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Section 3 Summary Section 3 summary Brian Krummel March 16, 2020 1 Terminology Transformation Domain Codomain or target space Image under a transformation Matrix transformation Linear transformation Standard matrix of a linear transformation Matrix multiplication Row-column rule Partitioned matrix Invertible matrix Nonsingular matrix Singular matrix Inverse matrix Invertible transformation Inverse transformation LU-factorization PLU-factorization Type-2 matrix Unipotent matrix 1 Permutation matrix Coordinate vector Point matrix Coordinate matrix Point transformation Coordinate transformation Linear isomorphism Isomorphic Matrix of a linear transformation with respect to bases on the domain and codomain 2 Theorems about linear transformations n m Theorem 1. Let TA : R ! R be the matrix transformation TA(X) = AX given by multiplication by an m × n matrix A. Then the image of a line segment under TA is either a point or a line segment. Theorem 2. Let T : V!W be any linear transformation. • T (0) = 0). • T (c1X1 + c2X2 + ··· + ckXk) = c1 T (X1) + c2 T (X2) + ··· + ck T (Xk) for all vectors Xi 2 V and all scalars ci 2 R. Theorem 3 (Matrix Representation Theorem). Let T : Rn ! Rm be a linear transformation. There exists a unique m × n matrix A such that T (X) = AX for all X 2 Rn. In particular, A = T (E1) T (E2) ··· T (En) n where Ej 2 R is the coordinate vector with j-th entry 1 and all other entries zero. (Equivalently, Ej is the j-th column of the n × n identity matrix I.) n m Theorem 4. Let TA : R ! R be the matrix transformation TA(Y ) = AY given by multiplication by an m × n matrix A. Let B be a basis for Rn and D be a basis for Rm. Then there exists a unique m × n matrix M of TA with respect to the bases B and D such that CDTA(Y ) = CDAY = MCBY for all Y 2 Rn; that is, the D-coordinates of AY are equal to M times the B-coordinate vector X of Y . In particular, −1 M = PD APB or equivalently −1 A = PDMPB : 2 Theorem 5. Let L : V!W be a linear transformation. Let B = fB1;B2;:::;Bng be a basis for V and D be a basis for W. Then there exists a unique matrix M of T with respect to the bases B and D such that C C TD (L(Y )) = MTB (Y ) for all Y 2 V; that is, the D-coordinates of L(Y ) are equal to M times the B-coordinate vector X of Y . In particular, C C C M = TD (L(B1)) TD (L(B2)) ··· TD (L(Bn)) so that the j-th column of M is the D-coordinates of L(Bj) and the (i; j)-entry of M is the Di- coefficient of L(Bj). 3 Properties of matrix multiplication Theorem 6. Let A; B; C be matrices of appropriate sizes and I denote the identity matrix of appropriate sizes. • A(BC) = (AB)C • A(B + C) = AB + AC • (A + B)C = AC + BC • r(AB) = (rA)B = A(rB) for every scalar r 2 R • IA = AI = A • (AB)T = BT AT Warnings! Let A; B; C be matrices of appropriate sizes. In general, • AB 6= BA • AB = AC does not imply B = C • AC = BC does not imply A = B • AB = 0 does not imply B = 0 or C = 0. Theorem 7. Let A; B be matrices of appropriate sizes. Then rank(AB) ≤ rank A and rank(AB) ≤ rank B. 4 Invertible matrices Definition 1. A is an invertible n × n matrix if there exists a matrix A−1 such that AA−1 = I andA−1A = I: Whenever A−1 exists, it is also unique. A singular n×n matrix is a matrix which is not invertible. 3 Algorithm for computing A−1. Apply Gaussian elimination A I −! I A−1 . Theorem 8. If A is an invertible n × n matrix, then for every B 2 Rn the equation AX = B has exactly one solution X = A−1B. Theorem 9. Consider the matrix transformation TA(X) = AX where A is an n×n matrix. TA is −1 −1 an invertible transformation if and only if A is an invertible matrix, in which case TA (X) = A X for all X 2 Rn. Theorem 10. • If A is an invertible n×n matrix, then A−1 is also an invertible n×n matrix and (A−1)−1 = A. • If A and B are invertible n × n matrices, then AB is also an invertible n × n matrix and (AB)−1 = B−1A−1. • If A is an invertible n × n matrix, then AT is also an invertible n × n matrix and (AT )−1 = (A−1)T . Theorem 11 (Invertible Matrix Theorem). Let A be an n × n square matrix. The following are logically equivalent: o (a) A is an invertible matrix. Main statement 9 (b) rank A = n. > => (c) A has n pivot positions. Echelon form > (d) A is row equivalent to the n × n identity matrix. ;> 9 (e) The homogeneous equation AX = 0 has only the trivial solution. > => (f) The columns of A form a linearly independent set. Uniqueness > (g) Nul A = f0g. ;> 9 n (h) The equation AX = B has at least one solution for each B 2 R . > => (i) The columns of A span Rn. Existence > (j) Col A = Rn. ;> 9 n (k) The equation AX = B has exactly one solution for each B 2 R . > => (l) The columns of A are a basis for Rn. Exists and unique > > (m) The matrix transformation TA(X) = AX is invertible. ; 9 (n) There is an n × n matrix C such that CA = I. = Definition of inverse (o) There is an n × n matrix D such that AD = I. ; 9 (p) AT is invertible. = Transpose (q) Row A = Rn. ; 4 5 LU-Factorization Theorem 12. Let A be a type-2 m × n matrix. Suppose that A I is row equivalent to U V for some m × n matrix U and m × m matrix V : A I −! U V : Then V is invertible and A = V −1U. Theorem 13. Let A be a type-2 m × n matrix. Then there exists an m × m lower trangular, unipotent matrix L (i.e. L is lower triangular and has entries one on the main diagonal) and there exists an m × n matrix U in echelon form such that A = LU: Algorithm for computing an LU-factorization. Assuming A is a type-2 n × n matrix, apply Gaussian elimination A I −! U V : At the same time, write down a lower-triangular n × n matrix L with diagonal entries 1. For each row replacement Ri − c Rj 7! Ri, which changes the (i; j)-entry of A to a zero, set the (i; j)-entry of L equal to +c (with the opposite sign). To solve AX = B, we can find the LU-factorization A = LU (if it exists). Then we use back-substitution to solve for Z such that LZ = B. Then we again back-substitution to solve for X such that UX = Z. Theorem 14. Let A be any non-type-2 m × n matrix. Then there exists an m × n matrix A0 with the same rows as A in a different order such that A0 is a type-2 matrix. Theorem 15. Let A be any m × n matrix. Then there exists an m × m permutation matrix P , m × m lower trangular, unipotent matrix L, and there exists an m × n matrix U in echelon form such that A = P LU: 5.
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