MATH 115A: SAMPLE FINAL SOLUTIONS

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1 MATH A: SAMPLE FINAL SOLUTIONS JOE HUGHES. Let V be the set of all functions f : R R such that f( x) = f(x) for all x R. Show that V is a vector space over R under the usual addition and scalar multiplication of functions. Solution: Let W be the set of all functions f : R R. We have seen that W is a vector space with respect to addition and scalar multiplication of functions, so it suffices to show that V is a subspace of W. Certainly 0 V. And if f, g V and c R, then (cf + g)( x) = cf( x) + g( x) = cf(x) + g(x) = (cf + g)(x) Therefore cf + g W, so W is a subspace.. Let V be a vector space, and L(V ) the space of all linear transformations on V. Show that V is finite-dimensional if and only if L(V ) is finite-dimensional. Solution: Suppose that V is a finite-dimensional vector space, of dimension n, and choose a basis {v,..., v n } for V. Then L(V ) has dimension n, since it has a basis given by the set {E i,j : i, j n}, where E i,j is defined by E i,j (v k ) = 0 if j k, and E i,j (v k ) = v i if j = k. (If we identify V with F n via the basis for V, then E i,j is the matrix having a in the i, j position and 0 everywhere else). Therefore L(V ) is finite-dimensional. For the other direction, suppose that L(V ) is finite-dimensional, and choose some basis {e i } i I for V. Note that we are not assuming this basis is finite, since after all that is what we are trying to prove. Fix an element e in this basis, and define a linear map T : V L(V ) as follows: set T (v) = f v, where f v (e i ) = v if e i = e, and f v (e i ) = 0 otherwise. Note that so T is linear. T (cv + w) = f cv+w = cf v + f w = ct (v) + T (w) I claim that T is one-to-one. Indeed, if T (v) = 0, then f v L(V ) is the zero map. In particular, f v (e ) = 0. But by definition f v (e ) = v, so v = 0. Therefore N(T ) = {0}, so T is one-to-one.

2 MATH A: SAMPLE FINAL SOLUTIONS Thus T is one-to-one, so T : V R(T ) is an isomorphism. But R(T ) is a subspace of the finite-dimensional vector space L(V ), so is also finite-dimensional. Thus V is finitedimensional because it is isomorphic to R(T ).. Let (V,, ) be a finite-dimensional inner product space of dimension n. Let S = {v,..., v k } be a linearly independent subset of V. Define T : V V by T (x) = x, v i v i a. Show that T is linear. Solution: Choose two vectors x, y V and a scalar c. Then ( ) T (cx + y) = cx + y, v i v i = c x, v i + y, v i v i = c x, v i v i + i= i= i= = ct (x) + T (y) using the properties of the inner product. Thus T is linear. b. What is the kernel of T? Solution: Suppose that T (x) = 0. By definition, this means that x, v i v i = 0 i= i= y, v i v i and as the set S is linearly independent, it follows that x, v i = 0 for all i. Therefore ker(t ) = {x V : x, v i = 0, i k} Recall the definition of the orthogonal complement of the subspace span(s): span(s) = {y V : y, z = 0 for all z S} I claim that the subspaces span(s) and {x V : x, v i = 0, i k} are equal. On the one hand, if x span(s), then x, v i = 0 for all i since each v i is in span(s). On the other hand, every element of span(s) is a linear combination of the v i, so if x, v i = 0 for all i then x, y = 0 for all y span(s). Therefore c. What is the image of T? ker(t ) = {x V : x, v i = 0, i k} = span(s) Solution: By the definition of T, for any x V we have T (x) = x, v i v i span(s) i= i=

3 MATH A: SAMPLE FINAL SOLUTIONS Thus R(T ) span(s). To show that this inclusion is an equality, it suffices to show that the two subspaces have the same dimension. Since S is a linearly independent set, span(s) has dimension k. And by the dimension theorem, dim R(T ) = dim V dim N(T ) = n dim span(s) = n (n k) = k Thus dim R(T ) = k = dim span(s), so R(T ) = span(s). 4. Let α = {v, v, v } be a basis of R, with the standard inner product. Find an invertible matrix A M (R) such that β = {Av, Av, Av } is an orthogonal basis for V. Solution: I think it s easiest to find the orthogonal basis we want first, then figure out what the matrix should be. To construct an orthogonal basis out of α, use the Gram-Schmidt process: set u = v, and u = v v, u u u = v v, u u v u = v v, u u u v, u u u = v v, u u v v, u ( u v v, u u v = v v, u u v + β = {u, u, u } is our desired orthogonal basis. ( v, u v, u u u v, u ) u v Now let B be the matrix changing β coordinates into α coordinates, so v,u v,u v,u u u u v,u u B = v 0,u u 0 0 and let Q be the matrix changing α coordinates into the standard coordinates. A = QBQ is the desired matrix. ) Then Let s consider an example to clarify what s going on. Let v = (,, 0), v = (, 0, ), v = (0, 0, ). Then u = v, u = v ( ) v =,, and The matrix B is u = v u = v v + v = (,, )

