Math 4153 Exam 2 Review with solutions

Transcription

1 The syllabus for Exam is the Gaussian Elimination supplement (found on the class website), Chapter 5 and Chapter 6 through page in Axler.. You should be sure to know precise definition of the terms we have used, and you should know precise statements (including all relevant hypotheses) for the main theorems proved. Know how to do all of the homework problems. 2. Outline of subjects for Exam 2: Gaussian elimination and applications Eigenvalues, eigenvectors, and eigenspaces Inner product spaces, norms Gram-Schmidt procedure 3. Definitions: The null space, row space, column space, and left null space of a matrix The Gaussian elimination algorithm for matrices The Gauss-Jordan elimination algorithm for matrices Row echelon form matrices Reduced row echelon form matrices Pivot columns of a matrix (not necessarily in row echelon form) Elementary matrices Permutation matrix The decomposition A = P LB for matrices using Gaussian elimination Eigenvalues and eigenvectors of linear transformations and matrices Eigenspaces of linear transformations and matrices Invariant subspace U V under a linear operator T L(V ) Inner Product and inner product space Norm associated to an inner product space The parallelogram equality (Theorem 6.4) Orthogonal and orthonormal lists and bases 4. Matrix computations: know how to compute A row echelon and reduced row echelon form matrix obtained from A by Gaussian or Gauss-Jordan elimination. The null space, row space, and column space of a matrix by means of row reduction.

2 The solution space Ax = c for A Mat(m, n, F), c Mat(n,, F), and x = x. x n. How to invert a matrix A A. Compute the P LB decomposition A = P LB where B is in row echelon form. Given λ F, compute the eigenspace of λ for a matrix A 5. Inner products: Use the Gram-Schmidt procedure to find an orthonormal basis from an arbitrary one. 6. Major results. Know the statements, but don t memorize them word for word know what they say mathematically. Especially important ones are in bold. Theorem 5.6, and how it implies Corollary 5.9. The existence of eigenvalues for operators on a complex vector space. (Theorem 5.) Block upper-triangular matrices for linear maps: Theorem 5.3 The eigenvalues of an upper triangular matrix are the entries on the diagonal. An operator T L(V ) has a diagonal matrix with respect to some basis of V if and only if V has a basis consisting of eigenvectors of T. (Page 88) Proposition 5.2. Cauchy-Schwarz inequality: Theorem 6.6; triangle inequality: Theorem 6.9. Gram-Schmit orthogonalization: 6.2 Review Exercises. Determine if each of the following statements is true or false. If true, explain why. If false give a counterexample. (a) Suppose V = {}. Then every linear transformation T L(V ) has an eigenvalue, i.e., there is λ F such that T v = λv for some v =. Solution. False. Exercise 5 (forward shift operator is a counterexample). (b) Same as the previous problem, but now assume that F = C. Solution. False. Exercise 5 (forward shift operator is a counterexample). (c) Same as the previous problem, but now assume that V is finite-dimensional (and F = C). 2 2

3 Solution. True. This is Theorem Consider the matrix (a) Find a row echelon form for A. Solution. Swap rows and 2, and then swap rows 2 and 3 to get 2 3 B = 2. The matrix B is in row echelon form. (b) Find a reduced row echelon form for A. Solution. Subtract 2 times row 3 from row 2, then subtract 3 times row 3 from row, and then add 2 times row 2 to row, to get I 3, the 3 identity matrix, which is in reduced row echelon form. (c) Find a basis of the row space of A. Solution. The 3 rows of the identity form a basis of the row space. (d) Find a basis of the null space of A. Solution. The null space of A is {}, which has the empty list () as a basis. (e) Find a basis for the column space using columns of the original matrix. Solution. All columns are pivot columns, so the 3 columns of A form a basis for the row space of A. (f) Compute A. Solution. Apply the same row operations in parts (a) and (b) to I 3 to get 7 2 A = 2. (g) Compute the A = P LB decomposition of A. 3 3

4 Solution. From part (a), P A = B where P is the permutation matrix that exchanges rows and 2 and then rows 2 and 3, and B is the row echelon matrix in part (b). That is, P =. If P = P =. Then A = P B. 3. Do the same problems as above for the matrix , except for finding A. Solution. (a) Swap rows and 2: Subtract 3 times row from row 3: Divide row 3 by 2: Subtract row 2 from row 3: Multiply row 3 by : The matrix B is in row echelon form B = 2 4 4

5 (b) Subtract 2 times row 3 from row 2 and then add 3 times row 3 to row : 3 5 Subtract 3 times row 2 from row : 2 R = The matrix R is in reduced row echelon form. (c) A basis of the row space of A is the three rows of R. (d) The null space of A is the same as the null space of the row equivalent matrix R that is in reduced row echelon form. The system of equations Rx = is x + 2x 3 = x 2 + x 3 = x 4 =. The only free variable is x 3. Solving for x, x 2 and x 4 in terms of x 3 gives x = 2x 3, x 2 = x 3, and x 4 =. Thus the null space of A, which is the same as the null space of R is { 2x 3 x 3 x 3 : x 3 is arbitrary 2 Letting x 3 = gives a basis (v) with v =. (e) The pivot columns are columns, 2, and 4, so columns, 2, and 4 of the original matrix A form a basis of the column space of A. (g) There was only one row exchange done in part (a) in finding the row echelon form B, namely, swapping rows and 2. This gives a row exchange matrix P =. The matrix L so that L P A = B is obtained by applying the row operations (in the same order) done in part (a) to the identity matrix I 3. Then A = (P ) (L ) B = P LB. (P ) = P, and the inverse of L is found by doing the inverse row operations in the reverse order needed for L. Thus, }. 5 5

