4 MT210 Notebook Eigenvalues and Eigenvectors Definitions; Graphical Illustrations... 3


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1 MT Notebook Fall / prepared by Professor Jenny Baglivo c Copyright 9 by Jenny A. Baglivo. All Rights Reserved. Contents MT Notebook. Eigenvalues and Eigenvectors Definitions; Graphical Illustrations Eigenspaces, Characteristic Polynomials, Characteristic Equations Eigenanalysis and Powers; Eigenvector Bases; Special Cases Fundamental Theorem of Algebra, Complex Numbers and Eigenvalues Algebraic and Geometric Multiplicity; More About Eigenvector Bases Similar Matrices, Diagonalizable Matrices Applications: Population Projections and Stochastic Matrices Orthogonality and Orthogonal Projections Inner Product, Length, Distance Properties of Inner Product Orthogonal, Orthogonal Sets, Orthogonal Complement Fundamental Theorem of Linear Algebra Orthogonal Spanning Sets Angles, Inner Products, and Orthogonal Projections GramSchmidt Orthogonalization Process Least Squares Analysis Best Approximate Solutions; Normal Equations Application: Least Squares Analyses of Data Footnote: Eigenvalues, Eigenvectors and Least Squares Analysis
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3 MT Notebook This notebook is concerned with further matrix concepts and their applications. In particular, we will study eigenvalues, eigenvectors, orthogonality and least squares. The notes correspond to material in Chapters and 6 of the Lay textbook.. Eigenvalues and Eigenvectors.. Definitions; Graphical Illustrations Let A be a square matrix of order n, and λ ( lambda ) be a scalar.. λ is said to be an eigenvalue of A if Ax = λx for some nonzero vector x.. If x O satisfies Ax = λx, then x is an eigenvector of A with eigenvalue λ. Note that if x is a nonzero eigenvector, then so is cx for each c R since A(cx) = c(ax) = c(λx) = λ(cx). Thus, the nonzero elements of Span(x) are all eigenvectors of A. Eigenvalues are exceptional values and eigenvectors are exceptional vectors. The prefix eigen comes from the German language meaning owned by or peculiar to. Example. Let A = [.. Then.. λ =. is an eigenvalue of A, with corresponding eigenvector e.. λ =. is an eigenvector of A, with corresponding eigenvector e. Consider the transformation with rule Then T (x) = Ax. T (x) contracts points in the xdirection; in particular, T (e ) =.e. T (x) expands points in the ydirection; in particular, T (e ) =.e. T (x) maps the unit circle to the ellipse shown in the plot.
4 Example. Let A = [.6.. Then..9. λ =. is an eigenvalue» of A, with corresponding eigenvector v = / /,. λ = is an eigenvector» of A, with corresponding eigenvector v = / 7 /. 7 Consider the transformation with rule Then T (x) = Ax. T (x) contracts points in the v direction; in particular, T (v ) =.v. T (x) leaves points in the v direction fixed; in particular, T (v ) = v. T (x) maps the unit circle to the ellipse shown in the plot. The complete analysis of Example will be carried out in the next section. Note that a square matrix of order n with values in the real numbers (a ij R for all i, j) may not have eigenvalues λ R, and eigenvectors x R n. For example, matrices A corresponding to rotations around the origin: [ cos θ sin θ A =, where θ mπ for some integer m, sin θ cos θ leave no direction in space fixed... Eigenspaces, Characteristic Polynomials, Characteristic Equations Let A be a square matrix of order n, and let λ be a scalar.. Eigenspace: If λ is an eigenvalue of A, then the eigenspace of λ is the set containing all x satisfying Ax = λx: Eigenspace(λ) = {x : Ax = λx}. The eigenspace of λ contains all eigenvectors with eigenvalue λ and the zero vector. Since Ax = λx = λi n x Ax λi n x = O (A λi n )x = O, we know that Eigenspace(λ) = Null(A λi n ). Further, since the eigenspace of λ must have positive dimension, we know that the matrix (A λi n ) must be singular.
