams, t=t = the storage time in weeks, and ε\varepsilon is a random error function whose mean value is 00. Suppose that an experiment is conducted, and the following data is obtained.

Time (t)12345678Loss (y).15.21.30.41.49.59.72.83 \begin{array}{c|cccccccc} \text{Time } (t) & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline \text{Loss } (y) & .15 & .21 & .30 & .41 & .49 & .59 & .72 & .83 \end{array}
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(a)

Determine the least squares estimates for the parameters α0\alpha_0 and α1\alpha_1.

(b)

Predict the mean weight loss for a pint of ice cream that is stored for 20 weeks.

After studying a certain type of cancer, a researcher hypothesizes that in the short run the number (yy) of malignant cells in a particular tissue grows exponentially with time (tt). That is, y=α0eα1ty = \alpha_0 e^{\alpha_1 t}. Determine least squares estimates for the parameters α0\alpha_0 and α1\alpha_1 from the researcher's observed data given below.

t (days)12345y (cells)16274574122 \begin{array}{c|ccccc} t \text{ (days)} & 1 & 2 & 3 & 4 & 5 \\ \hline y \text{ (cells)} & 16 & 27 & 45 & 74 & 122 \end{array}
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Using least squares techniques, fit the following data

x−5−4−3−2−1012345y27912131414131084 \begin{array}{c|ccccccccccc} x & -5 & -4 & -3 & -2 & -1 & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline y & 2 & 7 & 9 & 12 & 13 & 14 & 14 & 13 & 10 & 8 & 4 \end{array}

with a line y=α0+α1xy = \alpha_0 + \alpha_1 x and then fit the data with a quadratic y=α0+α1x+α2x2y = \alpha_0 + \alpha_1 x + \alpha_2 x^2. Determine which of these two curves best fits the data by computing the sum of the squares of the errors in each case.

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Consider the time (TT) it takes for a runner to complete a marathon (26 miles and 385 yards). Many factors such as height, weight, age, previous training, etc.can influence an athlete's performance, but experience has shown that the following three factors are particularly important:

x1=Ponderal index=height (in.)(weight (lbs.))1/3,x2=Miles run the previous 8 weeks,x3=Age (years). x_1 = \text{Ponderal index} = \frac{\text{height (in.)}}{\left(\text{weight (lbs.)}\right)^{1/3}}, \qquad x_2 = \text{Miles run the previous 8 weeks}, \qquad x_3 = \text{Age (years)}.

A linear model hypothesizes that the time TT (in minutes) is given by T=α0+α1x1+α2x2+α3x3+εT = \alpha_0 + \alpha_1 x_1 + \alpha_2 x_2 + \alpha_3 x_3 + \varepsilon, where ε\varepsilon is a random function accounting for all other factors and whose mean value is assumed to be zero. On the basis of the five observations given below, estimate the expected marathon time for a 43-year-old runner of height 74 in., weight 180 lbs., who has run 450 miles during the previous eight weeks.

Tx1x2x318113.16192319313.58034221213.82073122113.14093824812.548245 \begin{array}{c|ccc} T & x_1 & x_2 & x_3 \\ \hline 181 & 13.1 & 619 & 23 \\ 193 & 13.5 & 803 & 42 \\ 212 & 13.8 & 207 & 31 \\ 221 & 13.1 & 409 & 38 \\ 248 & 12.5 & 482 & 45 \end{array}

What is your personal predicted mean marathon time?

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For A∈Rm×n\mathbf{A} \in \mathbb {R}^{m \times n} and b→∈Rm\overrightarrow {b} \in \mathbb {R}^{m}, prove that x→2\overrightarrow {x}_2 is a least squares solution for Ax→=b→\mathbf{A}\overrightarrow {x} = \overrightarrow {b} if and only if x→2\overrightarrow {x}_2 is part of a solution to the larger system

