EXAM. Exam #3. Math 2360, Spring April 24, 2001 ANSWERS

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1 EXAM Exam #3 Math 2360, Spring 200 April 24, 200 ANSWERS

2 i

3 40 pts Problem In this problem, we will work in the vectorspace P 3 = { ax 2 + bx + c a, b, c R }, the space of polynomials of degree less than 3 Let P be the ordered basis P = [ x 2 x ] of P 3 and let Q be the ordered basis Q = [ 2x 2 + x + x 2 + 2x +, x 2 + x + ] of P 3 A Find the change of basis matrices S PQ and S QP The defining equation of the change of basis matrix S PQ is Q = PS PQ Reading off the coefficients, we have [ 2x 2 + x + x 2 + 2x +, x 2 + x + ] = [ x 2 x ] 2 2, }{{}}{{} Q P (a matrix multiplication), so S PQ = 2 2 () To find S QP, we use the fact that S QP = S PQ Thus, 0 S QP = 0, 3 where we have used a calculator to find the inverse of the matrix in () B Let p(x) = 2x 2 x + Find [p(x)] Q, the coordinate vector of p(x) with respect to the ordered basis Q Show how p(x) can be written as a linear combination to the elements of Q First we find [p(x)] P, since that s easy The defining equation of [p(x)] P is p(x) = P[p(x)] P We have p(x) = 2x 2 x + = [ x 2 x ] 2, }{{} P

4 and so [p(x)] P = 2 To find [p(x)] Q, we use the change of coordinate equation [p(x)] Q = S QP [p(x)] P, so [p(x)] Q = = 2, 3 2 using our value of S QP from above Thus, [p(x)] Q = 2 2 The defining equation of [p(x)] Q is p(x) = Q[p(x)] Q Thus, we have p(x) = Q[p(x)] Q = [ 2x 2 + x + x 2 + 2x +, x 2 + x + ] 2 }{{} 2 Q }{{} [p(x)] Q = (2x 2 + x + ) 2(x 2 + 2x + ) + 2(x 2 + x + ) Thus, we can write p(x) as a linear combination of the elements of Q as follows p(x) = (2x 2 + x + ) 2(x 2 + 2x + ) + 2(x 2 + x + ) 40 pts Problem 2 In this problem, we will work in the vectorspace P 3 = { ax 2 + bx + c a, b, c R }, the space of polynomials of degree less than 3 P = [ x 2 x ] of P 3 Let A be the matrix A = 3 0, Let P be the ordered basis which is invertible Let Q be the ordered basis defined by Q = PA Let T : P 3 P 3 be the linear transformation defined by T (p(x)) = p (x) + p(x) 2

5 A Find [T ] PP, the matrix of T with respect to the ordered basis P The defining equation of [T ] PP is T (P) = P[T ] PP (2) and T (P) = [ T (x 2 ) T (x) T () ] Since T (p(x)) = p (x) + p(x) we have T (x 2 ) = (x 2 ) + x 2 = 2x + x 2 T (x) = (x) + x = + x T () = () + = 0 + =, so T (P) = [ x 2 + 2x x + ] Reading off the coefficients, we have [ x 2 + 2x x + ] = [ x 2 x ] , }{{}}{{} 0 T (P) P (matrix multiplication) Comparing this with equation (2), we see that 0 0 [T ] PP = B Find [T ] QQ, the matrix of T with respect to the basis Q We will use the change of basis equation for linear transformations, ie, [T ] QQ = S QP [T ] PP S PQ Since S QP = S PQ, we can rewrite this equation as The defining equation of S PQ is [T ] QQ = S PQ [T ] PP S PQ (3) Q = PS PQ (4) Since the basis Q is defined by Q = PA, where A is given in the problem, comparison with equation (4) shows that S PQ = A Hence, plugging into (3) shows that [T ] QQ = A [T ] PP A 3