5 . Characteristic Polynomial: The expression det(a λi) is an n th degree polynomial in the variable λ, and is called the characteristic polynomial of A.. Characteristic Equation: The equation det(a λi) = is called the characteristic equation of A. To find the eigenvalues of A we solve the characteristic equation for λ. Example, continued. Let A = () Since A λi =»».6. λ..9 [.6., as above...9 =».6 λ., the characteristic polynomial is..9 λ det(a λi) = (.6 λ)(.9 λ) (.)(.) =..λ + λ. = λ.λ +.. Since det(a λi) = (λ.)(λ ), the solutions to the characteristic equation are the eigenvalues. and listed earlier. () Let λ =.. Since A λi = [..,.. [ A λi O = [.... [ [ x x = x [ = x, where x is free, we know that Note that v = j» Eigenspace(.) = Null(A.I) = Span [ / / = [ Eigenspace(.). ff. () Let λ =. Since A λi = [...., [ A λi O = [.... [ / [ x / x = x [ / = x, where x is free, we know that Note that v = j» / Eigenspace() = Null(A I) = Span [ / 7 / 7 = 7 [ ff Eigenspace(.). j» = Span ff.
6 [.8. Problem. Let A =..7 Determine the eigenvalues of A, and write each eigenspace as the span of a set of vectors.. 6
7 [ Problem. Let A =.... Determine the eigenvalues of A, and write each eigenspace as the span of a set of vectors.. 7
8 Problem. Let A =.8.. Determine the eigenvalues of A, and write each eigenspace as the span of a set of vectors. 8
9 .. Eigenanalysis and Powers; Eigenvector Bases; Special Cases Conducting an eigenanalysis (that is, finding eigenvalues and eigenvectors) can be challenging. The following is an initial list of useful theorems for eigenanalysis:. Powers: If λ is an eigenvalue of A with eigenvector x and k is a positive integer, then x is an eigenvector for A k with corresponding eigenvalue λ k.. Bases and Powers: Let A be a square matrix of order n. Suppose that v i is an eigenvector for A with corresponding eigenvalue λ i, for i =,,..., n, and the set {v, v,..., v n } is a basis for R n. Then, for every vector x and positive integer k, A k x can be computed quickly using the unique representation of x in the eigenvector basis {v, v,..., v n }. Specifically, if x = c v + + c n v n for unique constants c i, then A k x = A k (c v + + c n v n ) = c (A k v ) + + c n (A k v n ) = c (λ ) k v + + c n (λ n ) k v n.. Diagonal Matrices: Let A be a diagonal matrix of order n. Then e i is an eigenvector for A with eigenvalue a ii, for i =,,..., n. Thus, the standard basis for R n is an eigenvector basis for the diagonal matrix A, and the eigenvalues are the diagonal elements of A.. Distinct Eigenvalues: Let A be a square matrix of order n. If A has n distinct eigenvalues, then A has an eigenvector basis. To construct an eigenvector basis, choose one nonzero vector from each eigenspace.. Triangular Matrices: Let A be a triangular matrix of order n. Then det(a λi) = (a λ)(a λ) (a nn λ) and the eigenvalues of A are a, a,..., a nn. (There may be repeats in the list.) 9
10 General application: projections over time. A general application of eigenanalysis is to the analysis of projections over time. In this type of application,. x represents information at time,. x = Ax represents information at time,. x = Ax = A x represents information at time, and so forth. If A has an eigenvector basis, then information at time k is x k = A k x = c (λ ) k v + + c n (λ n ) k v n where x = c v + + c n v n. We will see an important application of this methodology in Section..7 (page 7). As a simple illustration, consider A = Now, v =»» /, v =».6. once again. Let..9, λ =., λ =, and x = c v + c v. x k = A k x = c (.) k v + c () k v c v as k. Thus, information at time k is approximately equal to the v component of information at time when k is large. Problem. Use the definitions of eigenvalue and eigenvector, and properties of matrices, to prove the following special case of the first theorem listed on the previous page: Let A be a square matrix of order n, and let x be an eigenvector of A with eigenvalue λ. Demonstrate that x is an eigenvector of A with eigenvalue λ.
11 Problem. Let A be a square matrix of order n, and assume that A has n distinct eigenvalues and let v i Eigenspace(λ i ) be a nonzero vector, for each i. Show that {v, v,..., v n } is a linearly independent set, thus forming a basis for R n.
12 Problem 6. The following triangular matrices each have eigenvalues,, : (a) A = ; (b) A = In each case, write Eigenspace() and Eigenspace() as spans of sets of vectors..
