1 Last time: leastsquares problems


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1 MATH Linear algebra (Fall 07) Lecture Last time: leastsquares problems Definition. If A is an m n matrix and b R m, then a leastsquares solution to the linear system Ax = b is a vector x R n such that b A x b Ax for all x R n. If Ax = b is a consistent linear system then every leastsquares solution will be an exact solution. For any linear system Ax = b, there is always at least one leastsquares solution. Theorem. Let A be an m n matrix and b R m. A vector x R n is a leastsquare solutions to Ax = b if and only if A T A x = A T b. The linear system A T Ax = A T b is always consistent. Theorem. Let A be an m n matrix. The following are then equivalent: (a) Ax = b has a unique leastsquares solution for each b R m. (b) The columns of A are linearly independent. (c) A T A is invertible. When these properties hold, the unique leastsquares solution to Ax = b is the vector x = (A T A) A T b (which is the unique exact solution to A T Ax = A T b). In applications, the matrix A usually has many more rows than columns, so m is much larger than n. In this case, the n n matrix A T A is much smaller than A, and we can find a leastsquares solution by row reducing [ A T A A T b to echelon form to find the exact solutions to A T Ax = A T b in the usual way. Example (Lines of best fit). Suppose we have n data points (x, y ), (x, y ),..., (x n, y n ) that appear to be close to a line. We want to find parameters β 0, β R such that [ y = β 0 + β x describes the line of β0 best fit for this data. If our points are collinear, then for some β = R β we would have exactly y i = β 0 + β x i for i =,,..., n, meaning that there is an exact solution to the linear system Xβ = Y where x y x y.. X =.. and Y =... x n y n If the given points are not already on a line, then no exact solution to Xβ = Y exists, and we should instead try to find a leastsquares solution to this linear system. To be concrete, suppose we have four points (, ), (5, ), (7, ), and (8, ) so that X = 5 7 and Y =. 8 The leastsquares solutions to Xβ = Y are the exact solutions to X T Xβ = X T Y. We have [ [ X T X = = 4 8
2 MATH Linear algebra (Fall 07) Lecture and X T Y = [ = [ 9 57 The matrix X T X is invertible. (Why?) It follows that a leastsquares solution β R is provided by β = [ β0 = β [ 4 4 [ 9 57 = 84 Thus our line of best fit for the data is y = x: [ 4 4. [ 9 57 = [ /7 5/4. Symmetric matrices A matrix A is symmetric if A T = A. This happens if and only if A is square and A ij = A ji for all i, j. [ 0 0 a b c 0 Example. and 5 8 and b d e are symmetric matrices c e f [ 4 0 [ and 4 and are not symmetric. 0 5 Proposition. If A is a symmetric matrix and k is a positive integer then A k is also symmetric. Proof. If A = A T then (A k ) T = (AA A) T = A T A T A T = (A T ) k = A k. Proposition. If A is an invertible symmetric matrix then A is also symmetric. Proof. This is because (A ) T = (A T ). Recall how we can diagonalise a matrix. Example. Let A =. 5 Then det(a xi) = (8 x)( x)( x) so the eigenvalues of A are 8,, and. By constructing bases for the null spaces of A 8I, A 5I, and A I, we find that the following are eigenvectors of A:
3 MATH Linear algebra (Fall 07) Lecture v = v = v = 0 with eigenvalue 8. with eigenvalue. with eigenvalue. These eigenvectors are actually an orthogonal basis for R. Converting these vectors to unit vectors gives an orthonormal basis of eigenvectors: / u = / /, u = / / 0 /, u = / /. We then have A = P DP where P = [ u u u and D = (Do you remember why this holds? It is enough to check that P DP v = Av for v {u, u, u }.) Since the columns of P are orthonormal, we actually have P T = P so A = P DP T. The special properties in this example will turn out to hold for all symmetric matrices. Theorem. Suppose A is a symmetric matrix. Then any two eigenvectors from different eigenspaces of A are orthogonal. In other words, if A is n n and u, v R n are such that Au = au and Av = bv for numbers a, b R with a b, then u v = 0. Proof. Let u and v be eigenvectors of A with eigenvalues a and b, where a b. Then au v = Au v = (Au) T v = u T A T v = u T Av = u Av = u bv. But au v = a(u v) and u bv = b(u v), so this means a(u v) = b(u v) and therefore Since a b 0, it follows that u v = 0. (a b)(u v) = 0. Definition. A matrix P is orthogonal if P is invertible and P = P T. Definition. A matrix A is orthogonally diagonalisable if there is an orthogonal matrix P and a diagonal matrix D such that A = P DP = P DP T. When A is orthogonally diagonalisable and A = P DP = P DP T, the diagonal entries of D are the eigenvalues of A, and the columns of P are the corresponding eigenvectors; moreover, these eigenvectors form an orthonormal basis of R n. In fact, it follows by the arguments in our earlier lectures about diagonalisable matrices that an n n matrix A is orthogonally diagonalisable if and only if there is an orthogonal basis for R n consisting of eigenvectors for A. Surprisingly, there is a much more direct characterisation of orthogonally diagonalisable matrices:
