Math 191 Applied Linear Algebra Lecture 16: Change of Basis Stephen Billups University of Colorado at Denver Math 191Applied Linear Algebra p.1/0
Rank The rank of A is the dimension of the column space of A. rank A = dim Col A =# of pivot columns of A = dim Row A. rank }{{ A} + dim }{{ Nul A} = }{{} n # of pivot # of nonpivot # of columns columns columns of A of A of A Math 191Applied Linear Algebra p./0
THEOREM 14 THE RANK THEOREM The dimensions of the column space and the row space of an m n matrix A are equal. This common dimension, the rank of A, also equals the number of pivot positions in A and satisfies the equation rank A + dim Nul A = n. NOTE: Since Row A = Col A T, rank A = rank A T. Math 191Applied Linear Algebra p./0
EXAMPLE: Suppose that a 5 8 matrix A has rank 5. Find dim Nul A, dim Row A and rank A T. Is Col A = R 5? Solution: rank A {z } + dim Nul A {z } = n {z} 5? 8 5 + dim Nul A = 8 dim Nul A = dim Row A = rank A = rank A T = rank = Since rank A = # of pivots in A = 5, there is a pivot in every row. So the columns of A span IR 5 (by Theorem 4, page 4). Hence Col A = IR 5. Math 191Applied Linear Algebra p.4/0
EXAMPLE: For a 9 1 matrix A, find the smallest possible value of dim Nul A. Solution: rank A + dim Nul A = 1 dim Nul A = 1 rank A }{{} largest possible value= smallest possible value of dim Nul A = Math 191Applied Linear Algebra p.5/0
EXAMPLE: Let A = the following: Basis for Nul A = a plane in IR. Visualizing Row A and Nul A [ 1 0 1 0 0 1 0, 1 0 1 ]. One can easily verify and therefore Nul A is Math 191Applied Linear Algebra p.6/0
Basis for Row A = line in IR. Basis for Col A = in IR. Basis for Nul A T = line in IR. {[ {[ 1 1 0 1 ]} 1 and therefore Row A is a and therefore Col A is a line ]} and therefore Nul A T is a Math 191Applied Linear Algebra p.7/0
x x 4 x x 1-4 - 4 x 1 - - 4 Subspaces Nul A and Row A Subspaces Nul A T and Col A Math 191Applied Linear Algebra p.8/0
The Rank Theorem provides us with a powerful tool for determining information about a system of equations. EXAMPLE: A scientist solves a homogeneous system of 50 equations in 54 variables and finds that exactly 4 of the unknowns are free variables. Can the scientist be certain that any associated nonhomogeneous system (with the same coefficients) has a solution? Math 191Applied Linear Algebra p.9/0
Solution: Recall that rank A = dim Col A = # of pivot columns of A dim Nul A = # of free variables In this case Ax = 0 of where A is 50 54. By the rank theorem, rank A + = or rank A =. So any nonhomogeneous system Ax = b has a solution because there is a pivot in every row. Math 191Applied Linear Algebra p.10/0
THE INVERTIBLE MATRIX THEOREM (continued) Let A be a square n n matrix. The the following statements are equivalent: m. The columns of A form a basis for IR n n. Col A = IR n o. dim Col A = n p. rank A = n q. Nul A = {0} r. dim Nul A = 0 Math 191Applied Linear Algebra p.11/0
Section 4.7: Change of Basis In Section 4.4, we introduced coordinates relative to a basis B and showed how to convert between coordinates relative to B and coordinates relative to the standard basis. We now look at how to change coordinates between two nonstandard bases B and C. We begin by assuming that we know the coordinates of the basis vectors of B relative to the basis C. Then we will show how to do it when you only know the coordinates of the two bases relative to the standard basis. Math 191Applied Linear Algebra p.1/0
EXAMPLE Consider two bases B = {b 1, b } and C = {c 1, c } for a vector space V, and suppose that b 1 = 4c 1 + c and b = 6c 1 + c. Suppose that [x] B = 4 1 5. Find [x] C. Solution: x] C = [b 1 + b ] C = [b 1 ] C + [b ] C = 4 4 1 5 + 4 6 1 5 = 4 4 6 1 1 5 4 1 5 = 4 6 4 5 Math 191Applied Linear Algebra p.1/0
EXAMPLE Consider two bases B = {b 1, b } and C = {c 1, c } for a vector space V, and suppose that b 1 = 4c 1 + c and b = 6c 1 + c. Suppose that [x] B = 4 1 5. Find [x] C. Solution: x] C = [b 1 + b ] C = [b 1 ] C + [b ] C = 4 4 1 = 5 + 4 6 1 4 4 6 1 1 5 5 4 1 5 = 4 6 4 P C B [x] B [x] C 5 Math 191Applied Linear Algebra p.1/0
Graphical Illustration Coordinates relative to B. Coordinates relative to C. Math 191Applied Linear Algebra p.14/0
Theorem 15 Let B = {b 1,..., b n } and C = {c 1,..., c n } be two bases of a vector space V. Then, there is a unique matrix P such C B that [x] C = P [x]. C B The columns of P are the C-coordinate vectors of the the C B vectors in B. That is [ ] P = [b 1 ] C [b ] C [b n ] C C B Math 191Applied Linear Algebra p.15/0
Graphical Illustration Math 191Applied Linear Algebra p.16/0
What if we don t know [b i ] C? Let V be a dimensional vector space with standard basis E. Suppose [b 1 ] E = [b ] E = 4 5 1 C = {c 1, c }. 5, [c 1 ] E = 4 1 4 5, [c ] E = 4 4 5 5, and let B = {b 1, b }, and 4 9 1 5, Find the change-of-coordinates matrix from B to C. Solution: We first find the coordinates of b 1 and b relative to C, by solving h i [c 1 ] E [c ] E 4 x h i 1 5 = [b 1 ] E and [c 1 ] E [c ] E 4 x 1 x x 5 = [b ] E. Math 191Applied Linear Algebra p.17/0
cont. Since both of these equations involved the same matrix, we can solve them simultaneously by row-reducing an expanded augmented matrix as follows: h [c 1 ] E [c ] E [b 1 ] E [b ] E i = 4 1 9 5 4 5 1 1 4 1 0 6 4 5 0 1 5 5 Thus, [b 1 ] C = 4 6 5 5 and [b ] C = 4 4 5 and P C B = 4 5 6 4 5 Math 191Applied Linear Algebra p.18/0
Forming the Change of Basis Matrix Recap: To find [ P C B, row reduce the matrix [c 1 ] E [c n ] E [b 1 ] E [b ] E ] Resulting in the matrix [ I P C B ]. Math 191Applied Linear Algebra p.19/0
Example 1 Let b 1 = 4 5, b = 4 5, c 1 = 4 7 5, c = 4 9 coordinates matrix from B = {b 1, b } to C = {c 1, c }. Solution: 4 5 7 5. Find the change of h [c 1 ] E [c n ] E [b 1 ] E [b ] E i = 4 1 7 5 4 9 7 4 1 0 5 0 1 6 4 5. 5 So P C B = 4 5 6 4 5. Math 191Applied Linear Algebra p.0/0