6 L =, and A = P LB = Find all of the eigenvalues and eigenvectors for the matrix 2 3 A =. 3 Solution. The matrix A is upper triangular, so the eigenvalues are just the diagonal entries. That is the eigenvalues are 2,, and 3. For each eigenvalue λ the eigenspace is found by computing null(a λi). 3 λ = 2: In this case A λi = A 2I = 2. Row reduce this matrix to get. The null space of this matrix is span, which is the eigenspace of the eigenvalue λ = : In this case A λi = A + I =. Row reduce this matrix to get 4 3. The null space of this matrix is span 3, which is the eigenspace of the eigenvalue. 3 λ = 3: In this case A λi = A 3I = 4. Row reduce this matrix /4 to get /4. The null space of this matrix is span, which is the 4 eigenspace of the eigenvalue Consider the forward shift operator T L(F ) given by T (a, a 2, a 3,...) = (, a, a 2,...). Show that T has no nonzero eigenvectors, and hence no eigenvalues. 6 6

7 Solution. Suppose the equation T v = λv is satisfied for some v = (a, a 2, a 3,...) =. Then λ(a, a 2, a 3,...) = (, a, a 2,...). This means that λa = λa 2 = a λa 3 = a 2. Since v = there is some index j such that a j =. Choose the smallest j such that a j =. If j = then λa = and a = implies that λ =. If j > then λa j = a j = so again λ =. Thus, if there is an eigenvector, then the associated eigenvalue must be λ =. But then, = λa j+ = a j =. This contradiction shows that the equation T v = λv has no solutions with v =. 6. Perform Gram-Schmidt orthogonalization on the list ((,, ), (,, 3), (,, 2)), using the standard inner product on R 3. Solution. e = 3 (,, ) v 2 = (,, 3) 3 (,, ) = (, 2, 2), 3 so that e 2 = 2 (,, ). v 3 = (,, 2) = 3 3 (,, ) 2 (,, ) = (,, ). 2 This means that v 3 span(v, v 2 ) so the list is not linearly independent, and Gram- Schmidt cannot produce an orthonormal basis 7. Perform Gram-Schmidt on the list (z, z 2 ) with respect to the inner product p, q = p(z)q(z) dz. Solution. z, z = zz dz = z2 dz = /3 so e = (/ 3)z. v 2 = z 2 z2, z z, z z = z2 /4 /3 z = z2 3 4 z. 7 7

8 Then v 2, v 2 = Thus, (z z)(z2 3 4 z) dz = (z z z2 ) dz = = 8. e 2 = 4 5(z z) = 5(4z 2 3z). Thus, the orthonormal list coming from (z, z 2 ) is (z/ 3, 5(4z 2 3z)). 8. Let T L(V ) for a finite-dimensional vector space V. Show that T is invertible if and only if is not an eigenvalue of T. Solution. By Theorem 3.2 (page 57), T L(V ) is invertible if and only if T is injective. By Proposition 3.2 (page 43) T is injective if and only if null(t ) = {}. But λ is an eigenvalue of T if and only if null(t λi) = {}, so is not an eigenvalue of T if and only if null(t ) = {}, which is true if and only if T is invertible. 9. Let T : R 3 R 3 be defined by T (x, y, z) = (y + z, x + z, x + y). (a) Find the matrix of T with respect to the standard basis. Solution. T (,, ) = (,, ), T (,, ) = (,, ), and T (,, ) = (,, ). Therefore, M(T ) =. (b) Is T invertible? Justify your answer. Solution. Row reducing the matrix M(T ) produces the matrix I 3 which shows that null(m(t )) = {}, which implies that null(t ) = {}. By proposition 3.2 T is injective, and then by Theorem 3.2, this implies that T is invertible. (c) What are the eigenvalues of T? Solution. λ is an eigenvalue of T if and only if T (x, y, z) = λ(x, y, z) for some (x, y, z) = (,, ). Thus, we must have a nonzero solution to the system of equations y + z = λx x + z = λy x + y = λz. 8 8

9 Rewrite this as a system of homogeneous equations: λx + y + z = x λy + z = x + y λz = This system of equations has a nonzero solution if and only if the rank of the coefficient matrix is less than 3. The coefficient matrix is λ M(T ) λi = λ λ Attempt to row reduce this matrix, being careful not to divide by : Swap rows and 2: λ λ λ Add λ times row to row 2 and subtract row from row 3: λ λ 2 λ + λ + λ Case : If λ + = then this matrix is which has rank and hence a null space of dimension 2. Therefore, λ = is an eigenvalue of T with eigenspace the plane x + y + z =, (which has a basis ((,, ), (,, ))). Case 2: If λ + =, then divide rows 2 and 3 by λ + to get λ λ Continue to row reduce this matrix by swapping rows 2 and 3: λ λ Add λ times row 2 to row 3 to get λ λ

10 If λ = 2 then the last row is and the rank of this matrix is 2, so 2 is an eigenvalue. If λ = 2 then the last row of the matrix can be divided by λ + 2 to get λ, which has rank 3. Therefore, null(t λi) = {} except for λ = 2,. Thus, the eigenvalues of T are 2 and -. (d) Describe the eigenspaces of T for each eigenvalue. Solution. We showed in the previous part that the eigenspace of λ = is the plane x + y + z = with basis ((,, ), (,, )). For the eigenvalue λ = 2, substitute λ = 2 in the next to last matrix of part (c) to get Continue row reducing this matrix to get. From this we can read off that null(t 2I) = {x 3 (,, ) : x 3 is arbitrary}. So 2 is an eigenvalue of T and the eigenspace is span((,, )). (e) Find a basis of R 3 consisting of eigenvectors of T, or show that no such basis exists. Solution. Combine the bases for the two eigenspaces from part (d) to get a basis ((,, ), (,, ), (,, )) of R 3 consisting of eigenvectors of T.