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1 Math 38 J - Spring 2 Final Exam - June 6 2 Name: Student ID no. : Signature: 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 .) (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 = 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 Z 8

3 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 has : R R 3 given by x (x ) = 2x 3x 3 range( ) = span 2 = span (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 ] = [ ]

4 3.) (2 points) Let : R 4 R 3 be the linear map with 3 5 [ ] = (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 w w 3 w 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] = [ ]. 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 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

5 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 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. [ ] [ ] which means x 3 is free x = 5x 3 = 4x 3. hus 5 (S + S 2 ) S 3 = x 3 = span = 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 ) =.

6 5.) (2 points) Let 3 4 A = (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 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 null(i A) = null = null = span 4 2. he eigenspace of 5 for A is given by null( 5I A) = null = null 2 = span.

7 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

8 .) 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.

9 (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.