(Im×mAAT0n×n)(x→1x→2)=(b→0→). \begin{pmatrix} \mathbf{I}_{m \times m} & \mathbf{A} \\ \mathbf{A}^T & \mathbf{0}_{n \times n} \end{pmatrix} \begin{pmatrix} \overrightarrow {x}_1 \\ \overrightarrow {x}_2 \end{pmatrix} = \begin{pmatrix} \overrightarrow {b} \\ \overrightarrow {0} \end{pmatrix}.
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Remark.
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It is not uncommon to encounter least squares problems in which A\mathbf{A} is extremely large but very sparse (mostly zero entries). For these situations, the system above will usually contain significantly fewer nonzero entries than the system of normal equations, thereby helping to overcome the memory requirements that plague these problems. Using this system also eliminates the undesirable need to explicitly form the product ATA\mathbf{A}^T \mathbf{A} — recall from Example 4.5.1 that forming ATA\mathbf{A}^T \mathbf{A} can cause loss of significant information.

In many least squares applications, the underlying data matrix Am×n\mathbf{A}_{m \times n} does not have independent columns — i.e., rk⁡(A)<n\operatorname {rk}\left(\mathbf{A}\right) < n — so the corresponding system of normal equations ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b} will fail to have a unique solution. This means that in an associated linear estimation problem of the form

y=α1t1+α2t2+⋯+αntn+ε y = \alpha _1 t_1 + \alpha _2 t_2 + \cdots + \alpha _n t_n + \varepsilon

there will be infinitely many least squares estimates for the parameters αi\alpha_i, and hence there will be infinitely many estimates for the mean value of yy at any given point (t1,t2,…,tn)(t_1, t_2, \ldots , t_n) — which is clearly undesirable. In order to remedy this problem, we restrict ourselves to making estimates only at those points (t1,t2,…,tn)(t_1, t_2, \ldots , t_n) that are in the row space of A\mathbf{A}. If

t→=(t1t2⋮tn)∈R(AT),and ifx→=(α^1α^2⋮α^n) \overrightarrow {t} = \begin{pmatrix} t_1 \\ t_2 \\ \vdots \\ t_n \end{pmatrix} \in R(\mathbf{A}^T), \quad \text{and if} \quad \overrightarrow {x} = \begin{pmatrix} \hat{\alpha }_1 \\ \hat{\alpha }_2 \\ \vdots \\ \hat{\alpha }_n \end{pmatrix}

is any least squares solution (i.e., ATAx→=ATb→\mathbf{A}^T \mathbf{A}\overrightarrow {x} = \mathbf{A}^T \overrightarrow {b}), prove that the estimate defined by

y^=t→Tx→=∑i=1ntiα^i \hat{y} = \overrightarrow {t}^T \overrightarrow {x} = \sum _{i=1}^{n} t_i \hat{\alpha }_i

is unique in the sense that y^\hat{y} is independent of which least squares solution x→\overrightarrow {x} is used.

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For matrix AA in Exercise 5.11.8, and for b=(−123−9)Tb = \begin{pmatrix} -12& 3& -9 \end{pmatrix}^{T}, find the solution of Ax=bAx=b that has minimum euclidean norm.

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Remark.
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Equation (5.12.17): when Ax=bAx=b is consistent, x=A†bx = A^{\dagger }b is the solution of minimal euclidean norm.

Explain why x^\hat{x} is a least squares solution for Ax=bAx=b if and only if ∥Ax^−b∥2=∥PN(AT)b∥2\left\| A\hat{x}-b\right\|_{2} = \left\| P_{\mathcal{N}(A^{T})}b\right\|_{2}.

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Prove that if ε=Ax^−b\varepsilon = A\hat{x}-b, where x^\hat{x} is a least squares solution for Ax=bAx=b, then ∥ε∥22=∥b∥22−∥PR(A)b∥22\left\| \varepsilon \right\|_{2}^{2} = \left\| b\right\|_{2}^{2} - \left\| P_{\mathcal{R}(A)}b\right\|_{2}^{2}.

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Let A=(10−112111−3011)A = \begin{pmatrix} 1 & 0 & -1 \\ 1 & 2 & 1 \\ 1 & 1 & -3 \\ 0 & 1 & 1 \end{pmatrix} and b=(1111)b = \begin{pmatrix} 1 \\ 1 \\ 1 \\ 1 \end{pmatrix}.