4 MATH Linear algebra (Fall 07) Lecture Theorem. A square matrix is orthogonally diagonalisable if and only if it is symmetric. We prove this after a sequence of lemmas. Lemma. If A is orthogonally diagonalisable then A is symmetric. Proof. Note that if X, Y, Z are n n matrices the (XY Z) T = Z T (XY ) T = Z T Y T X T. Suppose A = P DP T where D is diagonal. Then D = D T and (P T ) T = P, so A T = (P DP T ) T = (P T ) T D T P T = P DP T = A. Lemma. All (complex) eigenvalues of an n n symmetric matrix A with real entries belong to R. Proof. Suppose A is a symmetric n n matrix with real entries, so that A = A T = A. Let v C n. Then v T Av is some complex number. [ [ + i For example, if A = and v = then i v T Av = [ i + i [ = [ + i i [ + i i In fact, the number v T Av belongs to R since [ + i i = ( + i)( + i) + ( i)( i) = 4. v T Av = v T Av = ( v T Av ) T = v T Av. (The last equality holds since both sides are matrices, i.e., scalars.) Now suppose v C n is an eigenvector for A with eigenvalue λ C. Then v T Av = v T (λv) = λ(v T v) R. The complex number v T v always belongs to R (why?) so it must also hold that λ R. Lemma. An n n matrix A with all real eigenvalues can be written as A = URU T where U is an n n orthogonal matrix (i.e., has orthonormal columns) and R is an n n uppertriangular matrix. One calls A = URU T with U and R of this form a Schur factorisation of A. Proof. Suppose A is an n n matrix with all real eigenvalues. Let u R n be a unit eigenvector for A with eigenvalue λ R. Let u,..., u n R n be any vectors such that u, u,..., u n is an orthonormal basis for R n. (One way to construct these vectors: let u = x, x,..., x n be any basis, apply GramSchmidt to get u = v, v,..., v n, then convert each v i to a unit vector.) Define U = [ u u... u n so that U T = U. By considering the product U T AUe i for i =,,..., n, one finds that U T AU has the form [ U T λ AU = 0 B for some (n ) (n ) matrix B. Here, stands for n arbitrary entries. 4
5 MATH Linear algebra (Fall 07) Lecture The matrix U T AU = U AU has the same characteristic polynomial as A. This polynomial is (λ x) det(b xi), i.e., λ x times the characteristic polynomial of B. Since the characteristic polynomial of A has all real roots, the same must be true of the characteristic polynomial of B. Thus B must also have all real eigenvalues. By repeating the argument above, we deduce that there is an eigenvalue µ R for B, an (n ) (n ) orthogonal matrix V, and an (n ) (n ) matrix C with all real eigenvalues such that [ V T µ BV =. 0 C [ 0 The matrix is also orthogonal, and the product of orthogonal matrices is orthogonal. (Why?) 0 V [ 0 It follows for the orthogonal matrix W = U that 0 V W T AW = λ 0 µ 0 0 C By continuing in this way, we will eventually construct an orthogonal matrix X and an uppertriangular matrix R such that X T AX = R, in which case A = XX T AXX T = XRX T. Now we can prove the theorem. Proof of theorem. The first lemma shows that if A is orthogonally diagonalisable then A is symmetric. Suppose conversely that A is symmetric. Then A has all real eigenvalues, so there exists a Schur factorisation A = URU T. We then have A T = (URU T ) T = UR T U T but also A T = A = URU T. Since U T = U, it follows that R = R T. Since R is uppertriangular, this can only hold if R is diagonal. But if R is diagonal then A = URU T is evidently orthogonally diagonalisable.. To orthogonally diagonalise an n n symmetric matrix A, we just need to find an orthogonal basis of eigenvectors v, v,..., v n for R n. Then A = UDU T with U = [ u u... u n where ui = v v i i and D is the diagonal matrix of the corresponding eigenvalues. If all eigenspaces of A are dimensional, then any basis of eigenvectors will be orthogonal. If A has an eigenspace of dimension greater than one, then after finding a basis for this eigenspace, it is necessary to apply the GramSchmidt process to covert this basis to one which is orthogonal. Corollary. If A = UDU T where U = [ u u... u n has orthonormal columns and λ λ D =... λn is diagonal, then A = λ u u T + λ u u T + + λ n u n u T n. (*) Each product u i u T i is an n n matrix of rank. One calls (*) a spectral decomposition of A. 5
6 MATH Linear algebra (Fall 07) Lecture Example. Let A = A = [ 7 4. A spectral decomposition of A is given by [ / 5 / 5 / 5 / 5 [ [ / 5 / 5 / 5 / 5 [ / 5 = 8 / 5 [ /5 /5 = /5 8/5 [ [ / 5 / 5 / 5 + / 5 [ +. /5 /5 /5 /5 [ / 5 / 5
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