Practice Final Exam Solutions

Transcription

1 MAT 242 CLASS FALL 206 Practice Final Exam Solutions The final exam will be cumulative However, the following problems are only from the material covered since the second exam For the material prior to the second exam, I would recommend reviewing the previous exams (both the practice and actual exams - especially the problems on the exam that you missed points on) and the previous problem sets Problem : Determine if the matrix A [ ] is diagonalizable If A is diagonalizable, determine an invertible matrix P and diagonal matrix D such that A P DP If A is not diagonalizable, explain how you know that it is not Solution: To determine if A is diagonalizable, we need to determine bases for each eigenspace associated to A We begin by determining the eigenvalues of A The characteristic polynomial of A is given by ([ ]) x 4 p A (x) det 3 2 x ( x)(2 x) 2 2 3x + x 2 2 x 2 3x 0 (x 5)(x + 2) So we see that there are two eigenvalues, namely λ 2,5 Let s determine a basis for E 2 (A) All vectors in E 2 (A) are solutions to the homogenous linear system corresponding to [ ( 2) ( 2) 0 [ [ ] 4/

2 So every vector in E 2 (A) is of the form [ ] 4/3 a for some real number a This means {( 4/3,)} is a basis for E 2 (A) Next we ll determine a basis for E 5 (A) All vectors in E 5 (A) are solutions to the homogenous linear system corresponding to [ (5) (5) 0 So every vector in E 2 (A) is of the form [ a [ ] [ ] for some real number a This means {(,)} is a basis for E 5 (A) Since there are 2 linearly independent eigenvectors {( 4/3, ),(, )}, we know that A is diagonalizable Furthermore, from the above, we may take P [ ] 4/3, which means D [ ] Problem 2: Determine if the matrix 0 0 A 0 0 is diagonalizable If A is diagonalizable, determine an invertible matrix P and diagonal matrix D such that A P DP If A is not diagonalizable, explain how you know that it is not There was an important omission in the statement of this problem In this class we ve been ignoring the complex numbers So in particular, if you re asked if a matrix is diagonalizable, we really want to know if the matrix is similar to a real valued diagonal matrix It turns out that the above matrix A is NOT similar to a real valued diagonal matrix, but it is similar to a complex valued diagonal matrix Solution: To determine if A is similar to a real valued diagonal matrix (see the above), we need to determine bases for each eigenspace associated to a real eigenvalue of A We begin by determining these eigenvalues The characteris- 2

3 tic polynomial of A is given by 0 x 0 p A (x) det 0 x 0 x ([ ]) ([ ]) ( ) + x ( x)det + ( ) 2+ 0 ()det x x ( x)(x 2 x + ) + x 3 + x 2 x + (x )(x 2 + ) So we see that there is only one real eigenvalue, namely λ Let s determine a basis for E (A) All vectors in E (A) are solutions to the homogenous linear system corresponding to So every vector in E (A) is of the form a 0 for some real number a This means {(,0,)} is a basis for E (A) Since there is only linearly independent real eigenvector {(, 0, )}, we know that A is not similar to a real valued diagonal matrix (since A is 3 3, we would have needed 3 linearly independent real eigenvectors) Problem 3: Determine if each of the following functions are linear transformations If the function is a linear transformation, determine a basis for the kernel and range If the function is not a linear transformation, explain why (a) T : R 3 R 2 defined by T (a, a 2, a 3 ) (a a 2,2a 3 ) Solution: T is a linear transformation To see this, consider arbitrary vectors (a, a 2, a 3 ),(b,b 2,b 3 ) R 3 and an arbitrary scalar c R We have T (c(a, a 2, a 3 ) + (b,b 2,b 3 )) T (ca + b,ca 2 + b 2,ca 3 + b 3 ) (ca + b (ca 2 + b 2 ),2(ca 3 + b 3 )) c(a a 2,2a 3 ) + (b b 2,2b 3 ) ct (a, a 2, a 3 ) + T (b,b 2,b 3 ) 3