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(a)

Determine the rectangular QR factorization of AA.

(b)

Use the QR factors from part (a) to determine the least squares solution to Ax=bAx = b.

Use Householder reduction to compute the least squares solution for Ax=bAx=b, where A=(4−342−14−3−21401−715)A = \begin{pmatrix} 4 & -3 & 4 \\ 2 & -14 & -3 \\ -2 & 14 & 0 \\ 1 & -7 & 15 \end{pmatrix} and b=(5−15030)b = \begin{pmatrix} 5 \\ -15 \\ 0 \\ 30 \end{pmatrix}.

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Prove that a pseudosolution can be found as the solution of a new inhomogeneous system of equations formed using the Gram matrix G(a1,a2,…,an)G\left(a_{1}, a_{2}, \ldots , a_{n}\right):

((a1,a1)(a1,a2)…(a1,an)⋮⋮⋮(an,a1)(an,a2)…(an,an))(x1⋮xn)=((b,a1)⋮(b,an)) \left(\begin{smallmatrix} \left(a_{1}, a_{1}\right) & \left(a_{1}, a_{2}\right) & \ldots & \left(a_{1}, a_{n}\right) \\ \vdots & \vdots & & \vdots \\ \left(a_{n}, a_{1}\right) & \left(a_{n}, a_{2}\right) & \ldots & \left(a_{n}, a_{n}\right) \end{smallmatrix}\right)\left(\begin{smallmatrix} x_{1} \\ \vdots \\ x_{n} \end{smallmatrix}\right) =\left(\begin{smallmatrix} \left(b, a_{1}\right) \\ \vdots \\ \left(b, a_{n}\right) \end{smallmatrix}\right)

What is the geometric meaning of the constants x1,x2,…,xnx_{1}, x_{2}, \ldots , x_{n} found in this way?

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Using the method of least squares, find the pseudosolution of the following inconsistent systems of linear equations:

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(1)

{x+2y=12x+4y=1x+y=0\begin{cases} x+2 y = 1 \\ 2 x+4 y = 1 \\ x+y = 0 \end{cases}

(2)

{x=22x−y=1,x−2y=−1,x+y=2\begin{cases} x = 2 \\ 2 x-y = 1, \\ x-2 y = -1, \\ x+y = 2 \end{cases}

(3)

{2x1−x3=1,x2+x3=−1,x1−x2+x3=0,x1−x3=−1;\begin{cases} 2 x^{1}-x^{3} = 1, \\ x^{2}+x^{3} = -1, \\ x^{1}-x^{2}+x^{3} = 0, \\ x^{1}-x^{3} = -1 ; \end{cases}

(4)

{x1−x3=1,x2+x3−x4=−1,x1−x2=0,−x1+x3+2x4=1,x2+x3−x4=1.\begin{cases} x^{1}-x^{3} = 1, \\ x^{2}+x^{3}-x^{4} = -1, \\ x^{1}-x^{2} = 0, \\ -x^{1}+x^{3}+2 x^{4} = 1, \\ x^{2}+x^{3}-x^{4} = 1. \end{cases}

Prove that the solution Lmf(x)L_{m}^{f}(x) of the least-squares interpolation problem in the space Vn+1V_{n+1} exists and is unique, and that the coefficients ai,i=0,…,ma_{i}, i = 0, \ldots , m, are found from the system of inhomogeneous linear equations

G(φ0,φ1,…,φm)(a0⋮am)=((f,φ0)⋮(f,φm)) G\left(\varphi _{0}, \varphi _{1}, \ldots , \varphi _{m}\right)\left(\begin{smallmatrix} a_{0} \\ \vdots \\ a_{m} \end{smallmatrix}\right) =\left(\begin{smallmatrix} \left(f, \varphi _{0}\right) \\ \vdots \\ \left(f, \varphi _{m}\right) \end{smallmatrix}\right)

where G(φ0,φ1,…,φm)G\left(\varphi_{0}, \varphi_{1}, \ldots , \varphi_{m}\right) is the Gram matrix of the system of functions φ0(x),……,φm(x)\varphi_{0}(x), \ldots \ldots , \varphi_{m}(x) in the Euclidean space Vn+1V_{n+1}.

