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(c)2015 UM Math Dept licensed under a Creative Commons By-NC-SA 4.0 International License. Math 217: §2.3 Composition of Linear Transformations Professor Karen Smith1 Inquiry: Is the composition of linear transformations a linear transformation? If so, what is its matrix? T S A. Let R2 −→ R3 and R3 −→ R2 be two linear transformations. 1. Prove that the composition S ◦ T is a linear transformation (using the definition!). What is its source vector space? What is its target vector space? Solution note: The source of S ◦T is R2 and the target is also R2 . The proof that S ◦T is linear: We need to check that S ◦ T respect addition and also scalar multiplication. First, note for any ~x, ~y ∈ R2 , we have S ◦ T (~x + ~y ) = S(T (~x + ~y )) = S(T (~x)) + S(T (~y )) = S ◦ T (~x) + S ◦ T (~y ). Here, the second and third equal signs come from the linearity of T and S, respectively. Next, note that for any ~x ∈ R2 and any scalar k, we have S ◦ T (k~x) = S(T (k~x)) = S(kT (~x)) = kS(T (~x)) = kS ◦ T (~x), so S ◦ T also respects scalar multiplication. The second and third equal signs again are justified by the linearity of T and S, respectively. So S ◦ T respects both addition and scalar multiplication, so it is linear. 1 2 1 0 1 . 2. Suppose that the matrix of T is A = 0 −1 and the matrix of S is B = 0 −1 0 −1 3 x Compute explicitly a formula for S(T ( )). y x + 2y x x + 2y + (−x + 3y) 5y Solution note: S ◦ T ( ) = S( −y ) = = y y y −x + 3y 3. What is S ◦ T (~e1 )? S ◦ T (~e2 )? 0 5 Solution note: S ◦ T (~e1 ) = and S ◦ T (~e2 ) = . 0 1 4. Find the matrix of the composition S ◦ T . Compare to (2). 0 5 Solution note: . Note that the columns are the vectors we computed in (2). 0 1 1 Thanks to Anthony Zheng, Section 5 Winter 2016, for finding numerous errors in the solutions. 5. Compute the matrix product BA. Compare to (4). What do you notice? 0 5 Solution note: BA = . The same! 0 1 6. What about T ◦ S? What is its matrix in terms of A and B. Solution note: This is AB. 7. What is the general principle here? Say we have a composition of linear transformations T T A B Rn −→ Rm −→ Rp given by matrix multiplication by matrices A and B respectively. State and prove a precise theorem about the matrix of the composition. Be very careful about the order of multiplication! T T B A Rp are linear transformations given Rm −→ Solution note: Theorem: If Rn −→ by matrix multiplication by matrices A and B (on the left) respectively, then the composition TB ◦ TA has matrix BA. Proof: For any ~x ∈ Rn , we have TB ◦ TA (~x) = TB (TA (~x)) = TB (A~x) = BA~x = (BA)~x. Here, every equality uses a definition or basic property of matrix multiplication (the first is definition of composition, the second is definition of TA , the third is definition of TB , the fourth is the association property of matrix multiplication). 1 c 1 a . , and C = B. Let A = d 1 0 1 1. Compute AC. Compute CA. What do you notice? 2. Does matrix multiplication satisfy the commutative law? 3. TRUE or FALSE: If we have two linear transformations, S and T , both from Rn → Rn , then S ◦ T = T ◦ S. ad + 1 a + c 1 a+c Solution note: AC = , CA = . These are not equal in d 1 d ad + 1 general, so matrix multiplication does not satisfy the commutative law! In particular, linear transformations do not satisfy the commutative law either, so (3) is FALSE. 1 1 An explicit countexample is to let S be left multiplication by , and T be mul0 1 1 0 1 1 tiplication by . Then T S is multiplication by , but ST is multiplication 1 1 1 2 2 1 by . These are not the same maps, since for example, they take different values 1 1 on ~e1 . T C. The identity transformation is the map Rn −→ Rn doing nothing: it sends every vector ~x to ~x. A linear transformation T is invertible if there exists a linear transformation S such that T ◦ S is the identity map (on the source of S) and S ◦ T is the identity map (on the source of T ). 