4 To find a basis for ker(t ), we note that if (a, a 2, a 3 ) ker(t ), then a a 2 0 and 2a 3 0 In other words, a a 2 and a 3, which implies every vector in ker(t ) is of the form a 2 (,,0) for some a 2 R So the set {(,,0)} is a basis for ker(t ) To find a basis for range(t ), we note that every vector in range(t ) is of the form [ ] [ ] [ ] a a 2 0 (a 2a a 2 ) + a for some real numbers a, a 2, a 3 R So the set {(,0),(0,2)} is a basis for range(t ) (b) T : R 2 R 2 defined by T (a, a 2 ) (, a 2 ) Solution: T is not a linear transformation This function fails to respect vector addition and scalar multiplication For example, T (2(,0)) T (2,0) (,0) (2,0) 2T (,0) (c) T : R 2 R 3 defined by T (a, a 2 ) (a + a 2,0,2a a 2 ) Solution: T is a linear transformation To see this, consider arbitrary vectors (a, a 2 ),(b,b 2 ) R 3 and an arbitrary scalar c R We have T (c(a, a 2 ) + (b,b 2 )) T (ca + b,ca 2 + b 2 ) (ca + b (ca 2 + b 2 ),0,2(ca + b ) (ca 2 + b 2 )) (ca ca 2,0,2ca ca 2 ) + (b b 2,0,2b b 2 ) c(a a 2,0,2a a 2 ) + (b b 2,0,2b b 2 ) ct (a, a 2 ) + T (b,b 2 ) To find a basis for ker(t ), we note that if (a, a 2 ) ker(t ), then a + a 2 0 and 2a a 2 0 In other words, a a 2 and a 2 a 2 In order for these to both to hold, it must be the case that a 0 a 2, which implies ker(t ) {(0,0)} So the empty set is a basis for ker(t ) To find a basis for range(t ), we note that every vector in range(t ) is of the form a + a 2 0 a 0 + a 2 0 2a a 2 2 for some real numbers a, a 2 R So the set {(,0,2),(,0, )} is a basis for range(t ) 4

5 (d) T : R 2 R 2 defined by T (a, a 2 ) (a, a 2 ) Solution: This function is not a linear transformation For example, T ((,0)+(,0)) T (2,0) (2,4) (2,2) (,)+(,) T (,0)+T (,0) Problem 4: Suppose T : R 2 R 2 is linear with T (,0) (,4) T (,) (2,5) (a) What is T (2,3)? Solution: We know that T is associated to a unique 2 2 matrix A satisfying A [ ] 0 [ ] which implies A [ ][ ] [ ][ ] [ ] 4 Note that we could have also used the fact that A T (,0) T (0,) T (,0) T ((,) (,0)) T (,0) T (,) T (,0)) [ ] 4 Therefore, we have T (2,3) A [ ] 2 3 [ ][ ] [ ] 5 (5,) 5

6 (b) Is T one-to-one? Solution: Yes To show this we want to determine ker(t ) By part (a), we know that ker(t ) null(a), and every x null(a) is a solution to the homogeneous linear system associated to the augmented matrix [ 4 0 [ ] Hence, there is only one solution to this system x (0,0) Since ker(t ) {(0,0)} we know that T is one-to-one Problem 5: For each of the following linear transformations T, determine whether T is an isomorphism and justify your answer (Recall that M 2 2 (R) is the vector space of all 2 2 matrices with real number entries, and P 2 (R) is the vector space of all polynomials with real coefficients having degree at most 2) (a) T : R 2 R 3 defined by T (a, a 2 ) (a 2a 2, a 2,3a + 4a 2 ) Solution: T cannot be an isomorphism since rank(t ) 2 dim(ker(t )) 2 < 3 The above implies that T cannot onto (b) T : R 3 R 3 defined by T (a, a 2, a 3 ) (3a 2a 3, a 2,3a + 4a 2 ) Solution: Let s first determine ker(t ) In order for (a, a 2, a 3 ) to be in the ker(t ), it must be the case that 3a 2a 3 0 a 2 0 3a + 4a 2 0 It s not too hard to see that the only way all of these linear equations can be satisfied simultaneously is if a a 2 a 3 0 This means that ker(t ) {(0,0,0)}, which in turn implies that T is one-to-one To show that T is onto, it suffices to show that rank(t ) dim(range(t )) 3, since this would imply that the range of T is a subspace of R 3 having dimension 3 The only such subspace is R 3 itself By the Dimension Theorem we have rank(t ) 3 dim(ker(t )) 3 Hence, T is onto 6

7 Since T is one-to-one and onto, we know that T is an isomorphism ([ ]) a b (c) T : M 2 2 (R) P 2 (R) defined by T a + 2bx + (c + d)x 2 c d Solution: For this one we note that, for example, T ([ ]) Since ker(t ) is not the zero vector space, we know that T is not one-to-one 7