fall2010T2 - Do lo wn ad er ID 10 20 Te st ID 32 50 2 0 325...

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Unformatted text preview: Do lo wn ad er ID : 10 20 Te st ID : 32 50 2 0 325 t ID: Tes 02 University of Toronto Department of Mathematics 102 Dow r ID: 02 Do wn lo ad er ID : 102 20 2 r ID: 10 ade ade MAT223H1F Linear Algebra I Downloader ID: 10202 Dow nlo nlo 3250 ID: r ID: Test ade nlo Dow Midterm Examination II November 25, 2010 02 102 H. Kim, F. Murnaghan, B. Rowe, S. Uppal 50 Duration: 1 hour 20 minutes wn lo ad er ID : 10 20 Te st ID : 32 50 32 2 20 2 st ID : Last Name: 2 20 : 10 ID er lo ad wn Do ad e r Student Number: ID : 10 Given Name: 10202 ader ID: Downlo Do wn lo Tutorial Code: 50 10202 Downloader ID: Te st 32 50 32 ID : ID 0 Te 20 2 st 325 : No calculators or other aids are allowed. Tes t ID: : ID Mark 50 /5 /50 ID: 3250 Oxdia @ http://www.oxdia.com Downloader ID: 10202 1 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. nl oa /5 Test Do w 5 TOTAL 32 /10 6 : /10 4 ID /10 3 st /10 2 Te 1 de r Question 10 FOR MARKER USE ONLY 32 50 : ID Te st ID : 32 50 st Do wn lo ad er ID : 10 20 2 Te 1 1 0 0 −1 1 1. Consider the subspace W = span , , 0 0 0 −1 0 1 Test ID: 3250 of R4 . 0 325 t ID: Tes Downloa der ID: 10202 (a) Use the Gram-Schmidt process to find an orthonormal basis for W . 1 0 (b) Find the orthogonal projection of the vector onto W . 0 1 Down load er ID: 1020 2 Test ID: 3250 1 1 0 0 −1 1 (a) Let v1 = , v2 = , and v3 = . 0 0 0 −1 0 1 Te ID : 32 50 st : ID 32 20 2 st 50 Dow 10 nlo r ID: ade 102 : 02 u1 ||u1 || st Te ad e r ID – Let u1 = v1 and let e1 = : ID Do Do wn lo 50 32 – Let 1 √ 0 = 22 . 0 −1 wn Dow lo nlo ade er ad r ID: 102 : ID 02 : 32 50 2 20 10 u1 , v2 u2 = v2 − u1 u1 , u1 1 1 −1 1 0 = − 0 2 0 0 −1 1 wn lo ad er ID : 10 20 2 Test ID: 3250 Downloader ID: 10202 2 Do 20 32 50 st ade r ID: 102 02 st Te 10 : ID de nl oa Dow nlo r 2 −1 = , 0 Test ID: 3250 : ID 1 2 50 32 Do w : Te ID ID Te st Downloader ID: 10202 1 √ −2 = 66 0 1 and let e2 = u2 ||u2 || Test ID: 3250 Test ID: 3250 Downloader ID: 10202 Te st ID : 32 50 Test Oxdia @ http://www.oxdia.com 2 of 9 This test is copyrighted by the uploader and/or course instructor. Test ID: 3250 Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. Test ID: 3250 ID: 3250 Test ID: 3250 : ID 2 20 10 er ad lo wn 50 Do 32 32 50 Dow Downloader ID: 10202 Te ID : ID : st st Te nlo ade r ID: 102 – Let 02 20 ID ad : u2 , v3 u1 , v3 u3 = v3 − u2 − u1 u2 , u2 u1 , u1 2 2 1 = , 3 0 2 2 10 er lo Do wn 3250 Test ID: ad er 10 20 2 : ID : 2 lo 20 wn 10 Do Do wn lo ad er ID Test ID: 3250 Downloa 10202 der ID: 2 20 : 10 ID er ad lo wn Do Te st 50 1 √ 1 = 33 . 0 1 u3 ||u3 || 50 and let e3 = Dow nlo ade r ID: 102 02 20 2 32 Dow nlo ade 10 : r ID: : ID 3250 Test ID: ID 102 32 Do wn lo ad er 02 10 20 2 st ID : Then {e1 , e2 , e3 } form an orthonormal basis of W . 