13 .. Fundamental Theorem of Algebra, Complex Numbers and Eigenvalues Let A be a square matrix of order n. To find the eigenvalues of A we need to solve the characteristic equation, which requires that we factor the characteristic polynomial, det(a λi). By the fundamental theorem of algebra, the characteristic polynomial can always be factored into n linear terms if we allow both real and complex numbers: det(a λi) = (λ λ)(λ λ) (λ n λ), where each λ i C. The eigenvalues are λ, λ,..., λ n. In general, not all λ i s are distinct. For example, let A = 6 6. Since A λi = λ 6 6 λ λ = ( λ) λ 6 6 λ = ( λ)(λ + 6) = ( λ)(6i λ)( 6i λ) = implies λ =, 6i, 6i, and the eigenvalues of A are, 6i and 6i. Further, Eigenspace() Eigenspace(6i) Eigenspace( 6i) = Null(A I) = Null(A 6iI) = Null(A + 6iI) 6 + i 6 i = Span = Span + 6i = Span 6i Note that matrices with complex eigenvalues and complex eigenvectors are common in applied mathematics. Examples include population projection matrices (see Section..7, page 7)... Algebraic and Geometric Multiplicity; More About Eigenvector Bases Let A be a square matrix of order n and let λ be an eigenvalue of A. Then. Algebraic Multiplicity: The algebraic multiplicity of λ is the number of times (λ λ) appears as a factor of the characteristic polynomial.. Geometric Multiplicity: The geometric multiplicity of λ is the dimension of Eigenspace(λ ). Note that, by the fundamental theorem of algebra, the sum of the algebraic multiplicities of the eigenvalues of A must be n.
14 Problem 6, continued. Fillin the table below with information for the triangular matrices from the problem on page : (a) A = (b) A = Geometric Algebraic Geometric Algebraic Multiplicity Multiplicity Multiplicity Multiplicity of λ = of λ = of λ = of λ = More on finding eigenvector bases. doing eigenanalyses: Here are two additional theorems that are useful for. Algebraic and Geometric Multiplicity: Let A be a square matrix of order n and let λ be an eigenvalue of A. Then the algebraic and geometric multiplicities of λ must satisfy the following inequalities: Geometric Multiplicity of λ Algebraic Multiplicity of λ. (If the geometric multiplicity is strictly less than the algebraic multiplicity, then there is a deficiency of eigenvectors and we won t be able to find an eigenvector basis.). Pooling Eigenspace Bases: If A has p distinct eigenvalues (λ i for i =,,..., p) and B i is a basis for eigenspace of λ i for each i, then the set B B B p is a linearly independent set. (If you pool the bases, you get a linearly independent set.) Problem 6, continued. Do either of the matrices in Problem 6 have an eigenvector basis for R? If yes, explicitly write down a basis. If no, explain why.
15 ..6 Similar Matrices, Diagonalizable Matrices Similar matrices: Let A and B be square matrices of order n. Then A and B are said to be similar if there exists an invertible matrix P satisfying A = P BP (and B = P AP ). If A and B are similar matrices, then. they have the same determinant, det(a) = det(b),. they have the same eigenvalues, and. their k th powers satisfy A k = P B k P, for each positive integer k. The factorization A = P BP is useful when B is easier to work with than A. Diagonalizable matrices: Let A be a square matrix of order n. Then A is said to be diagonalizable if it is similar to a diagonal matrix. That is, if A = P DP where D is diagonal and P is invertible. The following theorem tells us exactly when A is diagonalizable. Theorem (Diagonalization). Let A be a square matrix of order n. Then A is diagonalizable if and only if A has n linearly independent eigenvectors. In fact, A = P DP iff the columns of P are n linearly independent eigenvectors and the diagonal entries of D are the corresponding eigenvalues. For example,. If A = [ If A = 6 6, then A = P DP where P = P = [ and D =, then A = P DP where 6 + i 6 i + 6i 6i [. and D =. 6i 6i.
