Math 4153 Exam 3 Review. The syllabus for Exam 3 is Chapter 6 (pages ), Chapter 7 through page 137, and Chapter 8 through page 182 in Axler.

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1 Math 453 Exam 3 Review The syllabus for Exam 3 is Chapter 6 (pages -2), Chapter 7 through page 37, and Chapter 8 through page 82 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 3: Eigenvalues, eigenvectors, and eigenspaces Inner product spaces, norms Orthogonal complements and projections Adjoint operators and matrices in orthonormal bases Self-adjoint and normal operators The spectral theorem for self-adjoint and normal operators on complex vector spaces The spectral theorem for self-adjoint operators on real vector spaces Generalized eigenvectors and generalized eigenspaces dim V The fact that the generalized eigenspace V (λ) = (T λi) Nilpotent operator The decomposition theorem (Theorem 8.23, Page 74). Multiplicity of an eigenvalue (Page 7) Characteristic polynomial of an operator T. Cayley-Hamilton Theorem Block decomposition of an operator into blocks which are upper triangular (Theorem 8.28) Square root of operators (Theorem 8.32) Minimal polynomial of an operator T 3. Definitions (there is a lot of repetition with the above section): Eigenvalues and eigenvectors of linear transformations and matrices Generalized eigenvectors and 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 Orthogonal and orthonormal lists and bases Orthogonal complement

2 Math 453 Exam 3 Review Projection and orthogonal projection Adjoint operator Self-adjoint and normal operator Characteristic polynomial of a complex operator or matrix in terms of uppertriangular form Nilpotent operators and matrices (T is nilpotent means T m = for some m ) Characteristic polynomial of an upper-triangular matrix or of a linear transformation with an upper triangular matrix form. That is, if V = λ V (λ), then χ T (x) = q(z) = λ (z λ)dim V (λ). Minimal polynomial (Page 79) 4. Computations: know how to compute Given an eigenvalue, compute the eigenspace using Gaussian elimination. Compute the orthogonal complement of a vector or a vector subspace Compute the orthogonal projection of a vector to a subspace Given the eigenvalues, compute the generalized eigenspaces using Gaussian elimination Compute the characteristic polynomial of a linear transformation or matrix 5. Major results. Know the statements, but don t memorize them word for word know what they say mathematically. Especially important ones are in bold. Orthogonal complements: Theorem 6.29 Existence of Adjoints: Theorem 6.45 and following discussion. The spectral theorem: Theorems 7.9 and 7.3. The decomposition theorem: Theorem Statement that dim(v (λ)) = d λ : Theorem 8.. Square roots: Theorem 8.32 and Exercise, page

3 Math 453 Exam 3 Review Review Exercises. True/False: If T is normal on a finite-dimensional complex inner product space, then all generalized eigenvectors of T are eigenvectors. Solution. True. This is because a normal operator has an orthonormal basis of eigenvectors, so that V = U U m where U i = null(t λ i I) is the eigenspace of T with eigenvalue λ i. By Theorem 8.2.3, V = V (λ ) V (λ m ) where V (λ i ) = null(t λ i I) dim V is the generalized eigenspace of T with eigenvalue λ i. Since U i V (λ ) for all i, the only way the two direct sums can equal the same space V is if U i = V (λ i ). That is, if every generalized eigenvector is in fact an eigenvector. 2. True/False: If (v, v 2 ) is an arbitrary basis of C 2 with the standard inner product, then M(T, (v, v 2 )) = M(T, (v, v 2 )) t. Solution. False. This is only true for orthonormal bases. 3. True/False: If T is self-adjoint, then T is normal. Solution. True If T = T, then certainly T T T 2. = T T since both are equal to 4. True/False: If T is normal, then T is self-adjoint. [ ] a b Solution. False. If T has the matrix M(T ) = A =, with respect to b [ a ] a b the standard orthonormal basis on C 2 then M(T ) = A =. If b =, then b a A = A so T is not self-adjoint. However, AA = A A so T is a normal operator. 5. True/False: If a linear operator T has a nonzero generalized eigenvector of eigenvalue λ, then it also has a nonzero eigenvector of eigenvalue λ. Solution. True. If v is a nonzero generalized eigenvector of T with eigenvalue λ, the (T λi) k v = for some k. Choose k so that (T λi) k v = but (T λi) k v =. If w = (T λi) k v then w = and (T λi)w = (T λi)(t λi) k v = (T λi) k v =. Thus w is a nonzero eigenvector of T with eigenvalue λ. 6. If V is a finite dimensional complex vector space and T L(V ), prove that T is nilpotent if and only if the characteristic polynomial of T is z m where m = dim V. 3 3