1 0 (b) Let v = . Then 0 1 Do wn lo : ad er ID : 10 20 Test ID: 3250 t ID: 0 325 projW (v ) = e1 , v e1 + e2 , v e2 + e3 , v e3 √ √ 23 6 e2 + e3 = 0e1 + 3 3 1 0 = 0 1 Te 32 50 Do wn Tes lo ad e r ID 2 st 32 50 ID : 10 Te 20 2 st ID : ID : Do w nl oa de r Remark: Part (b) shows that the vector v is actually in W . er : ID Test 2 20 10 ad lo wn Do : ID Te Test ID: 3250 Oxdia @ http://www.oxdia.com 3 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Downloa der ID: 10202 oader Downl ID: 10202 Solution (if any) is NOT audited, so use at your discretion. st 32 50 ID: 3250 Do wn lo ad er ID : 10 20 2 2. Find a basis for the orthogonal complement 1 −1 A = 1 −4 21 : 3250 : ID 20 10 2 Downloader ID: 10202 ad Do Do wn lo wn lo ad er ID : 10 20 2 Dow nlo ade r ID: 102 02 er ad lo wn Do of the nullspace of 32 7 8 . 2 −2 er ID : 10 20 2 The orthogonal complement of the nullspace is always the row space, so we have to find a basis of the rowspace of A. To do this, reduce A to row-echelon form using row reduction: 1 −1 3 2 R3 +R 0 −3 4 6 −→2 0 3 −4 −6 1 −1 3 2 0 −3 4 6 0 0 00 wn Do ID : er ad lo 1 0 −1 −3 Hence, { , } forms a basis for the orthogonal complement of the nullspace of 3 4 2 6 ID Dow nlo ade : r ID: 10 2 20 st 32 20 2 10 : A. r ad e lo : 50 32 50 32 : ID st Te Do wn lo ad er 10 20 2 : 2 20 ID ID : 50 32 10 st Te Te Test ID: 3250 st ID ID : : 32 50 ID Do st Te wn 02 0 102 325 r ID: t ID: ade Tes Dow nlo 102 02 50 : ID st Te ID Do wn lo ad er ID : 10 de r 20 2 50 : ID st 32 Do w Te nl oa Test ID: 3250 32 50 1 −1 3 2 1 ,R3 1 −4 7 8 R2 −R−→−2R1 2 1 2 −2 Do wn lo ad er ID : 10 20 2 Oxdia @ http://www.oxdia.com 4 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Downl oader ID: 10202 Solution (if any) is NOT audited, so use at your discretion. 32 50 ID : 102 02 Te st ade r ID: 2 20 10 : ID nlo er Dow lo ad Do wn 1 2 3. Let T : R2 → R2 be a linear transformation such that T T Since Test ID: 3250 3 4 −1 . Find T 1 = x y 7 , and 3 = x ∈ R2 . y for all 1 3 1 = −2 , it follows that 0 4 2 50 32 : ID st Te t ID: Tes 0 325 1 0 T 3 1 −2 4 2 =T =T 3 4 − 2T 1 2 2 er ID: 1020 load Down −1 7 −15 −2 = . 1 3 −5 st Te = : ID 50 32 Do wn 50 lo ad er ID : 10 20 32 2 0 1 13 3 31 − =T 22 24 2 37 1 −1 11 = − = . 3 1 4 2 2 325 0 10 t ID: Tes =T Te r ID T : er ID: 10202 Download , we have 20 2 st ID : 0 1 3 Similarly, since = 1/2 3 − 1 2 4 st ad e 3 4 1 0 + yT 0 1 wn 50 32 lo 1 −T 2 : ID 1 2 Test ID: 3250 Do Hence, for any x and y , Do wn Downl ID: 10202 er ad lo oader ID : 10 2 20 Do wn lo ad er ID : 10 20 2 x 1 0 +y 0 1 50 =T = xT 32 x y 20 2 st −15 11 −15x + 11y +y = −7 4 −5x + 4y Do w Te st ID : 32 50 2 20 : er ad lo wn Do ID: Test Oxdia @ http://www.oxdia.com 5 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader Downloader id is shown and also encrypted throughout the document. ID: 10202 Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. 