1. True/False: Circle the correct answer. No justifications are needed in this exercise. (1 point each)


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1 Math 33 AH : Solution to the Final Exam Honors Linear Algebra and Applications 1. True/False: Circle the correct answer. No justifications are needed in this exercise. (1 point each) (1) If A is an invertible 3 3 matrix, then its rows form a basis of R 3. T (2) For any matrix A, one has dim(im(a)) = dim(im(rrefa))). T We have ker(a) = ker(rref(a)), then apply the ranknullity theorem. (3) There exists a real 2 2 matrix A such that A 2 = I. T Rotation by 90. (4) If A and B are symmetric 3 3 matrices, then AB is symmetric. F A = [ ] 0 1, B = 1 0 [ ] (5) There exists an invertible 3 3 matrix, for which 7 of the 9 entries are identical. T A = (6) If T : R 2 R 2 is a linear transformation of determinant 1, and S R 2 is a line segment of length 1, then the image T(S) is a line segment of length 1 as well. F
2 This would only be true (for all segments S) if T would preserve distances, i.e. if it were orthogonal. (7) If A is an invertible 4 4 matrix, then the unique least squares solution to A x = b is A 1 b. T If so, A T is invertible, so we can multiply on the left with (A T ) 1 in A T A x = A T b. (8) There exist two invertible 2 2 matrices A and B such that det(a + B) = det(a) + det(b). T A = I and B = [ ] (9) The span of m orthonormal vectors is mdimensional. T (10) If A and B are invertible matrices, then AB and BA are similar. T AB = A(BA)A 1. (11) For any 3 3 real matrix there is a real number λ such that A+λI is not invertible.t The characteristic polynomial is of odd degree, so it has a real root a. Take λ = a. (12) Every n n matrix is diagonalizable over the complex numbers. F B = [ ] (13) The trace of a real skewsymmetric matrix is always equal to 0. T All diagonal elements are 0.
3 (14) The eigenvalues of an orthogonal matrix are always real. F No, e.g. rotation by 90. (15) For any matrix A, the product AA T is diagonalizable. T AA T is symmetric. (16) If A and B are square matrices with the same eigenbasis v 1,..., v n, then AB = BA. T (17) If the matrix A has an eigenbasis, then the matrix A 3 A 4I must also admit an eigenbasis. T The same basis works. (18) If A is invertible, then the quadratic form corresponding to A T A is positive definite. T v T A T Av = (Av) T Av = Av 2 > 0 for all v 0. (19) If A is an n n matrix with an eigenvalue λ of geometric multiplicity n, then A has to be a multiple of the identity matrix I. T (20) If A is a symmetric matrix, then its singular values coincide with its eigenvalues. F They are the absolute values of the eigenvalues.
4 In problems 2 through 9, you must justify your answers. 2. (10 points) (a) Is there a 3 3 matrix A such that im(a) = ker(a)? If yes, give an example. If not, justify your answer. (b) Is there a 6 6 matrix A such that im(a) = ker(a)? If yes, give an example. If not, justify your answer. (a) No. We have dim(im(a)) + dim(ker(a)) = 3, so the two cannot be equal. (b) Yes, for example the block matrix (with 3 3 blocks) [ ] 0 I3 A =. 0 0 Both the kernel and the image are the span of the last three standard vectors.
5 3.(10 points) (a) Find a basis for the plane in R 3 whose equation is x 1 + 2x 2 3x 3 = 0. (b) Find the matrix for reflection across the plane in part (a). (a) The columns of 1 2 A = Indeed, since the two vectors are linearly independent and the plane is two dimensional, they must form a basis. (b) The formula for orthogonal projection in that plane is A(A T A) 1 A T = Reflection is 2A(A T A) 1 A T I =
6 4.(10 points) (a) Let A = Find its rank and a basis for its image. (b) Find an orthonormal basis for ker(a). 1 (c) Find the orthogonal projection of the vector v = 0 0 onto ker(a). 0 (a) Rank two. Any two columns form a basis. (b) The kernel must be twodimensional, so to get a basis it suffices to find two linearly independent vectors in it, such as 1 2 1, 0 Then by applying GramSchmidt we get an orthonormal basis 3/10 (c) ( u 1 v) u 1 + ( u 2 v) u 2 = 4/10 1/10. 2/ / 6 u 1 = 2/ 2/ / 6, u 2 = 1/ 30 4/ /. 30
7 5. (10 points) Let A be an invertible n n matrix (with real coefficients) and adj(a) its classical adjoint. (a) Describe the relation between A 1 and adj(a). (b) Show that if adj(a) is symmetric, then A is symmetric. (c) Show that if adj(a) = I and n 2, then A = I or A = I. (a) A 1 = 1 det(a) adj(a). (b) We get (det(a) A 1 ) T = det(a)a 1. Since det(a) = det(a T ) 0, we can divide by it and get that A 1 is symmetric, which implies that so is A. (c) Let d = det(a). Then A 1 = 1 d I. Taking determinants we get d 1 = d n, so d n 1 = 1.Since n is real, we must have d = 1 or 1. We get A 1 = ±I, so A = ±I.
8 6.(10 [ points) ] Let A[ be] the 2 2 matrix with eigenvalues λ 1 = 2 and λ 2 = 1 for which 2 1 v 1 = and v 1 2 = are corresponding eigenvectors. 1 (a) Find A. (b) What are the eigenvalues of A + I? (c) Calculate (A + I) 100. (a) (b) 3 and 0, in fact A = [ ][ ][ ] 1 [ ] = [ ] [ ][ (A + I) = ] 1 (c) [ ][ ][ ] [ ] (A + I) = =
9 7.(10 points) Let A be a 3 3 matrix such that A 2 = 4A 4I. (a) What are all the possible eigenvalues of A? Justify your answer. (b) Give an example of a 3 3 matrix A 2I satisfying A 2 = 4A 4I. (a) If A v = λ v then 0 = (A 2 4A + 4) v = (λ 2 4λ + 4) v so (if v 0) λ = 2 is the only possibility (b) A =
10 8.(10 points) Consider the matrix A = Find an orthogonal matrix S and a diagonal matrix D such that D = S 1 AS. The eigenvalues are 2 and 2, the first with multiplicity two. The (+2)eigenspace is 0 1/ 2 spanned by the orthonormal vectors 1 and 0 0 1/. The ( 2)eigenspace is spanned by 2 1/ 2 the unit vector 0 1/. Thus we can take 2 0 1/ 2 1/ 2 S = / 2 1/ , D =
11 9.(10 points) For c > 0, let 0 c 0 A = c (a) Find the singular values of A. (b) Suppose we know that there exists a vector v R 3 such that v = 10 and A v = 100. What are the possible values of c? Remark: Given an ellipsoid centered at the origin with principal axes of lengths r 1 r 2 r 3, the set of possible distances from a point on the ellipsoid to the origin is the interval [r 1, r 3 ]. (a) 3, c and c + 1. (b) The image of the unit sphere is an ellipsoid with principal axes of lengths 3, c and c + 1. The vector v/10 is on the unit sphere, so A v/10 is on the ellipsoid (and has length 10). Since 10 > 3, the necessary and sufficient condition is that 10 c + 1, that is, c 9.
12 Do not write on this page. 1 out of 20 points 2 out of 10 points 3 out of 10 points 4 out of 10 points 5 out of 10 points 6 out of 10 points 7 out of 10 points 8 out of 10 points 9 out of 10 points Total out of 100 points
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