Final Exam Linear Algebra Summer Math SX (3) Corrin Clarkson August th, Name: Solutions Instructions: This is a closed book exam. You may not use the textbook, notes or a calculator. You will have 9 minutes to complete this exam. Unless otherwise stated you must show your work in order to receive full credit for your solution. Partial credit will be given for incomplete solutions. Problem Problem Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Total 6 points 4 points points points 6 points points 4 points 8 points
. ( points each) Determine which of the following statements are true. Circle T for true and F for false. You do not need to justify your conclusions. T F If v is an eigenvector of a matrix A, and B is similar to A, then v is also an eigenvector of B. False, similar matrices have the same eigenvalues, [ ] but [ may not ] have the same eigenvectors, for example and. T F If the linear system A x = b has a unique solutions for every vector b in the range of A, then A must be a square matrix. True, A must be an invertible matrix. T F If a matrix A is diagonalizable, then its transpose A T diagonalizable as well. must be True, A diagonalizable means A is similar to a diagonal matrix D. It follows that A T = (S DS) T = S T D T (S ) T = S T D(S T ) Thus A T is similar to D. T F There exists a linear transformation T from P to P such that the kernel of T is isomorphic to the image of T. False, P is three dimensional, so by the rank nullity theorem the rank of T can not equal the nullity of T. T F The matrix ] represents a rotation combined with a scaling. [ 5 6 6 5
[ ] [ ] 5 6 True, the columns and are orthogonal and have the 6 5 ([ ]) 5 6 same norm. Also, the determinant det = 6 is positive. 6 5 T F If A is a 3 3 matrix, then det(3a) = 3 det(a). False, det(3a) = 3 3 det(a). T F If V and W are linear spaces and T : V W is a linear map such that both im(t ) and ker(t ) are finite dimensional, then V must be finite dimensional. True, one can build a finite basis fro V by taking a basis for the kernel and appending elements that map to a basis of the image. T F The image of a cube in R 3 under an orthogonal map T : R 3 R 3 is another cube. True, orthogonal maps preserve lengths and angles. 3
. (4 points) Consider the matrix A = 3 4 5 (a) Find the eigenvalues of T and their algebraic and geometric multiplicities Solution: The characteristic polynomial of A is f A (λ) = det(a λi 3 ) 3 λ 4 = det 5 λ λ = (3 λ)(5 λ)( λ) 4(5 λ)( ) = (5 λ)[(3 λ)( λ) + 4] = (5 λ)(λ λ + ) = (5 λ)( λ) Therefore the eigenvalues of A are 5 and. Their algebraic multiplicities are and respective.y. The geometric multiplicity of 5 must be one, because the geometric multiplicity is bounded above by the algebraic multiplicity. To determine the geometric multiplicity of, consider the matrix 4 A I 3 = 4 This matrix has rank and nullity. It follows that he geometric multiplicity of is. (b) Find an eigenvector for each eigenvalue. Solution: is an eigenvector with eigenvalue 5 and is an eigenvector with eigenvalue. 4
(c) Is A diagonalizable? Solution: No, A is not diagonalizable, because the geometric multiplicity of is not the same as its algebraic multiplicity. (d) Compute the determinant of A. Solution: The determinant of A is 5, the product of its eigenvalues. 5
3. ( points) Let A be an skew symmetric n n matrix. Prove by induction that A k is symmetric for k even and antisymmetric for k odd. Proof. First, I will induct on k to show that (A k ) T = (A T ) k. The claim is trivial when k is one. Assuming (A k ) T = (A T ) k for some k, we must prove that (A k+ ) T = (A T ) k+. Therefore (A k ) T = (A T ) k for all k. (A k+ ) T = (A k A) T = A T (A k ) T = A T (A T ) k = (A T ) k+ A skew symmetric means that A T = A. It follows that { A (A T ) k = ( A) k = ( ) k A k k if k is even = A k if k is odd Thus A k is symmetric if k is odd and antisymmetric if k is even. 6
4. ( points) This is a short answer problem; you do not need to show your work. (a) Give an example a 3 3 matrix A such that 3 is in the kernel 5 of A and is in the image of A. Solution: 3 5 (b) Give an example of a matrix A such that im(a) im(rref(a)). Solution: [ (c) Give two different examples of 4 4 matrices with characteristic polynomial (3 λ) 4 such that the geometric multiplicity of the eigenvalue 3 is two. ] Solution: 3 3 3 3 and 3 3 3 3 (d) Give an example of an infinite dimensional linear space. Solution: Sequences of real numbers, continuous function from R to R, polynomials R[x]. 7
5. (6 points) Let C[π, π] be the space of continuous functions on the interval [π, π] equipped with its usual inner product, f, g = π f(t)g(t) dt. π π Consider the functions f(t) = 5 sin(t)+3 cos(5t)+ sin(5t) and g(t) = 9+ sin(t)+cos(t) 5 sin(5t) Compute the norm of f and determine whether or not f and g are orthogonal Solution: Recall that the functions sin(nt) and cos(mt) form an orthonormal set in C[ π, π]. Thus we can compute the necessary inner products by considering the coefficients of the sines and cosines in f and g. f = 5 + 9 + 4 = 38 and g = 8 + 4 + + 5 = The are orthogonal, because f, g = + + + = 8
6. ( points) Fit a linear function of the form f(x) = mx + b to the data (, ), (, ), (, ). Graph your solution with the data points and explain why it is the best fit. Solution: The three data points give the linear system: = m() + b = m() + b = m() + b Thus we are looking for a vector v = [ m b ] R such that v = The least squares solution to this linear system is v = (A T A) A T where A = Thus we can compute v as follows v = = [ ] = [ = 3 [ ] [ 3 [ 3 ] [ ] 5 ] [ ] ] We conclude that the best fit linear function is f(x) = x + 3. 9
7. (4 points) Consider the function F : P R 3 that maps a polynomial f(3) P to the vector f (3) f (3). (a) Show that F is a linear transformation. Solution: This function is linear, because evaluation a polynomial at a point and taking derivatives are both linear operation. To check this, prove that F satisfies the conditions for being a linear function: i. F (f + g) = F (f) + F (g) for all polynomials f and g. ii. F (af) = af (f) for all polynomials f and scalars a R. (b) Compute the determinant of F. Solution: First evaluate F on each of the elements of the basis {, t, t } 3 9 F () = and F (t) = and F (t ) = 6 The determinant of F is then detf = det 3 9 6 = (c) Determine whether or not F is an isomorphism. Solution: F is invertible, because it has a nonzero determinant.