4 Math 453 Exam 3 Review Solution. If T is nilpotent, then we claim that every eigenvalue of T is. To see this, suppose that λ is an eigenvalue of T. Then there is a vector v = with T v = λv. Since T is nilpotent, there is some positive integer k such that T k v =. Choose k so that T k v =, but T k v =. Then apply T k to the equation T v = λv to get = T k v = T k T v = T k (λv) = λt k v. Since T k v = this implies that λ =. Thus, the only eigenvalue of a nilpotent dim V T is, and the multiplicity of this eigenvalue is dim null T = dim V = m so the characteristic polynomial is z m. Conversely, suppose that the characteristic polynomial of an operator is z m. This means that the only eigenvalue is with multiplicity m. That is, the dimension of the generalized eigenspace of the eigenvalue λ = is m = dim V. Thus, every vector is a generalized eigenvector with eigenvalue, and by Theorem 8.23 (c), this means that T is nilpotent. 7. Let V be a finite-dimensional inner product space, and let V = U U. Show that the orthogonal projection operator P U, U is self-adjoint. Solution. To show that T is self-adjoint, it is sufficient to show that T v, w = v, T w for all v, w V. Let v = v + v 2 and w = w + w 2 where v, w U and v 2, w 2 U. Then P U, U v, w = v, w + w 2 = v, w + v, w 2 = v, w since v, w 2 = because v U and w 2 U. Also, v, P U, U w = v + v 2, w = v, w since v 2, w =. Thus, P U, U v, w = v, P U, U w for all v, w V, so P U, U is self-adjoint. Alternative Solution. Choose an orthonormal basis (v,..., v k ) of U and an orthonormal basis (w,..., w l ) of U. Then B = (v,..., v k, w,..., w l ) is an orthonormal basis of V since V = U U. Then P U, U v j = v j for j k and P U, U w j = for j l. Thus, the matrix of P U, U [ ] is the symmetric matrix Ik with the first k diagonal entries equal to and the last l diagonal entries equal to. Since the matrix of P U, U with respect to an orthonormal basis is symmetric, it follows that the operator is self-adjoint. (See Proposition 6.47.) 8. For the matrix A = compute the eigenvalues, the characteristic polynomial, the multiplicity of each eigenvalue, and bases for each generalized eigenspace. 4 4

5 Math 453 Exam 3 Review Solution. Since the matrix A is upper triangular, the eigenvalues and their multiplicities can be read from the diagonal entries. Thus is an eigenvalue of multiplicity 3 and is an eigenvalue of multiplicity. Hence, the characteristic polynomial is q(z) = z(z ) 3. λ = : Since the multiplicity of the eigenvalue is, the generalized eigenspace is just the eigenspace, that is null A. To find a basis for the nullspace of A, simply row reduce A to get: From this we find that the null space of A is span(v) where v = 2 3. λ = : Start by finding the eigenspace with eigenvalue by computing null(a I): A I = 2 6 3,., from which we see that null(a I) = span(w ) where w = null(a I) 2 : 4 4 (A I) 2 = 2 3,. Now compute, 5 5

6 Math 453 Exam 3 Review from which we see that null(a I) = span(w, w 2 ) where w w 2 =. Now compute null(a I)3 : 26 (A I) 3 = 2 3, is given above and, from which we see that null(a I) = span(w, w 2, w 3 ) where w and w 2 are given above and w 3 =. Since the multiplicity of the eigenvalue is 3, we have now found all of the generalized eigenspace of A with eigenvalue and a basis of this space is (w, w 2, w 3 ). 9. The matrix 2 A = 3 3 has characteristic polynomial q(z) = (z 2) 3 (z 3). Determine the eigenvalues, the multiplicity of each eigenvalue, and bases for each generalized eigenspace. Solution. From the characteristic polynomial we see that 2 is an eigenvalue with multiplicity 3 and 3 is an eigenvalue with multiplicity. Thus, the generalized eigenspace of the eigenvalue 2 has dimension 3 and the generalized eigenspace of the eigenvalue 3 has dimension, so in particular, it is just the eigenspace of the eigenvalue 3. λ = 3: A 3I = 2, 6 6

7 Math 453 Exam 3 Review, from which we see that null(a 3I) = span(v) where v =. Now for λ = 2: A 2I =, which row reduces to, from which we see that null(a 2I) = span(v, v 2 ) where w = and w 2 =. Now compute null(a 2I) 2 : 2 (A 2I) 2 =, 2 which row reduces to /2 /2, from which we see that null(a 2I) 2 = span(w, w 2, w 3 ) where w =, w 2 = 2, w 3 =. 2 Since the multiplicity of the eigenvalue 2 is 3, this is a basis of the generalized eigenspace. 7 7

Contents. Preface for the Instructor. Preface for the Student. xvii. Acknowledgments. 1 Vector Spaces 1 1.A R n and C n 2

Contents. Preface for the Instructor. Preface for the Student. xvii. Acknowledgments. 1 Vector Spaces 1 1.A R n and C n 2 Contents Preface for the Instructor xi Preface for the Student xv Acknowledgments xvii 1 Vector Spaces 1 1.A R n and C n 2 Complex Numbers 2 Lists 5 F n 6 Digression on Fields 10 Exercises 1.A 11 1.B Definition