16 Problem, continued. The square matrix A =.8 is diagonalizable.. Use the work you did on page 8 to find matrices P and D so that A = P DP. Diagonalization and Transformations. Suppose that A = P DP, where λ P = [ λ v v... v n and D =..., λ n and let B = {v, v,..., v n } be the basis whose elements are the columns of P. Then the action of A is to ( ): change from the standard basis of R n to basis B; ( ): operate as a diagonal matrix in basis B, and ( ): change back to the standard basis for interpretation. x = n i= c iv i A c c [x B = c n Ax = n i= λ ic i v i D [Ax B = λ c λ c 6. 7 λ nc n Similarly, the action of A k is to ( ): change from the standard basis of R n to basis B; ( ): operate as the k th power of a diagonal matrix in basis B, and ( ): change back to the standard basis for interpretation. x = n i= c iv i A k c c [x B = c n Ak x = n i= λk i c iv i D k [Ak x B = λ k c λ k c 6. 7 λ k nc n 6
17 ..7 Applications: Population Projections and Stochastic Matrices This section contains two applications of eigenanalysis. I. Population projections, and the northern spotted owl (Source: Lay text, p.6). Researchers used demographic data for the northern spotted owl to develop a stagematrix model using life stages (juvenile, subadult and adult). Their goal was to track the population growth/decline of the owl in a particular old growth forest in the Pacific northwest. If j i is the number of juveniles, s i is the number of subadults and a i is the number of adults in the population at time i, then x i = j i s i is the population vector at time i, a i and the total population at time i is the sum of the components (j i + s i + a i ). The matrix that allows you to project one year is. A = Given x i, the population vector at time (i + ) is. j i.a i x i+ = Ax i =.8 s i =.8j i =.7.9 a i.7s i +.9a i j i+ s i+ a i+. Matrices used in population problems are generally diagonalizable, with both real and complex eigenvalues. For this problem, we can write A = P DP, where.98 D =. +.6i..6i and P =.8.6.9i.6 +.9i i..6i The diagonal elements of D are the eigenvalues of A. Since (.98) k, (. +.6i) k and (..6i) k as k, we know that the population will eventually crash given any initial population vector. 7
18 The author tells us that, if the (,)entry of the A matrix was.6 (the proportion that would be appropriate for this species in a different location), then the population would grow. The new matrix (the one with the new (,)entry) would be diagonalizable with i.77.i D =.6 +.8i and P = i i i Since (.6) k, (.6 +.8i) k and (.6.8i) k as k, the statement the author makes is correct. I simulated population growth over years starting with an initial population of individuals (j + s + a = ) and using both the true matrix and the matrix with the (,)entry changed. The results are summarized in the plot below.. Left Plot: Using the true matrix, the total population declined over time.. Right Plot: Using the altered matrix, the total population increased over time with approximate geometric growth. Geometric growth kicks in when k is large enough so that k th powers of the last two eigenvalues are close to zero: (.6 +.8i) k and (.6.8i) k. II. Stochastic matrices, moving cars, and searching the internet. A probability vector is one whose entries are nonnegative real numbers with sum. A stochastic matrix is a square matrix whose columns are probability vectors. The following matrices are examples of stochastic matrices: [ Stochastic matrices are used to model population movement over time, where individuals move among n different locations. The following simple example considers the movement of rental cars over time. 8
19 Example. A car rental agency has three rental locations (,, ). A customer may rent a car from any of the three locations and return the car to any of the three locations. From past experience, management observes that:. Location : Cars rented from location are returned to locations,, with probabilities.,. and., respectively;. Location : Cars rented from location are returned to locations,, with probabilities.,.8 and, respectively; and. Location : Cars rented from location are returned to locations,, with probabilities.,. and., respectively. Suppose that we would like to determine the probabilities that a car initially rented from a given location (either, or ) will be returned to locations,, after k rental periods. Let A be the matrix whose columns are the probabilities listed above, let a i, b i and c i be the probabilities that the car is at locations,, after i rental periods, and let x i be the vector whose components are the probabilities a i, b i and c i :... a i A =..8., x i = b i... The matrix A can be used to project one rental period. That is, x i+ = Ax i for each i. The starting location vectors (x ) are for location, for location, for location. c i We can write A = P DP, where D =., P = [... v v v =.6.,.... and the first column of P has nonnegative terms with sum. Note that k, (.) k and (.) k as k. If x corresponds to the location vector, for example, then x = v (/)v + (/)v and x k = A k x v for large k. In fact, x k v after only time periods: k = k = k = k = k = k = k = 6 k = 7 k = 8 k = 9 k = k = a k b k c k
20 Similarly, if x corresponds to the location vector, then x = v + v v and x k = A k x v for large k, and if x corresponds to the location vector, then x = v + v + v and x k = A k x v for large k. Thus, if k is large, the probabilities that a car initially at any one of the three locations will be returned to locations,, after k rental periods are (approximately).,.6 and.. Surfing the web. Now, imagine yourself surfing the web starting from some initial location and randomly following hyperlinks. Assuming an appropriate A matrix can be created and analyzed as above, the probability that you will be at a given location after a sufficient number of steps can be determined. The designers of Google use the eventual probabilities to determine the order in which the results of a search are reported; specifically, webpages with higher probabilities are listed before those with lower probabilities. Their A matrix uses the hyperlink structure of the web and some proprietary information.. Orthogonality and Orthogonal Projections.. Inner Product, Length, Distance Let v and w be vectors in R k. Then. Inner product: The inner product (dot product) of v and w is the number v w = v T w = [ w v v v k 6 7. = v w + v w + + v k w k. w k. Length: The length of v is the number v = v v = w v + v + + v k.. Distance: The distance between v and w is the length of the difference vector v w: dist(v, w) = v w = (v w ) + (v w ) + + (v k w k ). Further, a unit vector is a vector of length one. If v O, then u = v is the unit vector in the direction of v. v
21 For example, if v = 8 and w =, then () v w = () The length of v is () The unit vector in the direction of v is () The distance between v and w is.. Properties of Inner Product Let u, v, w R k, c R. Then. Commutative: v w = w v.. Scalars: (cv) w = v (cw) = c(v w).. Distributive: (u + v) w = (u w) + (v w) and w (u + v) = (w u) + (w v).. Nonnegative: v v. Further v v = iff v = O. Problem. Let v, v and w be vectors in R k, and suppose that v w = and v w =. Use the properties of inner product to demonstrate that v w = for every v Span {v, v }.
