Matrix transformations #4 - Questions and Answers

Question 1 (a) Find the transition matrix from 𝐵 B to 𝐵 ′ B ′ . (b) Find the transition matrix from 𝐵 ′ B ′ to 𝐵 B. (c) Verify that the two transition matrices are inverses of each other. (d) Solve the coordinate matrix [ 𝑥 ] 𝐵 [x] B ​ , given the coordinate matrix [ 𝑥 ] 𝐵 ′ [x] B ′ ​ . 𝐵 = { ( 1 , 3 ) , ( − 2 , − 2 ) } B={(1,3),(−2,−2)}, 𝐵 ′ = { ( − 12 , 0 ) , ( − 4 , 4 ) } B ′ ={(−12,0),(−4,4)}, [ 𝑥 ] 𝐵 ′ = [ − 1 3 ] . [x] B ′ ​ =[ −1 3 ​ ]. Answer (a) Start by creating the augmented matrix ( 𝐵 ′ 𝐵 ) (B ′ B): ( 𝐵 ′ 𝐵 ) = [ − 12 − 4 1 − 2 0 4 3 − 3 ] . (B ′ B)=[ −12 0 ​ −4 4 ​ 1 3 ​ −2 −3 ​ ]. Using Gauss-Jordan elimination, we obtain the above in the form ( 𝐼 𝑃 − 1 ) (IP −1 ): ( 𝐼 𝑃 − 1 ) = [ 1 0 − 1 / 3 1 / 3 0 1 3 / 4 − 1 / 2 ] . (IP −1 )=[ 1 0 ​ 0 1 ​ −1/3 3/4 ​ 1/3 −1/2 ​ ]. Then, the transition matrix from 𝐵 B to 𝐵 ′ B ′ is: 𝑃 − 1 = [ − 1 / 3 1 / 3 3 / 4 − 1 / 2 ] . P −1 =[ −1/3 3/4 ​ 1/3 −1/2 ​ ]. (b) First form the matrix ( 𝐵 𝐵 ′ ) (BB ′ ): ( 𝐵 𝐵 ′ ) = [ 1 − 2 − 12 − 4 3 − 2 0 4 ] . (BB ′ )=[ 1 3 ​ −2 −2 ​ −12 0 ​ −4 4 ​ ]. By Gauss-Jordan process, we get: ( 𝐼 𝑃 ) = [ 1 0 6 4 0 1 9 4 ] . (IP)=[ 1 0 ​ 0 1 ​ 6 9 ​ 4 4 ​ ]. The transition matrix from 𝐵 ′ B ′ to 𝐵 B is then: 𝑃 = [ 6 4 9 4 ] . P=[ 6 9 ​ 4 4 ​ ]. (c) Calculating 𝑃 − 1 𝑃 P −1 P and 𝑃 𝑃 − 1 PP −1 , we obtain: 𝑃 − 1 𝑃 = [ − 1 / 3 1 / 3 3 / 4 − 1 / 2 ] [ 6 4 9 4 ] = [ 1 0 0 1 ] . P −1 P=[ −1/3 3/4 ​ 1/3 −1/2 ​ ][ 6 9 ​ 4 4 ​ ]=[ 1 0 ​ 0 1 ​ ]. 𝑃 𝑃 − 1 = [ 6 4 9 4 ] [ − 1 / 3 1 / 3 3 / 4 − 1 / 2 ] = [ 1 0 0 1 ] . PP −1 =[ 6 9 ​ 4 4 ​ ][ −1/3 3/4 ​ 1/3 −1/2 ​ ]=[ 1 0 ​ 0 1 ​ ]. Thus, the matrices are inverses of each other. (d) Given the basis 𝐵 B and the coordinates [ 𝑥 ] 𝐵 ′ [x] B ′ ​ : [ 𝑥 ] 𝐵 = 𝑃 [ 𝑥 ] 𝐵 ′ = [ 6 4 9 4 ] [ − 1 3 ] = [ 6 3 ] . [x] B ​ =P[x] B ′ ​ =[ 6 9 ​ 4 4 ​ ][ −1 3 ​ ]=[ 6 3 ​ ]. Question 2 Find the change-of-coordinates matrix from 𝐵 B to the standard basis in 𝑅 3 R 3 : 𝐵 = { [ 3 − 1 4 ] , [ 2 0 − 5 ] , [ 8 − 2 7 ] } . B= ⎩ ⎨ ⎧ ​ ​ 3 −1 4 ​ ​ , ​ 2 0 −5 ​ ​ , ​ 8 −2 7 ​ ​ ⎭ ⎬ ⎫ ​ . Answer 1 The change-of-coordinates matrix from 𝐵 B to the standard basis is: 𝑃 𝐵 = [ 𝑏 1   𝑏 2   𝑏 3 ] = [ 3 2 8 − 1 0 − 2 4 − 5 7 ] . P B ​ =[b 1 ​ b 2 ​ b 3 ​ ]= ​ 3 −1 4 ​ 2 0 −5 ​ 8 −2 7 ​ ​ . Question 3 Construct a matrix whose column space contains ( 1 , 1 , 5 ) (1,1,5) and ( 0 , 3 , 1 ) (0,3,1) and whose nullspace contains ( 1 , 1 , 2 ) (1,1,2). Answer The matrix 𝐴 A has columns: 𝐴 = [ 1 0 𝑎 13 1 3 𝑎 23 5 1 𝑎 33 ] . A= ​ 1 1 5 ​ 0 3 1 ​ a 13 ​ a 23 ​ a 33 ​ ​ ​ . The nullspace condition gives: [ 1 0 𝑎 13 1 3 𝑎 23 5 1 𝑎 33 ] [ 1 1 2 ] = 0. ​ 1 1 5 ​ 0 3 1 ​ a 13 ​ a 23 ​ a 33 ​ ​ ​ ​ 1 1 2 ​ ​ =0. Solving: 𝑎 13 = − 1 2 ,   𝑎 23 = − 2 ,   𝑎 33 = − 3. a 13 ​ =− 2 1 ​ ,a 23 ​ =−2,a 33 ​ =−3. Thus: 𝐴 = [ 1 0 − 1 2 1 3 − 2 5 1 − 3 ] . A= ​ 1 1 5 ​ 0 3 1 ​ − 2 1 ​ −2 −3 ​ ​ . Question 4 Find the characteristic equation, eigenvalues, and eigenvectors of the matrix: 𝐴 = [ 2 − 2 3 0 3 − 2 0 − 1 2 ] . A= ​ 2 0 0 ​ −2 3 −1 ​ 3 −2 2 ​ ​ . Answer a) The characteristic equation: 𝜆 3 − 7 𝜆 2 + 14 𝜆 − 8 = 0. λ 3 −7λ 2 +14λ−8=0. b) Eigenvalues: 𝜆 1 = 1 ,   𝜆 2 = 2 ,   𝜆 3 = 4. λ 1 ​ =1,λ 2 ​ =2,λ 3 ​ =4. c) Eigenvectors: For 𝜆 = 1 λ=1: Solve ( 𝐴 − 𝜆 𝐼 ) 𝑥 = 0 (A−λI)x=0. For 𝜆 = 2 λ=2: Solve ( 𝐴 − 𝜆 𝐼 ) 𝑥 = 0 (A−λI)x=0. For 𝜆 = 4 λ=4: Solve ( 𝐴 − 𝜆 𝐼 ) 𝑥 = 0 (A−λI)x=0. Question 5 Is a matrix multiplied with its transpose something special? Answer Yes, 𝐴 𝑇 𝐴 A T A is symmetric and positive semi-definite. It has useful properties in covariance computations and spectral theorems. If 𝐴 A is a column matrix, it has rank 1 with one nonzero eigenvalue.