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Using the method of least squares, find the interpolation (best mean-square approximation) of the function ff given by the values f(0)=1,f(1)=2,f(2)=3,f(3)=5f(0) = 1, f(1) = 2, f(2) = 3, f(3) = 5:

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(1)

by a linear polynomial L1f(x)=b1x+b0L_{1}^{f}(x) = b_{1} x+b_{0} (i.e., interpolate using the system of functions φ0(x)=1,φ1(x)=x\varphi_{0}(x) = 1, \varphi_{1}(x) = x);

(2)

by a quadratic polynomial L2f(x)=b2x2+b1x+b0L_{2}^{f}(x) = b_{2} x^{2}+b_{1} x+b_{0};

(3)

by a polynomial of degree 3, L3f(x)=b3x3+b2x2+b1x+b0L_{3}^{f}(x) = b_{3} x^{3}+b_{2} x^{2}+b_{1} x+b_{0}.

Find the quadratic deviation ∥Lif−f∥4\left\| L_{i}^{f}-f\right\|_{4} in each of these cases.

Using the method of least squares, find, in the Euclidean space of continuously differentiable functions with the inner product

(f,g)=∫02πf(x)g(x)dx (f, g) = \int _{0}^{2 \pi } f(x) g(x) d x

the interpolation (best mean-square approximation) of a continuously differentiable function ff using the system of functions φ0(x)=1,φ1(x)=cos⁡x,φ2(x)=sin⁡x\varphi_{0}(x) = 1, \varphi_{1}(x) = \cos x, \varphi_{2}(x) = \sin x.

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Prove that the set of pseudosolutions of the system Ax=bA x = b forms an affine subspace (linear manifold)

P=x′+Ker⁡A P = x^{\prime }+\operatorname {Ker} A

in Rn\mathbb {R}^{n} (x′x^{\prime } is some pseudosolution).

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Let P=x′+Ker⁡AP = x^{\prime }+\operatorname {Ker} A be the affine subspace of pseudosolutions of the system Ax=bA x = b, and let x′′∈Px^{\prime \prime } \in P be some pseudosolution. Prove that the following statements are equivalent:

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(1)

x′′⊥Ker⁡Ax^{\prime \prime } \perp \operatorname {Ker} A;

(2)

∣x′′∣=min⁡x∈P∣x∣\left|x^{\prime \prime }\right| = \min_{x \in P}\left|x\right|;

(3)

x′′∈Im⁡A∗x^{\prime \prime } \in \operatorname {Im} A^{*}.

Prove that there exists a unique vector x′′x^{\prime \prime } determined by conditions 1-3. It is called the normal pseudosolution.

Consider the system Ax=bA x = b, where AA is the matrix of the operator A:Rn→Rm\mathrm{A}: \mathbb {R}^{n} \rightarrow \mathbb {R}^{m}. The spaces Rn\mathbb {R}^{n} and Rm\mathbb {R}^{m} are equipped with the standard scalar products. Let e1,e2,…,ene_{1}, e_{2}, \ldots , e_{n} be the canonical basis of the nonnegative self-adjoint operator A∗A\mathbf{A}^{*} \mathbf{A}, and let λ1,λ2,…,λr\lambda_{1}, \lambda_{2}, \ldots , \lambda_{r} be its positive (i.e. nonzero) eigenvalues, corresponding to the eigenvectors e1,e2,…,ere_{1}, e_{2}, \ldots , e_{r} respectively. Prove that the general pseudosolution x′x^{\prime } of the system Ax=bA x = b has the form

x′=(A∗b,e1)λ1e1+…+(A∗b,er)λrer+αr+1er+1+…+αnen x^{\prime } = \frac{\left(\mathbf{A}^{*} b, e_{1}\right)}{\lambda _{1}} e_{1}+\ldots +\frac{\left(\mathbf{A}^{*} b, e_{r}\right)}{\lambda _{r}} e_{r}+\alpha _{r+1} e_{r+1}+\ldots +\alpha _{n} e_{n}

moreover the normal pseudosolution

xnorm ′=(A∗b,e1)λ1e1+…+(A∗b,er)λrer x_{\text{norm }}^{\prime } = \frac{\left(\mathbf{A}^{*} b, e_{1}\right)}{\lambda _{1}} e_{1}+\ldots +\frac{\left(\mathbf{A}^{*} b, e_{r}\right)}{\lambda _{r}} e_{r}

