Math 38 J - Spring 2 Final Exam - June 6 2 Name: Student ID no. : Signature: 2 2 2 3 2 4 2 5 2 6 2 28 otal his exam consists of problems on 9 pages including this cover sheet. Show all work for full credit. You may use a scientific calculator during this exam but it is by no means required. Graphing calculators and other electronic devices are not allowed. You may use 2 double-sided 8.5 by sheets of handwritten notes. You have minutes to complete this exam. You may write on the back of a page but indicate that you have done so. You may ask questions for clarification but there is no guarantee I will answer them. If you find parts of this exam difficult that is as intended.
.) (2 points) Alice ob and Charlie go to the candy store. he candy store sells packages X Y Z which have the following contents: Candy Reeses Snickers X Y 3 4 Z 2 Alice ob and Charlie each buy some number of packages X Y and Z. (a) (5 points) Alice opens her packages and counts her candy to find she got 58 Reeses and 4 Snickers. Describe the possible amounts of each package Alice bought. Let x y z be the number of packages of X Y Z Alice buys respectively. We solve [ ] [ ] [ ] [ ] 3 58 x + y + z = 4 2 4 which reduces to [ ] 2 6 so x = + 2z y = 6 z and z is free. (b) ( points) ob and Charlie got 58 Reeses and 4 Snickers as well. If Alice got 3 times as many of package X than ob who got twice as many of package Y as Charlie who got 2 less of packace Z than Alice how many of each package did Alice ob and Charlie get? Let z z 2 z 3 be the amount of package Z Alice ob and Charlie buy respectively. hen the amounts of packages X Y that each of them bought are given by x = + 2z y = 6 z for each. hen these 3 facts give the equations + 2z = 3( + 2z 2 ) 6 z 2 = 2(6 z 3 ) z 3 = z 2. Solving this system gives z = z 2 = z 3 = 8. Hence Person Alice ob Charlie X 3 26 Y 6 6 8 Z 8
2.) (2 points) Give the following examples if possible justifying that your answers satisfy the conditions. If it is not possible why not? (a) (4 points) A linearly dependent set of vectors where one of the vectors specify which one is NO in the span of the others. is linearly dependent and [ ] / span{ }. { [ ]} 3 (b) (4 points) A one-to-one linear map : R n R 3 for some n with range( ) = span 2 6. 3 9 has : R R 3 given by x (x ) = 2x 3x 3 range( ) = span 2 = span 2 6. 3 3 9 (c) (4 points) A non-diagonal (at least one of the nonzero entries is off the main diagonal) 2 2 matrix P which satisfies P 2 = P. Any projection map P : R 2 R 2 will satisfy P 2 = P. We just have {[ to ]} pick a projection so that [P ] is not diagonal such as the projection P onto S = span with respect to S 2 = span with {[ ]}. Since [ ] x + S and [ x ] = [ ] x + + [ ] x2 S 2 we have P ([ x ]) = [ ] x + [ x2 ] = [P ] = [ ]
3.) (2 points) Let : R 4 R 3 be the linear map with 3 5 [ ] = 2 2 2 4 8 (a) (6 points) Find a linear map U : R 3 R m for some m so that ker(u) = range( ). We find the relation form of range( ) by row reducing 3 5 w w 3 + w 3 2 2 2 w 2 2 2 4 w 3 w 3 2. 2 4 8 w 3 2w 3 + w 3 2 w 2 3 hus 2w 3 + w 3 2 w 2 3 or 4w 2w 2 + 3w 3 = is the relation defining range( ). So w range( ) = w 2 w 3 4w 2w 2 + 3w 3 = = ker(u) where U : R 3 R is given by w U w 2 = 4w 2w 2 + 3w 3 or [U] = [ 4 2 3 ]. w 3 (b) (3 points) Express ker( ) as the span of a finite set of vectors. Plugging in w = w 2 = w 3 = we have that ker( ) is the solutions to 2 4 which is given by x 3 x 4 are free x = x 3 x 4 = 2x 3 4x 4. Hence x x 3 x 4 x 3 = 2x 3 4x 4 x 3 = x 2 3 + x 4 4 = ker( ) = span 2 4. x 4 x 4 (c) (3 points) Find a basis for range( ). We take columns of [ ] corresponding to the pivot columns or its RREF or EF so 2 is a basis for range( ). 2
4.) (2 points) Consider the subspaces S S 2 S 3 of R 3 given by 3 x S = span 2 S 2 = span 4 S 3 = 5 x 3 x 4 + 3x 3 =. (a) (9 points) Express (S + S 2 ) S 3 as the span of a finite set of vectors. First 3 S + S 2 = span 2 4. 5 o express this in relation form we find 3 w 3 w 2 4 w 2 2w + w 2 5 w 3 2w + w 2 2w 3 So S + S 2 is defined by 2w + w 2 2w 3 = or 2w w 2 + 2w 3 =. o intersect this with S 3 we find when both relations hold simultaneously. [ ] [ ] 4 3 5 2 2 4 which means x 3 is free x = 5x 3 = 4x 3. hus 5 (S + S 2 ) S 3 = x 3 = span 4 5 4 5 = span 4 (b) (3 points) Express (S S 3 ) + (S 2 S 3 ) as the span of a finite set of vectors. c o have x S S 3 we need x = 2c and c 4(2c) + 3() = = c = = c = 3c which means x = so S S 3 = { }. Also to have x S 2 S 3 we need x = 4c 5c and 3c 4(4c) + 3(5c) = = 4c = = c = which means x = so S S 3 = { }. herefore (S S 3 ) + (S 2 S 3 ) =.