3250 ID 10 20 2 10 : ID er ad wn lo Do nl oa de r ID : 10 Te =x ID : T 325 0 Tes t ID: 4. Let T : R3 → R3 be the linear transformation whose standard matrix is 1 34 A = 3 4 7 . −2 2 0 Test ID: 3250 50 : ID 32 st Te wn Do ad lo er ID : 10 20 2 Dow nlo ade r ID: 102 02 325 0 t ID: Tes 50 ID : 32 st 50 Te 32 10202 der ID: ID : Downloa Te st (a) Show that the kernel of T is a line through the origin and find parametric equations of the line. (b) Show that the range of T is a plane through the origin and find an equation for the plane. Do (a) Reduce A to row-echelon form using row operations. wn lo ad er ID : 10 20 2 r ID: nlo 102 02 ade Dow 13 4 134 1 −5),R 3 −R 0 −5 −5 R2 /(−→ 3 /8 0 1 1 R−→2 0 08 8 011 0 134 x Hence, Ax = 0 if and only if 0 1 1 y = 0. Setting z = t, we have y 000 z x = −3(−t) − 4t = −t, so the kernel of A is the set of vectors of the form −t −t , t 1 34 3 4 7 R2 −3R1 ,R3 +2R1 −→ −2 2 0 34 1 1 00 : 32 = −t and Do 32 50 Do wn lo ad e r ID : 10 20 2 50 ID : load Down ID st 1020 st Te 32 50 2 er ID: wn lo ad er which is a line through the origin with parametric equations ID : 10 ID 2 : 20 Downloader ID: 10202 02 102 10 r ID: ade nlo Dow Te 20 2 st x = −t, y = −t, z = t. st : ID 50 32 Do w Te nl oa de r ID : (b) Since the reduced row-echelon form of A has leading ones in the first and second column, 1 3 { 3 , 4} −2 2 st Te forms a basis of the column space of A, which is the range of T . Since the range is two dimensional, it is a plane through the origin. To find an equation describing im(T ), we a need to find the normal vector n = b . Since n is normal to both elements, it must c be that a + 3c − 2b = 0 and 3a + 4b + 2c = 0. One particular solution to this is ID : 50 32 Do wn lo ad er ID : 10 20 2 Te a = −14, b = 8, c = 5, st : ID 50 32 so the image is is the plane given by the equation −14x + 8y + 5z = 0. Oxdia @ http://www.oxdia.com Downloa der ID: 10202 6 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Do wn lo ad er ID : 10 20 2 Unauthorized reproduction/distribution is strictly prohibited. er ID: 10202 Download Solution (if any) is NOT audited, so use at your discretion. Test ID: 3250 ID: 3250 Test 32 50 2 ID : er Test ID: 3250 Te st ad lo 20 10 EXTRA PAGE FOR QUESTION 4 - do not remove. : ID wn Do er ad lo wn Do : ID 2 20 10 : ID 2 20 : wn Do 10 lo ad 20 10 2 er ID er ad lo wn Do Test ID: 3250 2 20 10 : ID er ad lo wn Do Downloader ID: 10202 Te st ID : 32 50 Downloader ID: 10202 50 Downl Test st ID: ID : 3250 32 50 ID : 32 Te Dow nlo ade r ID: 102 02 ad e r ID : 10 20 2 st Test ID: 3250 Do wn lo ad er lo ID : 10 20 2 wn lo st ID 20 Do ad er ID : 10 20 2 : 10 Te Te st ID : 32 50 Do wn 2 50 32 10202 der ID: Downloader ID: 10202 Do wn lo ad er ID : Te ID : 32 50 t ID: Tes 0 325 Test ID: 3250 wn Do ad lo er ID : 10 2 20 02 102 r ID: ade nlo Oxdia @ http://www.oxdia.com 7 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. 