If we have an inconsistent system Ax=bA x = b of linear inhomogeneous equations, then from the point of view of the theory of linear operators this is a situation in which the operator A\mathbf{A} is not invertible, and the vector bb does not lie in its image Im A. With the help of the scalar product it is possible to define the so-called pseudoinverse operator A+\mathbf{A}^{+}, which will "in the best way replace" the nonexistent inverse operator A−1\mathbf{A}^{-1}.

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Let the matrix AA of a system of equations define some linear map A:Rn→Rm\mathbf{A}: \mathbb {R}^{n} \rightarrow \mathbb {R}^{m}. The spaces Rn\mathbb {R}^{n} and Rm\mathbb {R}^{m} may be regarded as Euclidean, by equipping them with corresponding (e.g. standard) scalar products. Define the pseudoinverse operator A+\mathrm{A}^{+}as follows:

A+(x)=A~−1(x),x∈Im⁡AA+(x)=0,x∈(Im⁡A)⊥ \begin{array}{ll} \mathbf{A}^{+}(x) = \widetilde{\mathbf{A}}^{-1}(x), & x \in \operatorname {Im} \mathbf{A} \\ \mathbf{A}^{+}(x) = 0, & x \in (\operatorname {Im} \mathbf{A})^{\perp } \end{array}

where A~=A∣(Ker⁡A)⊥\widetilde{A} = \left.\mathrm{A}\right|_{(\operatorname {Ker} \mathbf{A})^{\perp }} is the operator constructed in the previous problem. Prove that:

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(1)

if the operator A is invertible, then A+=A−1\mathbf{A}^{+} = \mathrm{A}^{-1};

(2)

the vector x′=A+bx^{\prime } = \mathrm{A}^{+} b is the normal pseudosolution of the system Ax=bA x = b.

Let A:Rn→Rm\mathrm{A}: \mathbb {R}^{n} \rightarrow \mathbb {R}^{m} be a linear map of Euclidean spaces given by the matrix AA. Define the map B:Rm→Rn\mathbf{B}: \mathbb {R}^{m} \rightarrow \mathbb {R}^{n} as follows: let y∈Rmy \in \mathbb {R}^{m}, then define the value B(y)∈RnB(y) \in \mathbb {R}^{n} as the (unique) normal pseudosolution of the system Ax=yA x = y. Prove that:

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(1)

B is a linear operator;

(2)

B coincides with the pseudoinverse operator A+\mathbf{A}^{+}.

Prove that the matrix A+A^{+} of the pseudoinverse operator A+\mathbf{A}^{+} (defined with respect to the standard scalar product) equals the normal pseudosolution of the matrix equation AX=EA X = E (considered as an inhomogeneous system of linear equations with nmn m unknowns).

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Let AA be an m×nm \times n matrix over C\mathbb {C}. The Moore-Penrose pseudo inverse matrix A+A^{+} is the unique n×mn \times m matrix which satisfies

AA+A=A,A+AA+=A+,(AA+)∗=AA+,(A+A)∗=A+A. AA^{+}A = A, \quad A^{+}AA^{+} = A^{+}, \quad (AA^{+})^{*} = AA^{+}, \quad (A^{+}A)^{*} = A^{+}A.

We also have that

x→=A+b→(1) \overrightarrow {x} = A^{+}\overrightarrow {b} \tag {1}

is the shortest length least square solution to the problem Ax→=b→A\overrightarrow {x} = \overrightarrow {b}.

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(i)

Show that if (A∗A)−1(A^{*}A)^{-1} exists, then A+=(A∗A)−1A∗A^{+} = (A^{*}A)^{-1}A^{*}.

(ii)

Let

A=(132435). A = \begin{pmatrix} 1 & 3 \\ 2 & 4 \\ 3 & 5 \end{pmatrix}.