6 Using the given value of A and the value of [T ] PP from the previous part of the problem, we have [T ] QQ = Using a calculator for the matrix computations, we find 3 7 [T ] QQ = 2 5/2 /2 0 /2 3/2 60 pts Problem 3 Let U = [ u ] u 2 be the ordered basis of R 2 where [ [ 2 2 u =, u ] 2 = 2] Let T : R 2 R 2 be the linear transformation whose matrix with respect to the standard basis E is [ ] 2 A =, 0 in other words, T (x) = Ax A Find the change of basis matrices S EU and S UE The defining equation of S EU is U = ES EU The corresponding matrix equation is mat(u) = mat(e)s EU = IS EU = S EU Thus, we have Thus, S EU = mat(u) = S EU = [ ] [ ] We know that S UE = S EU Using a calculator for the computation of the inverse, we have [ ] S UE = /2 4

7 B Find [T ] UU, the matrix of T with respect to U We are given the matrix of T with respect to the standard basis, so [ ] 2 [T ] EE = A = 0 To find [T ] UU, we use the change of coordinate equation for linear transformations, ie, [T ] UU = S UE [T ] EE S EU [ = /2 [ ] 2 4 = 0 ] [ 2 0 Thus, the answer to this part of the problem is [T ] UU = [ ] ] [ ] C Find the scalars c and c 2 such that T (2u 3u 2 ) = c u + c 2 u 2 Let v = 2u 3u 2 Then, of course, v = [ [ ] 2 u u 2, }{{} 3] U The defining equation of [v] U is v = U[v] U, so we have [v] U = [ 2 3], The equation for the action of the linear transformation T in the U-coordinates is [T (v)] U = [T ] UU [v] U Thus, we have [T (v)] U = [ ] [ ] = 0 3 [ ] 8 3 5

8 The defining equation of [T (v)] U is T (v) = U[T (v)] U So we have T (2u 3u 2 ) = T (v) = [ [ ] ] 8 u u 2 = 8u }{{} 3 + 3u 2, U }{{} [T (v)] U So the answer is c = 8, c 2 = 3 40 pts Problem 4 Let A = [ ] Find the characteristic polynomial of A and the eigenvalues of A We have A λi = [ ] [ ] [ ] 3 0 λ 3 λ = λ Thus, the characteristic polynomial p(λ) of A is Thus, we have p(λ) = det(a λi) = λ 3 2 λ = ( λ)( λ) 6 = λ + λ 2 6 = λ 2 λ 6 p(λ) = λ 2 λ 6 The eigenvalues of A are the roots of the characteristic polynomial The characteristic polynomial factors as p(λ) = (λ 3)(λ) + 2), so the roots are 3 and 2 Thus, Eigenvalues of A = 2, 3 60 pts Problem 5 In each part you are given an matrix A and its eigenvalues Find a basis for each of the eigenspaces of A Determine if A is diagonalizable, and if it is, find a matrix P and a diagonal matrix D so that P AP = D 6

9 A A = 2 2, Eigenvalues =, 2 Consider first the eigenvalue λ = We have A λi = A I = The RREF of this matrix is (by calculator) R = The eigenspace E is the nullspace of A I, which is the same as the nullspace of R, ie, the solution space of Rx = 0 Calling the variables x, x 2 and x 3, we see that x 3 is a free variable, say x 3 = α The second row of R gives the equation x 2 x 3 = 0, so x 2 = α and the first row gives the equation x x 3 = 0, so x = α Thus, the vectors in the nullspace of R are x x 2 = α α = α α x 3 Thus, a basis of the eigenspace E is Basis of E = so E has dimension one Next, consider the eigenvalue λ = 2 We have A λi = A 2I = 2 0, which has the RREF R = 0 /2 0 3/

10 The eigenspace E 2 is the nullspace of A 2I, which is the same as the nullspace of R Abbreviating the computation of the nullspace of R, we have x 3 = α x x 3 = 0 = x 2 = 3 2 α x 2 x 3 = 0 = x = 2 α so the nullspace is parametrized by x 3 2 x 2 = α 2 α = α 3/2 /2 α x 3 Thus, we have Basis of E 2 = 3/2 /2 Alternatively, we could multiply this vector by 2 and say Basis of E 2 = 3 2 and so avoid the agony of dealing with fractions We have only produced 2 independent eigenvectors, and there are no additional eigenvectors that are independent of these two Thus, A does not have a basis of eigenvectors, we we conclude A is not diagonalizable B A = 2 5 6, Eigenvalues = 2, First consider the eigenvalue λ = 2 Then we have A λi = A + 2I = 2 3 6,