22 .. Orthogonal, Orthogonal Sets, Orthogonal Complement The concept of orthogonality is important in applications.. Orthogonal: Let v and w be vectors in R k. Then v and w are said to be orthogonal if their inner product is zero: v w =.. Orthogonal Set: Let v, v,..., v p be vectors in R k. Then {v, v,..., v p } is said to be an orthogonal set if v i O for all i, and v i v j = when i j.. Orthogonal Complement: Let V be a subspace of R k. The orthogonal complement of V, denoted by V ( V perp ), is the collection of all vectors orthogonal to V : Problem. Let v = a V = {w : w is orthogonal to each v V }., v =, v = b Find values of a, b so that {v, v, v } is an orthogonal set..
23 Properties of orthogonal complements. orthogonal complement. Then Let V be a subspace of R k and let V be its. Subspace: The orthogonal complement of V is a subspace of R k. Further, V V = {O} since O is the only vector satisfying x x =.. Orthogonal Complement of V : The orthogonal complement of V is V : ( V ) = V.. Spanning Sets: Suppose that V = Span{v, v,..., v p }. Then w V if and only if w v i = for i =,,..., p.. Pooling Bases: If B is a basis for V and B is a basis for V, then the union of the bases, B B, is a basis for R k. To illustrate orthogonal complements, let {[ V = Span{v} = Span, } and w = [ w w. Since = w v = w + w w = w [ where w is free, we know that {[ V = Span, }, as shown in the plot.
24 Problem. In each case, write V as a span. (a) V = Span 8 < :, 9 = ; (b) V = Span 8 < : 9 = ;
25 .. Fundamental Theorem of Linear Algebra Let A be an m n matrix and let A T be its transpose. The following theorem, known as the fundamental theorem of linear algebra, gives important relationships among the four subspaces related to A and its transpose. Fundamental Theorem of Linear Algebra. Let A be an m n matrix. Then. Null(A) and Row(A) = Col(A T ) are orthogonal complements in R n.. Null(A T ) and Row(A T ) = Col(A) are orthogonal complements in R m. Further, if rank(a) = r, then. dim(col(a)) = dim(col(a T )) = r,. dim(null(a)) = n r and. dim(null(a T )) = m r. Proof of first statement: It is instructive to demonstrate the first statement in the theorem. Let A T = [ α α α m and A = Now (complete the proof), Ax = O α T.. Then T α m T α α x T α. x = α x. = O. T α m α m x
26 Problem. Find bases for Null(A), Col(A), Null(A T ) and Col(A T ), where A =
27 .. Orthogonal Spanning Sets The following theorem tells us that a set of mutually orthogonal nonzero vectors is linearly independent. Further, the coordinates of a vector w with respect to a basis of mutually orthogonal nonzero vectors can be found quickly using dot products: Theorem (Orthogonal Spanning Sets). Let {v, v,..., v p } be an orthogonal set of vectors in R k and let V = Span{v, v,..., v p }. Then. {v, v,..., v p } is a basis for V.. If w V, then w = c v + + c p v p where c i = w v i v i v i for each i. Problem. In each case, use dot products to find the coordinates of the vector w with respect to the given orthogonal basis. 8 < (a) V = Span{v, v } = Span :, 9 = ;, and w =. 7