Find the Moore-Penrose matrix inverse A+A^{+} of AA.

Show that the curve fitting problem

j01234tj−1.0−0.50.00.51.0yj1.00.50.00.52.0 \begin{array}{c|ccccc} j & 0 & 1 & 2 & 3 & 4 \\ \hline t_{j} & -1.0 & -0.5 & 0.0 & 0.5 & 1.0 \\ y_{j} & 1.0 & 0.5 & 0.0 & 0.5 & 2.0 \end{array}

by a quadratic polynomial of the form p(t)=a2t2+a1t+a0p(t) = a_{2} t^{2}+a_{1} t+a_{0} leads to an overdetermined linear system.

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Consider the overdetermined linear system Ax→=b→A\overrightarrow {x} = \overrightarrow {b}. Find an x→^\hat{\overrightarrow {x}} such that

∥Ax→^−b→∥2=min⁡x→∥Ax→−b→∥2≡min⁡x→∥r→(x→)∥2 \left\| A\hat{\overrightarrow {x}}-\overrightarrow {b}\right\| _{2} = \min _{\overrightarrow {x}} \left\| A\overrightarrow {x}-\overrightarrow {b}\right\| _{2} \equiv \min _{\overrightarrow {x}} \left\| \overrightarrow {r}(\overrightarrow {x})\right\| _{2}

with the residual vector r→(x→):=b→−Ax→\overrightarrow {r}(\overrightarrow {x}) := \overrightarrow {b}-A\overrightarrow {x} and ∥⋅∥2\left\| \cdot \right\|_{2} denotes the Euclidean norm.

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Consider the overdetermined linear system Ax→=b→A\overrightarrow {x} = \overrightarrow {b} with

A=(111213141516171819110),x→=(x1x2),b→=(444458478493506516523531543571). A = \begin{pmatrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \\ 1 & 4 \\ 1 & 5 \\ 1 & 6 \\ 1 & 7 \\ 1 & 8 \\ 1 & 9 \\ 1 & 10 \end{pmatrix}, \qquad \overrightarrow {x} = \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}, \qquad \overrightarrow {b} = \begin{pmatrix} 444 \\ 458 \\ 478 \\ 493 \\ 506 \\ 516 \\ 523 \\ 531 \\ 543 \\ 571 \end{pmatrix}.

Solve this linear system in the least squares sense (see the previous problem) by the normal equations method.

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An underdetermined linear system is either inconsistent or has infinitely many solutions. Consider the underdetermined linear system Hx→=y→H\overrightarrow {x} = \overrightarrow {y}, where HH is an n×mn \times m matrix with m>nm>n and

x→=(x1x2⋮xm),y→=(y1y2⋮yn). \overrightarrow {x} = \begin{pmatrix} x_{1} \\ x_{2} \\ \vdots \\ x_{m} \end{pmatrix}, \qquad \overrightarrow {y} = \begin{pmatrix} y_{1} \\ y_{2} \\ \vdots \\ y_{n} \end{pmatrix}.

Assume that Hx→=y→H\overrightarrow {x} = \overrightarrow {y} has infinitely many solutions. Let QQ be the n×mn \times m matrix

Q=(10…00…001…00…0⋮⋮⋱⋮⋮⋮00…10…0). Q = \begin{pmatrix} 1 & 0 & \ldots & 0 & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 & 0 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & & \vdots \\ 0 & 0 & \ldots & 1 & 0 & \ldots & 0 \end{pmatrix}.

We define x→^:=Qx→\hat{\overrightarrow {x}} := Q\overrightarrow {x}. Find

min⁡x→∥Qx→−y→∥22 \min _{\overrightarrow {x}} \left\| Q\overrightarrow {x}-\overrightarrow {y}\right\| _{2}^{2}

subject to the constraint ∥Hx→−y→∥22=0\left\| H\overrightarrow {x}-\overrightarrow {y}\right\|_{2}^{2} = 0. Assume that (λHTH+QTQ)−1(\lambda H^{T} H+Q^{T} Q)^{-1} exists for all λ>0\lambda >0. Apply the Lagrange multiplier method.

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