11 which has the RREF /4 /2 R = For the abbreviated computation of the nullspace we have x 2 = α x 3 = β x 4 x 2 2 x 3 = 0 = x = 4 α + 2 β The parametrization of the nullspace is x 4 x 2 = α + 2 β /4 /2 α = α + β 0 β 0 Thus, we get x 3 /4 Basis of E 2 =, 0 /2 0 (5) As matter of personal taste, I ll get rid of the fractions by multiplying the first vector by 4 and the second by 2 That gives me Basis of E 2 = 4, 0 0 (6) 2 We see that E 2 has dimension 2 Next, consider the eigenvalue λ = We have A λi = A I = 2 3 9, which has has the RREF R = Note that we have already computed the nullspace of this matrix R in the first part of the problem Thus, we know Basis of E = (7) 9

12 Since we ve found three linearly independent eigenvectors, A has a basis of eigenvectors and A is diagonalizable Using the basis (6) of E 2 as a matter of taste (I could equally well use (5)), and the basis (7) of E, set P = 4 0, D = then P is invertible, D is a diagonal matrix and P AP = D (check it if you don t believe me) 40 pts Problem 6 In each part, determine if the given transformation T is linear If it is linear, find a matrx so that T (x) = Ax If T is not linear, justify your answer A T : R 2 R 2 is given by ([ ]) [ ] x x x T = 2 x 2 x 2 + x 2 2 For T to be linear, it must preserve scalar multiplication, ie, it must be true that T (αx) = αt (x) for all scalars α and all x R 2 We have ( [ ]) ([ ]) [ ] [ ] x αx (αx T α = T = )(αx 2 ) α x 2 αx 2 (αx ) 2 + (αx 2 ) 2 = 2 x x 2 α 2 (x 2 + x 2 (8) 2) ([ ]) [ ] [ ] x x x αt = α 2 αx x x 2 x 2 + x 2 = 2 2 α(x 2 + x 2 (9) 2) If T was linear, the right hand sides of equations (8) and (9) would have to be equal for all values of α, x and x 2 In particular, we would have to have α 2 x x 2 = αx x 2 for all values of α, x and x 2 This is plainly not the case (eg, α = 2, x = x 2 = ), so T is not linear 0

13 B T : R 3 R 2 is given by T x = x 2 x 3 [ ] 2x 3x 2 + 5x 3 x x 3 We can rewrite the right hand side of the definition of T as a matrix multiplication T x [ ] x = x x 0 2 x 3 Thus, T (x) = Ax for all x R 3, where A is the matrix A = [ 2 3 ] 5 0 Since T is given by matrix multiplication, T must be linear T is linear x 3 50 pts Problem 7 Let S be the subspace of R 4 spanned by the vectors 2 8 v = 2, v 2 = 3, v 3 = 0 5, v 4 = A Cut down the list of vectors above to a basis for S What is the dimension of S? B For each of the following vectors, determine if the vector is in S and, if so, express it as a linear combination of the basis vectors you found in the previous part of the problem 3 5 w = 5 2, w 2 =

14 We can do all the computation in one RREF calculation on the calculator Make up a matrix containing all these vectors A = v v 2 v 3 v 4 w w The RREF of this is R = Columns, 2 and 4 of R form a basis for the span of the columns left of the bar, so the same is true of A Thus, v, v 2 and v 4 are a basis of S Since S has a basis with three elements, The dimension of S is 3 In R, we have col 5 (R) = col (R) + 2 col 2 (R) + col 4 (R) The same relation must hold among the columns of A, so we have w = v + 2v 2 + v 4 Since w is a linear combination of vectors in S, w S The last column of R is not a linear combination of the columns to the left of the bar (because of the last row), so the same is true of A Thus, w 2 / S 2

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