5.) (2 points) Let 3 4 A = 4 2. 5 (a) (4 points) Find the eigenvalues of A and specify their algebraic multiplicities. λ 3 4 det(λi A) = 4 λ + 2 λ λ + 5 = (λ + 5)(λ ) λ 3 4 4 λ + = (λ + 5)(λ )((λ 3)(λ + ) 6) = (λ + 5)(λ )((λ + 4λ 5) = (λ + 5) 2 (λ ) 2 so the eigenvalues of A are 5 both with algebraic multiplicity 2. (b) (8 points) Find a basis for each eigenspace of A. he eigenspace of for A is given by 2 4 2 2 2 null(i A) = null 4 8 2 = null = span 4 2. he eigenspace of 5 for A is given by 8 4 2 null( 5I A) = null 4 2 2 4 = null 2 = span.
6.) (2 points) Consider the basis = given by in -coordinates. (a) (5 points) Find ([ x {[ ] [ ]} 2 for R 3 2 and the linear map : R 2 R 2 ] ) [ ] x + 2x = 2 3x + 4 ([ ]) and express your answer in standard coordinates. S-basis: [ ] [ ] 3 (b) (5 points) Find -basis: [ ] /2 3/2 [ ] /2 5/2 ([ ]) and express your answer in standard coordinates. S-basis: [ ] [ ] 4 2 -basis: [ ] [ ] 2 4 (c) (2 points) Find [ ]. We conclude [ ] 4 [ ] =. 3 2
.) Proof Questions (28 points): (a) (6 points) Suppose A are n n matrices that satisfy A = A. Show that if v is an eigenvector of A with eigenvalue λ then v is also an eigenvector of A with eigenvalue λ. We have A v = λ v. hen A( v) = (A v) = (λ v) = λ( v) so v is also an eigenvector of A with eigenvalue λ. (b) (6 points) Suppose S is a subspace of R n and : R n R m is a linear map. Show that the image of S under or the set of all vectors that are of something in S: is a subspace of R m. (a) ecause S = ( ) V. (b) Next for any x y V we have V = { ( x) x S} x = ( u) y = ( v) for some u v S. hen u + v S so x + y = ( u) + ( v) = ( u + v) V. (c) Next for any x V an scalar c we have x = ( u) for some u S. hen c u S so c x = c ( u) = (c u) V.
(c) (8 points) Let S S 2 be subspaces of R n and let { u... u k } be a basis for S and { v... v m } be a basis for S 2. Show that S S 2 = { } IF AND ONLY IF { u... u k v... v m } is linearly independent. First suppose that S S 2 =. Suppose herefore c u + + c k u k + d v + + d m v m =. c u + + c k u k = d v d m v m Since the left hand side lies in S the right hand side lies in S 2 and S S 2 = this can only happen when c u + + c k u k = d v d m v m =. ecause { u... u k } is linearly independent and { v... v m } is linearly independent we must have c... c k d... d m =. Hence { u... u k v... v m } is linearly independent. On the other hand suppose { u... u k v... v m } is linearly independent. hen consider x S S 2. hen we can write x = c u + + c k u k = d v + + d m v m for some scalars c... c k d... d m. hen we have c u + + c k u k d v d m v m =. Since { u... u k v... v m } is linearly independent c... c k d... d m = so x = u + + u k =. Hence S S 2 =. (d) (8 points) Suppose : R n R m is an onto linear map. Show that there exists a linear map U : R m R n so that ( U)( x) = x for all x R m. ecause is onto for each j =... m there exists a j R n so that ( a j ) = e j. hen define U : R m R n to be the linear map with It then follows that for each x R m [U] = [ ] a... a m. x x m ( U)( x) = U. = (x a + + x m a m ) = x ( a ) + + x m ( a m ) = x e + + x m e m = x.