Math 308 Final, Autumn 2017

Transcription

1 Math 308 Final, Autumn 2017 Name: ID#: Signature: All work on this exam is my own. Instructions. You are allowed a calculator and notesheet (handwritten, two-sided). Hand in your notesheet with your exam. Other notes, devices, etc are not allowed. Unless the problem says otherwise, show your work (including row operations if you row-reduce a matrix) and/or explain your reasoning. You may refer to any theorems, facts, etc, from class. All the questions can be solved using (at most) simple arithmetic. (If you find yourself doing complicated calculations, there might be an easier solution...) Raise your hand if you have any questions. Good luck! 1 /20 5 /15 2 /20 6 /15 3 /20 7 /20 4 /20 Bonus /7

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3 A virus, influenza levinsonia, has escaped from Levinson s Linear Laboratory. Anyone affected becomes obsessed with solving linear equations (until cured). (1) Let S R 4 be the set of solutions to the equations: Find a basis for each of S and S. [10 points each] a + b + c + d = 0 a + b c + 5d = 0. S: Solution. To solve the equations, we first reduce them to echelon form: [ ] [ ] [ ] We have free variables x 3 = s 1 and x 4 = s 2 and we solve to get x 2 = 3s 2 and x 1 = s 1 + 2s 2. So, the general solution is 1 2 { 1 2 } x = s s 3 2 0, and the basis is 0 1, S : (Hint: You were given some elements of S above.) Solution. (Method 1) We know from class that a set of defining equations for S corresponds to a spanning set for S, so S = span{ 1 1 } 1 1, 1 1. These two 1 5 vectors are clearly linearly independent, so they form a basis for S. (Method 2) We know that x S if and only if 1 2 x 0 1 = 0 and x 3 0 = In other words, x satisfies x 1 + x 3 = 0 and 2x 1 3x 2 + x 4 = 0. So, after reducing this system of equations to echelon form and solving for x, we get 1 0 { 1 0 } x = s 1 2/3 1 + s 1/3 2 0, and the basis is 0 1, 1/

4 1 3 0 (2) Let A = The characteristic polynomial of A is λ 3 + 4λ 2 4λ (a) [15 points] Find the eigenvalues of A and a basis for each eigenspace. Solution. The eigenvalues are the solutions to λ 3 + 4λ 2 4λ = 0, which factors as λ(λ 2) 2 = 0. So, we get λ = 0, 2. For λ = 0, we find the null space of A 0I, which is just A. After row reducing, we get A , corresponding to the equations x 1 + 3x 2 = 0 and x 3 = 0. So, the null space has a basis consisting of just the vector For λ = 2, we find the null space of A 2I = 1 1 1, which row reduces to , so the basis for the null space is just the vector (b) [5 points] Is A diagonalizable? If so, give an invertible matrix P and a diagonal matrix D so that A = P DP 1. If not, explain why not. Solution. A is not diagonalizable, because the λ = 2 eigenspace has dimension 1, which is strictly less than the multiplicity of that eigenvalue (multiplicity two). (So, we only found two linearly independent eigenvectors in all, not three). Since A does not have a basis of eigenvectors, A is not diagonalizable.

5 (3) Influenza levinsonia is spreading. Let s represent the population as a vector: x = x 1 # at-risk people x 2 = # sick people R 3. # recovered people Each week, the vector x changes as follows: x 3 1 the at-risk group becomes sick; the other 1 stays healthy % of the sick group stays sick; the other 30% recovers and becomes immune. For example, if x = today, then in one week it will be (a) [10 points] Let T ( x) be the population vector one week after it was x. Express T ( x) either as T ( x) = A x or by giving a formula in terms of x 1, x 2, x 3. Solution. ( T x ) 1 1 x 2 = x x x 2 = x 1 x 2 2. x 3 0.3x 2 + x x 3 Note: One version of this exam asked whether T is invertible. Yes, it is invertible because det(t ) = (This problem continues on the next page.)

6 (b) [5 points] Determine the eigenvalues of T. Is T diagonalizable? (This should not require extended computation.) Solution. Since T is lower-triangular, so is A λi, so det(a λi) = ( 1 λ)(0.7 λ)(1 λ). So, the eigenvalues of T are λ = 1, 0.7, Since there are three distinct, real eigenvalues, T is diagonalizable. (c) [5 points] In terms of the epidemic, what is the meaning of computing T 52 ( x)? Solution. T 52 ( x) represents the number of healthy, sick and immune people after 52 weeks (one year) of the epidemic.

7 (4) Let T : R 5 R 3 and S : R 3 R 4 be linear transformations. Let Q : R 5 R 4 be the composition, Q( x) = S(T ( x)). (a) [5 points] If x ker(t ), what is Q( x)? Solution. Q( x) = S(T ( x)) = S( 0) (since x ker(t )) = 0 (since S is a linear transformation.) (b) [5 points] Explain why ker(q) has dimension 2. (Hint: based on part (a), compare ker(q) and ker(t ).) Solution. Based on part (a), we know that ker(q) contains ker(t ). By the rank-nullity theorem, ker(t ) has dimension 5 rank(t ) 5 3 = 2 (since T has rank at most 3). Therefore dim(ker(q)) 2. (c) [5 points] If T is onto and S is one-to-one, what is the dimension of range(q)? Solution. Since T is onto, the range of T is all of R 3. Since S is one-to-one, S maps R 3 to a 3-dimensional plane in R 4. Therefore S T maps R 5 to this 3-dimensional plane in R 4. In other words, range(q) has dimension 3. (d) [5 points] Either give an example of S, T that results in Q having rank 4, or explain why this is not possible. Solution. This is not possible. By the rank-nullity theorem, the rank of Q is 5 dim(ker(q)) 5 2 = 3, so the rank cannot be 4.