28 8 >< (b) V = Span{v, v, v } = Span 6 >: 7, 6 7, 6 9 >= 7, and w = 6 >; Angles, Inner Products, and Orthogonal Projections Angle between v and w. Let v and w be vectors in R or R represented as directed line segments beginning at the origin. The angle between v and w, θ, is the smaller of the two angles at the origin determined by v and w. The angle θ lies in the interval [, π. The following plots illustrate angles satisfying < θ < π (left) and π < θ < π (right): Analytic geometry can be used to demonstrate that v w = v w cos(θ). Note, in particular, that if θ = π, then v w =. 8
29 Orthogonal projection. For vectors in R or R, the (orthogonal) projection of w onto v, denoted by projv(w), is the vector highlighted in each diagram below for angles θ π :. Left Plot ( θ < π ) : The projection of w onto v is the vector that points in the direction of v and whose length is w cos(θ).. Right Plot ( π < θ π) : The projection of w onto v is the vector that points in the direction opposite to v and whose length is w cos(π θ). When θ = π, the projection of w onto v is the zero vector. Geometry, trigonometry and the relationship v w = v w cos(θ) can be used to demonstrate that the projection can be computed as follows: ( w v ) projv(w) = v. v v That is, the projection is the scalar multiple cv, where c is the ratio ( w v ) v v. [ 6 For example, let v = and w = Then () projv(w) = [. () w projv(w) = () The inner product of () and () is. 9
30 Orthogonal projection in kspace. Let v and w be vectors in R k, and let V = Span{v}. Then the (orthogonal) projection of w onto v (equivalently, the projection of w onto the subspace spanned by v) is defined as follows: ( w v ) projv(w) = proj V (w) = v. v v The projection, ŵ = projv(w) = proj V (w), is an element of the vector space V and satisfies the following properties:. Minimum Distance: ŵ is the unique vector in V closest to w.. Orthogonality: ŵ is the unique vector in V for which w ŵ is orthogonal to V. Thus, we can find the distance between w and V by computing w ŵ. Problem 6. (a) w = In each case, find the distance between w and V = Span{v}. 8 8 < and V = Span : 9 = ; (b) w = 7 8 < and V = Span : 9 = ;
31 Orthogonal projection onto V. Let V = Span{v, v,..., v p } be the span of an orthogonal set of vectors (a set of mutually orthogonal nonzero vectors) in R k and let w R k. Then the (orthogonal) projection of w onto V is defined as follows: ŵ = proj V (w) = c v + c v + + c p v p where c i = w v i v i v i for each i. This definition generalizes the case for projection onto the span of a single vector in R k, and requires that the spanning set is an orthogonal set. Note also that. If V i = Span{v i } for each i, then ŵ is the sum of projections in each coordinate direction:. If w V, then ŵ = w. ŵ = proj V (w) = proj V (w) + proj V (w) + + proj Vp (w). Properties of orthogonal projections are stated in the following theorem, and illustrated to the right. In the plot, the horizontal axis represents vector space V and the vertical axis represents the orthogonal complement, V ; and w is decomposed into the part of w in V and the part in V. Theorem (Orthogonal Projections). Let V be a subspace of R k, w be any vector in R k, and ŵ be the projection of w onto V. Then. Orthogonal Decomposition: The difference (w ŵ) is a vector in V, and the sum w = ŵ + (w ŵ) is the unique representation of w as the sum of a vector in V and a vector in V. (Thus, we have an orthogonal decomposition of w into the part of the vector in V and the part of the vector in V.). Best Approximation: The vector ŵ is the closest point in V to w.