10 20 2 Do w st nl oa de r ID : Downloa Dow oader ID: 10202 Downloader ID: 1 Downloa der 0 325 t ID: Tes 20 : 2 10 ID er lo 50 : 32 Do ad wn ID st Te 50 5. Let V be an inner product space. If x, and y are vectors in V , show that Te st ID : 32 50 Te st ID : 32 Downl Te st ID : 32 50 1 1 < u, v >= ||u + v ||2 − ||u − v ||2 . 4 4 0 325 Tes t ID: Test ID: 3250 Recall that, for the inner product u, v , we have, for any c ∈ R and for any vectors u, v, w: ID: 3250 Test • u, v = v , u ; • u, u = ||u||2 ; Do wn lo ad er ID : 10 20 2 0 325 t ID: Tes • u, v + w = u, v + u, w ; Tes t ID: 325 0 32 50 • cu, v = c u, v = u, cv . 2 20 : 10 ID er ad lo wn Do ID : Using these facts, we see that Te st ID : 20 2 st 32 50 : 20 10 2 ID Te er ad st lo wn Do : ID 10 1 1 1 1 ||u + v ||2 − ||u − v ||2 = u + v, u + v − u − v, u − v 4 4 4 4 1 1 1 1 = u + v, u + u + v, v − u − v, u − u − v, v 4 4 4 4 1 1 1 1 = u + v, u + u + v, v − u − v, u + u − v, v 4 4 4 4 1 1 1 1 = u, u + v , u + u, v + v , v 4 4 4 4 1 1 1 1 − u, u − v , u + u, v − v , v 4 4 4 4 = u, v . wn lo ad e r ID : 50 32 0 Do Tes Te t ID: 325 st : 32 50 oader 32 50 ID ID: 10202 load er ID: 1020 2 2 : ID 20 10 er ad lo wn Do Dow nlo ade r ID: 102 02 Te st ID : 32 50 Oxdia @ http://www.oxdia.com 8 of 9 This test is copyrighted by the uploader and/or course instructor. Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. Do w Test ID: 3250 nl oa de r ID : 10 Te 20 Down 2 st ID : Downl 50 : ID st Te 32 oader ID: 10202 r ID: 102 02 Test ID: 3250 Dow nlo ade Downloader ID: 10202 Do wn lo ad er 10 20 2 lo ad er ID : 10 20 2 ID : Do wn 6. Suppose an m × n matrix A has a nullspace of dimension d. What is the dimension of the nullspace of AT . (Your answer should be in terms of m, n and d). 2 er ID: load Down Tes t ID: 1020 325 0 Since A is an m × n matrix, it corresponds to a linear transformation T : Rn → Rm . Hence, dim(imT ) + dim(ker T ) = n. Do wn lo ad er ID : 10 20 2 Since rank(A) = dim(colA) = dim(imT ), we have rank(A) + dim(nullA) = n. er ID: load Down 32 50 Hence, rank(A) = n − dim(nullA) = n − d. Since AT is an n × m matrix, it corresponds to a linear transformation S : Rm → Rn . Hence, ID : dim(imS ) + dim(ker S ) = m, st Test ID: rank(AT ) + dim(nullAT ) = m. Downloader ID: 10202 or 3250 10 20 2 Since rank(A) = rank(AT ), we have ID : dim(nullAT ) = m − rank(AT ) = m − rank(A) = m − (n − d) = m + d − n. 20 : 3250 Test ID: 2 10 ID ad er lo r wn Do wn lo ad e Do lo ad er ID : 10 32 2 20 50 wn Do nlo : Dow Do w nl oa de r ID : 10 Te 20 2 st 02 102 ID r ID: ade Do wn lo ad er ID : er ID: 10202 10 20 Download 2 ID: 3250 Test Oxdia @ http://www.oxdia.com 9 of 9 This test is copyrighted by the uploader and/or course instructor. Dow nlo ade r ID: 102 02 Downloader id is shown and also encrypted throughout the document. Unauthorized reproduction/distribution is strictly prohibited. Solution (if any) is NOT audited, so use at your discretion. 2 1020 ...
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