Math 110: Worksheet 13 Solutions

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1 Math 110: Worksheet 13 Solutions November 29 Thursday Nov. 16, Tuesday Nov. 21: 6.3 & Let V = {f : R R f is infinitely differentiable and is zero outside a bounded subset of R}. Equip V with the inner product f, g = f(x)g(x) dx. Let D be the differentiation operator on V. Assuming it exists, find D. Note that D (f), g = f, D(g) = f(x)g (x) dx = [f(x)g(x)] f (x)g(x) dx = f (x)g(x) dx ( integration by parts) where we used the fact that h vanishes at ± if h V. As this holds for all g, it follows that D (f) = f for all f V so D = D. 2. Let V = M n n (C) be equipped with the inner product A, B = tr(b A). Let P be a fixed invertible matrix in V, and let T P be the linear operator on V defined by T P (A) = P 1 AP. Find the adjoint of T P. We have T P (A), B = A, T P (B) = A, P 1 BP = tr((p 1 BP ) A) = tr(p B (P ) 1 A) = tr(b (P ) 1 AP ) ( cyclic property of trace) = (P ) 1 AP, B. As this holds for all B V, it follows that T P (A) = (P ) 1 AP for all A V so T P = T P. 1

2 3. Suppose V is a finite-dimensional real vector space and T L(V ). Prove that V has a basis consisting of eigenvectors of T if and only if there is an inner product on V that makes T into a self-adjoint operator. Observe first that if T is self-adjoint with respect to some inner product on V, then it is orthogonally diagonalizable; in particular, it is diagonalizable, that is, V has a basis consisting of eigenvectors of T. Conversely, assume that V has a basis {v 1,..., v n } consisting of eigenvectors of T. Define a i v i, b j v j = a i b i. It is straightforward to check that, is an inner product on V. Note that ( ) T a i v i, b j v j = a i λ i v i, b j v j = λ i a i b i and ( ) a i v i, T b j v j = a i v i, b j λ j v j = λ i a i b i so that T x, y = x, T y for all x, y V. We conclude that T is self-adjoint with respect to this inner product. 4. Let V be a finite-dimensional inner-product space. Suppose P L(V ) is such that P 2 = P. Prove that ker(p ) = Im(P ) if and only if P = P. (Such a linear transformation is called an orthogonal projection) Assume first that P = P. Observe then that x ker(p ) P x = 0 P x, y = 0 for all y V x, P y = 0 for all y V x, P y = 0 for all y V x Im(P ). Conversely, assume that ker(p ) = Im(P ). This implies that V = ker(p ) Im(P ). Let x, y V ; we can then write x = x 1 + P (x 2 ) and y = y 1 + P (y 2 ), where x 1, y 1 ker(p ). We then have P (x), y = P (x 1 + P (x 2 )), y 1 + P (y 2 ) = P 2 (x 2 ), y 1 + P (y 2 ) = P (x 2 ), P (y 2 ) where we used the fact that P 2 (x 2 ) = P (x 2 ) is orthogonal to y 1. Similarly, we have x, P (y) = x 1 + P (x 2 ), P (y 1 + P (y 2 )) = x 1 + P (x 2 ), P 2 (y 2 ) = P (x 2 ), P (y 2 ). Thus, P (x), y = x, P (y) so we conclude that P = P. 2