32 Problem 7. In each case, find ŵ and (w ŵ). Note that each V has been written as the span of an orthogonal set. j» (a) V = Span 8 < (b) V = Span : 8 >< (c) V = Span 6 >: ff», w =, 7, 6 7, 6 9 = ;, w = 9 7 >= 7, w = 6 >;
33 ..7 GramSchmidt Orthogonalization Process Let V R k be a pdimensional subspace and let {x, x,..., x p } be a basis for V. The Gram Schmidt orthogonalization process allows us to construct an orthogonal basis {v, v,..., v p } for V starting with {x, x,..., x p }. The method is as follows:. Let v = x.. Let v = x proj V (x ) where V = Span{v } = Span{x }.. Let v = x proj V (x ) where V = Span{v, v } = Span{x, x }.. Let v = x proj V (x ) where V = Span{v, v, v } = Span{x, x, x }. And, so forth. The final set, {v, v,..., v p }, is an orthogonal basis for V. Problem 8. In each case, find an orthogonal basis for V. 8 < (a) V = Span :, 9 = ; 8 >< (b) V = Span 6 >: 7, 6 7, 6 9 >= 7 >;
34 . Least Squares Analysis.. Best Approximate Solutions; Normal Equations Let A be an m n coefficient matrix and assume that Ax = b is inconsistent. We propose to find approximate solutions to the system as follows: () Find the projection of b onto Col(A), b, and () Report solutions to the consistent system Ax = b. Observation : Since b is as close to b as possible, each approximate solution x satisfies b b = b Ax is as small as possible. The difference vector is b (a, x + a, x + + a,n x n ) b (a, x + a, x + + a,n x n ) b Ax =. b m (a m, x + a m, x + + a m,n x n ) and the square of the length of the difference vector is m (b i (a i, x + a i, x + + a i,n x n )). i= Each approximate solution x will minimize the above sum of squared differences. reason the approximate solutions are called least squares solutions. For this Observation : Since the difference vector (b b) = (b Ax) (Col(A)) and the orthogonal complement of the column space of A is the null space of the transpose of A, (Col(A)) = Null ( A T ), by the Fundamental Theorem of Linear Algebra, we know that A T (b b) = A T (b Ax) = O.
35 Further, A T (b Ax) = O A T b A T Ax = O A T Ax = A T b. Thus, least squares solutions can be found by solving the consistent system on the right (called the normal equation of the system). By using the normal equation, we do not need to find the projection of b on the column space of A. The following theorem gives the properties of this process: The Least Squares Theorem. Under the conditions above,. x is a least squares solution to Ax = b iff x is a solution to A T Ax = A T b.. A T A is invertible iff the columns of A are linearly independent. Thus, there is a unique least squares solution iff the columns of A are linearly independent. 6 For example, consider the inconsistent system Ax = b where A = and b =. Then [ A T A A T b = [ 6 [ 6 [ is the unique least squares solution to the inconsistent system above. [ x = Problem. In each case, find the least squares solution(s) to Ax = b. (a) A = and b = 6.
36 (b) A = (c) A = and b =. and b =
37 .. Application: Least Squares Analyses of Data The methodology from the last section can be applied to finding curves of best fit (as minimizing a sum of squared differences). As a simple illustration, consider the four data pairs (, ), (, ), (8, ) and (, ). These points lie close to a straight line with equation ŷ = a + bx, as illustrated in the left plot below. The intercept and slope of the line can be found by the method of least squares. Specifically, we start with a by system of linear equations, convert the system to a matrix equation Ax = b, and find the least squares solution(s) by solving the normal equation A T Ax = A T b. a + b = a + b = a + b 8 = a + b = [ a 8 = b [ [ a = b Thus, the least squares regression line is ŷ = ( ) ( 9 + 9) x. +.x, as shown on the left above. Let ŷ i = ( ) ( 9 + 9) xi and e i = y i ŷ i for i =,,, : ŷ i e i A plot of (ŷ i, e i ) pairs is shown on the right above. [ 96 [ a = b [ /9 /9 The following pages contain several applications of this general strategy to real data. 7
38 Example: Olympic winning times (Source: Hand et al, 99). Consider the following data pairs, where x i is the time in years since 9 and y i is the Olympic winning time in seconds for men in the final round of the meter event. i x i y i i x i y i The data cover all Olympic events held between 9 and 988. (Note that Olympic games were not held in 96, 9, and 9.) The twenty data pairs lie approximately on a straight line with equation ŷ = a + bx, whose intercept and slope can be estimated by the method of least squares. a + b =.8 a + b =. a + b88 = 9.9 [.8 [ [ a =. 9 a b = b [ [ a which implies that b [ Thus, the least squares regression line is ŷ =.898.x, as illustrated on the left above. Note that the origin of the plot is not (, ). The results suggest that the winning times have decreased at the rate of about. seconds per year during the 88 years of the study. Let ŷ i =.898.x i and e i = y i ŷ i for i =,,...,. A plot of (ŷ i, e i ) pairs is shown on the right above. 8