1. What is the matrix of the identity transformation? Prove it! T 2. If Rn −→ Rm is invertible, what can we say about its matrix? Solution note: The matrix of the identity transformation is In . To prove it, note that the identity transformation takes ~ei to ~ei , and that these are the columns of the identity matrix. So the identity matrix is the unique matrix of the identity map. If T is invertible, then the matrix of T is invertible. Math 217: §2.3 Block Multiplication Professor Karen Smith D. In the book’s Theorem 2.3.9, we see that we can think about matrices in “blocks” (for example, a 4 × 4 matrix may be thought of as being composed of four 2 × 2 blocks), and then we can multiply as though the blocks were scalars using Theorem 2.3.4. This is a surprisingly useful result! 1 1 1 1 0 1 0 1 C11 C12 . If we write this as a block matrix, C = , 1. Consider the matrix C = C21 C22 0 0 1 0 0 0 0 1 where all the blocks are the same size, what are the blocks Cij ? 1 1 , C21 = 02×2 (the 2 × 2 zero matrix), Solution note: One way is: C11 = C12 = 0 1 and C22 = I2 (the 2 × 2 identity matrix). D1 2. Suppose we want to calculate the product CD, where D is the block matrix D = , with D2 D1 and D2 each being a 2 × 2 block. Write the product in terms of Cij and Dk by multiplying the blocks as if they were scalars, as suggested by Theorem 2.3.9. Solution note: We have C11 C12 CD = C21 C22 D1 C11 D1 + C12 D2 = . D2 C21 D1 + C21 D2 Calculate 3. Compute the product 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 A= 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 AB where 0 1 0 0 a p x 0 b 0 0 1 0 q y 0 1 0 0 1 r z 0 c 1 0 1 0 0 2 3 1 0 0 1 0 and B = 1 2 3 0 1 2 3 0 1 0 0 1 −a −p −x 0 0 1 0 0 −b −q −y 0 0 0 1 0 −c −r −z 0 1 0 0 1 [Hint: Be clever, take advantage of lurking identity matrices and block multiplication.] 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 . 1 0 0 1 Solution note: Break this up, sudoku-like, into nine 3 × 3 blocks. Note, in A, almost all these are identity matrices or zero matrices, so the multiplication is especially easy. 4. With C as in (1), find another way to break C up as = C11 C12 , but where the blocks are C21 C22 not the same size. Solution note: You could C11 to be1 × 1 and take C22 to be 3 × 3. So C11 = 1, 0 1 0 1 C12 = 1 1 1 , C21 = 0 , and C22 = 0 1 0. It’s easiest to see what I mean 0 0 0 1 by drawing in two lines cutting the matrix up into blocks. A B F. Consider a matrix M = where A is 3 × 5 and D is 7 × 1 (so M is in block form). C D 1. What are the sizes of B and C? What is the size of M ? Solution note: M is 10 × 6, B is 3 × 1 and C is 7 × 5. A0 B 0 ? What are the possible dimensions of N so that the product M N is C 0 D0 defined? In this case, what are the dimensions of the smaller blocks A0 , B 0 , C 0 and D0 so that the product can be computed using a block-product? 2. Suppose N = Solution note: The only restrictions are as follows: A0 and B 0 must have 5 rows, and C 0 and D0 must have one row. Also, A0 and C 0 must have the same number of columns, as must B 0 and D0 . So A0 is 5 × n, B 0 is 5 × m, C 0 is 1 × n and D0 is 1 × m. 3. What are the possible dimensions of N so that the product N M is defined? In this case, what are the dimensions of the smaller blocks A0 , B 0 , C 0 and D0 that would allow us to compute this as a block product? Solution note: For this, the number of columns of N must equal the number of rows of M , so N must by p × 10. For the block multiplication to work, we must have A0 is a × 3, B 0 is a × 7, C 0 is b × 3 and D0 is b × 7. Here a + b = p. 4. If A is m × d and B is d × n, how can we think of the product AB as a column (of row vectors) times a row (of column vectors)? R1 R2 Solution note: Write A as a column of row vectors: A = . where each Ri is a .. Rn C1 C2 1 × m matrix (row vector). Write B as a row of columns vectors: B = . . . Cm where each Cj is a n × 1 matrix (columns vector). Then R1 R2 AB = . C1 C2 . . . Cm , . . Rn which is the n × m matrix with Ri · Cj in the ij-spot.