4 and similarly A computation gives MATH A: SAMPLE FINAL SOLUTIONS 4 Q = Q = A = QBQ = and to check that this works, note that and finally Av = Av = Av = = = u = 0 0 = = u = u. Let T be a linear transformation of a vector space V over R. a. Let λ be an eigenvalue of T. Show that λ is an eigenvalue of T. Solution: Since λ is an eigenvalue of T, there is a non-zero vector v V such that T v = λv. Then by linearity, T v = T (T v) = T (λv) = λt (v) = λ v which shows that λ is an eigenvalue of T. b. If T = I, then show that the only possible eigenvalues of T are ±. Solution: Suppose that λ is an eigenvalue of T, and choose a non-zero vector v such that T v = λv. Then by part (a) we know that T v = λ v. On the other hand, T = I, hence v = Iv = T v = λ v Therefore ( λ )v = 0, and since v 0, this implies that λ =, or λ = ±.. Let A = M (R). a. Find all eigenvalues of A.

5 MATH A: SAMPLE FINAL SOLUTIONS Solution: The eigenvalues of A are the roots of the polynomial det(xi A). To compute this determinant, we use expansion by minors in the third column: x + x x = x x + + (x ) x + x = 9 x+ ( x +9)+(x )(x 4x +9) = (x )(x 4x+4) = (x )(x ) Therefore the eigenvalues of A are and. b. Find bases for the eigenspaces of A, and show that A is diagonalizable. Solution: The eigenspaces of A are the kernels of A I and A I. have A I = 4 0 For λ =, we so the vector (,, ) is in the kernel. Moreover, the kernel has dimension one since the algebraic multiplicity of λ = is. Therefore (,, ) is a basis for the kernel. For λ =, A I = and the vectors (,, 0), (0,, ) are a basis for the kernel. The set {(,, ), (,, 0), (0,, )} is a basis of R consisting of eigenvectors for A, so A is diagonalizable. c. Find an invertible matrix Q and a diagonal matrix D such that D = Q AQ. Solution: Take Q to be the matrix Then a computation gives Q = 0 0 Q = 0 and Q AQ = 0 =

6 which is the desired diagonal matrix D. MATH A: SAMPLE FINAL SOLUTIONS = Let V = P (R) be the space of all polynomials with coefficients in R having degree at most. Define an inner product on V by f, g = a. Show that W = P (R) is a subspace of V. 0 f(x)g(x) dx Solution: W contains the zero polynomial, and if f, g are two polynomials of degree then the degree of cf +g is also at most. Therefore cf +g W, so W is a subspace. b. Let α = {, X, X }. Use the Gram-Schmidt process on α to find an orthogonal basis α of W. Solution: Let u =. Next, define Finally, set u = X X, u u u = X u = X X, u u u X, u u u = X X + Then α = {u, u, u } is an orthogonal basis for W. c. Extend α to an orthogonal basis β of V. Solution: β = {, X, X, X } is a basis for V, so we continue the Gram-Schmidt process one more step to get an orthogonal basis β. So take u 4 = X X, u u u X, u u u X, u u u = X X + X 0 Then β = {u, u, u, u 4 } is an orthogonal basis for V. 8. Let V be a three-dimensional vector space and T a linear operator on V. Show that there is a basis β for V such that [T ] β is upper triangular if and only if there are subspaces V, V V such that (i). V and V are stable under T (i.e. T (V i ) V i ); (ii). {0} V V V.

7 MATH A: SAMPLE FINAL SOLUTIONS 7 Solution: Suppose that there is a basis β = {v, v, v } for which [T ] β is upper triangular, say [T ] β = a a a 0 a a 0 0 a Let V = span{v } and V = span{v, v }. Then {0} V V V since V is a one-dimensional vector space and V is a two-dimensional vector space. Moreover, we know that [v ] β = 0 [v ] β = Therefore by using the matrix representation of T, we see that T v = a v and T v = a v + a v. Thus V and V are stable under T. For the converse, suppose that there are subspaces V, V of V such that V and V are stable under T and {0} V V V. By counting dimensions, we see that V must have dimension and V must have dimension. Choose a non-zero vector v V. Then V = span{v } since V has dimension, and since V is stable under T, we know that T v V = span{v }, i.e. T v = a v for some scalar a. Next, since V has dimension we may choose v V such that v span{v }. Then {v, v } is a basis for V. And T v V = span{v, v } since V is stable under T, hence we can write T v = a v + a v for some scalars a, a. Finally, choose v V not in V = span{v, v }. Then β = {v, v, v } is a basis for V, and T v = a v + a v + a v for scalars a, a, a. Thus [T ] β = a a a 0 a a 0 0 a is upper triangular, as desired.