39 Example: Brainbody study (Source: Allison & Cicchetti, 976). As part of a study on sleep in mammals, researchers collected information on the average body weight (in kilograms) and average brain weight (in grams) for different species. Let x i = ln (Average Body Weight i ) and y i = ln (Average Brain Weight i ) for i =,,...,. The (x i, y i ) pairs lie approximately on a line with equation ŷ = a + bx, whose intercept and slope can be estimated by the method of least squares. Starting with the normal equation A T Ax = A T b: [ [.86 a = b [ [ a b [...78 Thus, the least squares regression line is ŷ =. +.78x. As with the earlier examples, the left plot above is a plot of (x i, y i )pairs superimposed on the least squares regression line and the right plot is a plot of (ŷ i, e i ) pairs, for i =,,...,. It is instructive to examine the estimated relationship between average brain and body weights on their original scales. The graph of this relationship is shown to the right. The formula for this curve is (please complete). Note that Man s brain weight is much larger than expected given the modest body weight. The Asian Elephant has an enormous body weight and a correspondingly large brain weight. 9
40 Example: Timber yield study (Source: Hand et al, 99). As part of a study designed to estimate the volume of a tree (and therefore its yield) given its diameter and height, data were collected on the volume (in cubic feet), diameter at inches above the ground (in inches), and height (in feet) of black cherry trees in the Allegheny National Forest. Let x,i = ln(diameter i ), x,i = ln(height i ) and y i = ln(volume i ) for i =,...,. The (x,i, x,i, y i ) triples lie approximately on a plane with equation ŷ = a + bx + cx, whose coefficients can be estimated using the method of least squares. Starting with the normal equation A T Ax = A T b: a b = c a 6.6 b.98. c.7 Thus, the least squares regression equation is ŷ = x +.7x. The regression equation is plotted on the left above, along with the (x i,, x,i, y i ) triples. Triples lying under the surface appear slightly lighter in color. The right plot is a plot of (ŷ i, e i ) pairs, for i =,,...,. It is instructive to examine the estimated relationship among diameter, height and volume in their original scales. The graph of this relationship is shown to the right. The formula for this curve is (please complete). Any comments?
41 Example: Body fat study (Source: Johnson, 996). As part of a study to determine if the percentage of body fat can be predicted accurately using only a scale and measuring tape, data were collected on men. Let x,i, x,i, x,i and x,i be the abdomen, wrist, hip and neck circumferences (in centimeters) of the i th individual, and let y i be the man s percent body fat measured using an accurate underwater technique. Here is some summary information: Average Value Minimum Value Maximum Value Abdomen (x ) Wrist (x ) Hip (x ) Neck (x ) Body Fat (y) Consider fitting a linear function of the form ŷ = a + bx + cx + dx + ex using the method of least squares to estimate the coefficients (a through e). Starting with the normal equation A T Ax = A T b: a b 7 6c 7 d = 6 e a b 6c 7 d 6 e Thus, the least squares regression formula is ŷ = x.89x.6x.x. The (ŷ i, e i ) pairs are shown on the left below. Individuals with. the largest ŷ i and smallest e i, and measurements (Abdomen, Wrist, Hip, Neck, Body Fat) = (8.,., 7.7,.,.), and. the largest e i, and measurements (Abdomen, Wrist, Hip, Neck, Body Fat) = (9.9, 7.,., 9.,.9), have been highlighted in the plot. Any comments? If these two individuals are removed and a new least squares solution is computed, the pattern of errors does not change dramatically, as shown in the right plot above.
42 .. Footnote: Eigenvalues, Eigenvectors and Least Squares Analysis Let M = A T A be the matrix used in the normal equation for finding least squares solutions to inconsistent systems. M is a symmetric matrix, satisfying the following properties:. M is a diagonalizable matrix,. the eigenvalues of M are nonnegative real numbers,. M has an eigenvector basis that is an orthogonal set, and. M is invertible if and only if all eigenvalues are positive. If M is invertible, then the least squares solution is unique. Further, as long as the smallest eigenvalue is not too close to zero, then the computer will have no trouble finding the unique solution accurately. Body fat study example, continued. Consider again the body fat study from the last section. M = A T A can be written as M = P DP, where D and P The eigenvalues of M are written in decreasing order along the diagonal of D; all eigenvalues are comfortably greater than zero, implying that the computer had no trouble finding accurate least squares estimates of the coefficients of the prediction formula.. Finally, to improve the accuracy of least squares estimates in situations where eigenvalues may be close to zero (and M may be close to singular ), practitioners use a singular value decomposition of M before trying to find the estimates.
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