3 Tuesday Nov. 28: 6.4 & Prove that a normal operator on a complex inner-product space is self-adjoint if and only if all its eigenvalues are real. Observe first that if an operator is self-adjoint, all its eigenvalues are necessarily real by one of the results from class. For the converse, note first that a normal operator is orthogonally diagonalizable over C. As all the eigenvalues are real, it follows that the operator is in fact orthogonally diagonalizable over R and hence self-adjoint. 6. Prove or give a counterexample: if S L(V ) and there exists an orthonormal basis {e 1,..., e n } of V such that S(e j ) = 1 for each j, then S is unitary. We give a counter-example. Let V = R 2 with the standard inner product and let {e 1, e 2 } be the standard basis. Define S L(V ) by S(e 1 ) = e 1 and S(e 2 ) = (3e 1 + 4e 2 )/5. Note then that S(e j ) = 1 for all j but S is not unitary as S(e 1 ), S(e 2 ) = e 1, (3e 1 + 4e 2 )/5 = 3/5 0 = e 1, e Let W be a subspace of a finite-dimensional inner product space V, and let T be a reflection across W : in other words, if w W and w W, then T (w+w ) = w w. (a) Draw a picture illustrating such a transformation in R 2. If W is a line, then it s reflection orthogonally across the line. If W = {0}, then every vector gets sent to its negative. (b) Prove that T is self-adjoint. For any w + w and any z + z in V, we have T (w+w ), z+z = w w, z+z = w, z w, z = w+w, z z = w+w, T (z+z ). (c) Prove that T is normal. Self-adjoint operators are normal. (d) Prove that T is unitary. Since T is self-adjoint, it follows for any w + w that T T (w + w ) = T 2 (w + w ) = T (w w ) = w + w. Therefore, T T = I. 3

4 (e) What are the eigenvalues and eigenvectors of T? Since T 2 = I, the eigenvalues of T are all ±1. Any vector in W is an eigenvector for 1 and any vector in W is an eigenvector for Suppose you pick up a (perfectly spherical) basketball, spin it around for a while, then put it down in its original spot. Prove that the surface of the ball will have at least two fixed points that is, two points that are in the exact same location that they first were. (a) Argue that the act of spinning the basketball can be described by an orthogonal linear transformation Q on R 3. We ll skip over the argument that it s a linear transformation and focus on orthogonality. If we take the origin to be the center of the ball, then the fact that the ball is a perfect sphere implies that Qv = v for every point v on the surface (and therefore, by linearity, every point in R 3 ). Thus Q must be orthogonal. (b) Using the characteristic polynomial, prove that the matrix Q has at least one real eigenvalue λ. The characteristic polynomial det(q ti) is a real degree-3 polynomial, which tends to as t and as t. By the Intermediate Value Theorem, there exists some t R such that det(q ti) = 0. In other words, Q has a real eigenvalue. (c) Prove that either λ = 1 or λ = 1. Since Q is orthogonal, we know that λ = 1 for any eigenvalue λ of Q. Any real eigenvalue must therefore be 1 or 1. (d) Prove that λ = 1. It may help to consider the relation between 3 3 matrix determinants and triple products. Let q 1, q 2, q 3 be the columns of Q, so that q 1 = Q(e 1 ), q 2 = Q(e 2 ), and q 3 = Q(e 3 ). Then since Q is a rigid rotation the triple product (q 1 q 2 ) q 3 should (intuitively?) be the same as the triple product (e 1 e 2 ) e 3. Therefore, det(q) = (q 1 q 2 ) q 3 = (e 1 e 2 ) e 3 = 1. Since Q is real, its complex eigenvalues must come in conjugate pairs. If Q does not have 1 as an eigenvalue, then its only possible eigenvalue combinations are ( 1, 1, 1) and ( 1, µ, µ). Both of these have determinant 1, so if det(q) = 1 then it must have 1 as an eigenvalue. 4

5 (e) Conclude that the basketball has at least two fixed points. Since Q has 1 as an eigenvalue, there are at least two eigenvectors v and v on the sphere. So Q(v) = v and Q( v) = v. (f) Why does a similar result not hold for spinning a frisbee (R 2 )? Find a rotation with no fixed points, and explain why the above proof does not work. Any non-trivial rotation of the circle will have no fixed points. The above proof does not carry over because we needed the characteristic polynomial of Q to have odd degree in order to show that Q had a real eigenvalue. (g) If the surface of the basketball has more than two fixed points, prove that Q = I and so the ball is exactly in its original orientation. If the surface has more than two fixed points, the eigenspace corresponding to 1 must be at least 2-dimensional. So the only eigenvalue options are (1, 1, 1) and (1, 1, 1). Since only the first lets Q have determinant 1, Q must be the identity matrix. 5