8 (5) In this problem, v 1 and v 2 are vectors in R n with v 1 v 1 = 3, v 2 v 2 = 5, and v 1 v 2 = 3. (a) [5 points] Find the values of a so that 2 v 1 + a v 2 is orthogonal to a v v 2. Solution. The vectors will be orthogonal if (2 v 1 + a v 2 ) (a v v 2 ) = 0. Expand out the product (remember the rules ( u + v) w = u w + v w and v w = w v): 0 = 2a( v 1 v 1 ) + 6( v 1 v 2 ) + a 2 ( v 2 v 1 ) + 3a( v 2 v 2 ) = 6a a a = 3(a + 1)(a + 6). So the solutions are a = 1, 6. (b) [5 points] Find vectors v 1, v 2 in R 2 satisfying the equations above. [ [ x z Solution. If v 1 = and v y] 2 =, then the three equations are w] x 2 + y 2 = 3, z 2 + w 2 = 5, xz + yw = 3. There are four variables but only three equations, so it won t be possible to solve for all four variables uniquely. Instead, we can try setting one variable to a simple number (like 0 or 1) and then solve for the other three. For instance, setting y = 0, we get [ ] [ ] 3 3 v 1 =, v 0 2 =. 2 Many other answers are possible, but this was probably the easiest one. (c) [5 points] Either give an example of v 1, v 2 (in any R n ) that are linearly dependent, or explain why they must be independent. Solution. Many explanations are possible. One is to compute proj v2 ( v 1 ) = v 1 v 2 v 2 v 2 v 2 = 3v 5 2. If v 1 and v 2 are linearly dependent, then the projection should not change the vector, so 3 v 5 2 = v 1. But the length of v 1 is 3 and the length of v2 is 5, which is not consistent with this scalar multiple. A second explanation is to consider a dependence relation, i.e. a v 1 + b v 2 = 0. If we take the dot product with v 1, we see that 3a + 3b = 0. On the other hand, if we take the dot product with v 2, we see that 3a + 5b = 0. But then a = b = 0. So, there are no nontrivial linear relations between v 1 and v 2.

9 (6) A scientist, (your name here), is preparing an influenza levinsonia antibody shaped like a parallelogram. A 5-dimensional parallelogram (a) [10 points] The volume is given by det(a), where A = s Find s so that det(a) = 2. Solution. By cofactor expansion, we see that det(a) = 2 1 det s Using either row operations or the diagonals shortcut for a 3 3 matrix, this smaller matrix has determinant 3s + 6. So, we get the equation det(a) = 2( 3s + 6) = 2, with solution s = 5. 3 (b) [5 points] Oops, your microscope settings were backwards. The volume is actually det(a 1 ). Find s so that det(a 1 ) = 2. Solution. Since det(a 1 ) = 1, we see that det(a) = 1. Using our equation det(a) 2 from above, det(a) = 2( 3s + 6) = 1 1, which has the solution s = 2 =

10 ( ) (7) (a) [10 points] Let S = span 0 2, 0 4, 1 0, 3 4, 0 0, 1 1, Find a basis for S. 2 0 Solution. The vectors 0 4 and 0 0 are clearly redundant. Some people noticed that 3 4 was also redundant, since it s a combination of the 1st and 0 3rd vectors. To check for any remaining relations, we row-reduce the remaining vectors: Since the 1st, 2nd and 3rd columns have pivots, a basis is given by the first three columns, , 1 0, (b) [10 points] It s time to be cured of influenza levinsonia. You have to be injected with the vaccine you prepared earlier. Give a one-to-one linear transformation 1 T : R n R 4, so that range(t ) is the subspace S above. (Be careful to choose what n should be.) Solution. Since dim(s) = 3, we will need the domain of T to be R 3. We ll have T ( x) = A x for some matrix A, and we want range(t ) = col(a) to be S. So, the columns of A should span S, and since there will only be three columns, the columns will have to be a basis for S. It s simplest to just use the basis we found in part (a), so A = an injection. :-)

11 (Bonus) (+7 points) Suppose S R n is a k-dimensional subspace, and proj S : R n R n is the projection onto S. Note: the kernel of proj S is S, of dimension n k. (a) What are the eigenspaces of proj S for the eigenvalues λ = 0 and λ = 1? Solution. The λ = 0 eigenspace is the kernel, which is S since proj S ( x) = 0 if and only if x S. The λ = 1 eigenspace is the set of vectors satisfying the equation proj S ( x) = x, the vectors unchanged by the transformation. These are the vectors in S itself, so the eigenspace is S. (b) The characteristic polynomial of a linear transformation T : R n R n is always a polynomial of degree n. With this in mind, find the characteristic polynomial of proj S. (Hint: By part (a), the factors λ and (λ 1) must occur at least a certain number of times. Are there any missing factors?) Solution. Since the λ = 0 eigenspace has dimension n k, there is a factor of at least λ n k. Since the λ = 1 eigenspace has dimension k = dim(s), there is a factor of at least (λ 1) k. But these two factors already give a polynomial of degree n, so there are no other factors. So, the characteristic polynomial is λ n k (λ 1) k. Side note (for anyone reading this far): this implies that proj S is always diagonalizable by finding bases for S and S.

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