( 9x + 3y. y 3y = (λ 9)x 3x + y = λy 9x + 3y = 3λy 9x + (λ 9)x = λ(λ 9)x. (λ 2 10λ)x = 0

Transcription

1 Math 46 (Lesieutre Practice final ( minutes December 9, 8 Problem Consider the matrix M ( 9 a Prove that there is a basis for R consisting of orthonormal eigenvectors for M This is just the spectral theorem: the matrix for M is symmetric, so it is self-adjoint, and the real spectral theorem says that there s an orthonormal basis of eigenvectors b Compute the eigenvectors and eigenvalues for M Since we never covered determinants, we need to do this by hand Let s suppose that (x, y is a λ-eigenvector, and solve for these variables ( ( ( 9 x x λ ( 9x + y x + y y y ( λx λy y (λ 9x x + y λy 9x + y λy 9x + (λ 9x λ(λ 9x (λ λx λ(λ x λx λ x 9λx So we must have either λ or λ For λ, we get (x, y (,, or any multiple of this For λ, we get (,, or any multiple Problem Let M (R denote the vector space of matrices with real entries, and define a map T : M (R R by ( T (M M a Prove that T is a linear map We have ( T (M + M (M + M ( T (λm (λm λ That s all we need to check ( ( M + M T (M + T (M ( ( M λ(t (M

2 b What is the dimension of M (R? Describe a basis for this space It s four dimensional, with basis: ( ( ( (,,, c Prove that the nullspace of T is not just {} This is just for dimensional reasons: T is a map from a 4-dimensional space to a -dimensional one, and according to the fundamental theorem, the nullspace has dimension at least d Compute bases for null(t and range(t This is one where we want to use what we know about row reduction We re going to find the matrix for T Notice that ( ( ( ( T ( ( ( ( T ( ( ( ( T The matrix M(T is thus T ( M(T ( ( ( What luck! It s already in rref Let s do range(t first The pivots are columns and That means these span the range, so a basis for range(t is (, and (, I made this easy, but on the real exam don t forget: you need to find rref to figure out what the pivot columns are, but your basis is given by those columns for the original matrix For null(t, the deal is that the free variables x and x 4 can be anything, and we solve for the pivot variables in terms of them: ( x x x x 4 ( ( x + x x + x 4 So x x and x x 4 The solutions are then x x x 4 x + x 4 x 4

3 The two vectors are a basis for the nullspace Switching back to write them as matrices, (( ( null(t span, Problem Let C ([, ] denote the vector space of real-valued continuous function on [, ] Define an inner product by f, g f(xg(x dx a Let U P (R C ([, ] denote the subspace of linear functions Compute an orthonormal basis for P (R We need to use the Gram-Schmidt process There are only two vectors to deal with, so it s not going to be that painful Start with the not-orthonormal basis v, v x Then e v v ( v v, e e x v v, e e ( x x dx x dx e v v, e e v v, e e x / ( x b Compute P U f, where f(x e x (Sorry, this one turns out messier than I intended! We know that P U f f, e e + f, e e First notice that e x x dx ( xe x e x dx e (e P U f f, e e + f, e e ( ( e x dx + e x ( x ( dx ( x ( e x ( + (e ( ( x ( (e + ( e ( ( x

4 Problem 4 Suppose that V and W are two vector spaces a Given a function T : V W, what axioms do you need to check to show that T is linear? Not much: just T (v + w T v + T w for any v, w V and T (λv λ T v for any λ F and v V b Give an example of a linear map which is surjective but not injective Let V be the set of infinite real-valued sequences (a, a, a, Let T : V V be the forget the first entry map T (a, a, a, (a, a, a, This is not injective, since (,,,, is in the kernel It is surjective, since given any (a, a, a,, it s T (, a, a, a, c Give an example of a linear map on a real vector space which has no eigenvalues You can use the rotation map ( ( x y T y x Problem 5 Consider the linear map T : R R whose matrix with respect to the standard basis is a Prove that T is not surjective This is again just for dimensional reasons A map from a -dimensional space to a - dimensional space can t be surjective b Find a basis for range(t We should row-reduce the matrix: We see that both columns are pivot columns, and so the basis is columns and from the original matrix (as opposed to the row-reduced one, 4

5 c Find a basis for range(t This is going to be the nullspace of ( Row reduction is easy: ( ( The pivots are the first two columns, so the third is free ( ( x x x x x x + x So x x and x x The nullspace is x x x x That s our basis vector Problem 6 Suppose that V is a finite-dimensional vector space a Suppose that T L(V is a linear map Define what it means for T to be invertible It means that there is another linear map S so that ST Id V and T S Id V b Suppose that λ is an eigenvalue of T and that T is invertible Prove that /λ is an eigenvalue of T If λ is an eigenvalue of T, there is a nonzero V so that T v λv This shows that /λ is an eigenvalue T v λv T T v T λv v λt v T v λ v c Prove that T and T have the same eigenvectors We essentially just did this We showed that if v is an eigenvector for T with eigenvalue λ, then v is an eigenvector for T with eigenvalue /λ (notice that λ can t be, since then T wouldn t be invertible Problem 7 Suppose that V is an inner product space over R Given an element u V, define a map φ u : V R by setting φ u (v u, v 5

6 a Prove that φ u : V R is linear Fix u V Suppose that v, v V and c is a scalar Then φ u (cv u, cv c u, v cφ u (v φ u (v + v u, v + v u, v + u, v φ u (v + φ u (v, which shows it s linear Notice that since we said it s over the reals, and not the complex numbers, we don t have to take any conjugates This is actually important here b Define Φ : V V be Φ(u φ u Prove that Φ is a linear map Suppose that u, u V, and c R We need to show that Φ(u + u Φ(u + Φ(u, ie that φ u + φ u φ u +u Remember that all of these thinks are functions, ie linear maps V R How do you show two functionals are equal? Well, you have to show that they give the same result on any input v V Suppose that v V Then: (Φ(u + u (v φ u +u (v u + u, v u, v + u, v φ u (v + φ u (v (Φ(u (v + (Φ(u (v Since this holds for any v, we conclude that Φ(u + u Φ(u + Φ(u Similarly, (Φ(cu(v φ cu (v cu, v c u, v cφ u (v (cφ(u(v, and we conclude that Φ(cu cφ(u c Prove that Φ is injective Conclude that if V is finite-dimensional, Φ : V V is an isomorphism Suppose that u null(φ Then Φ(u, which means that φ u, ie that φ u (v for all v But then φ u (u u, u, which implies that u by definiiton of an inner product So Φ is injective If V is finite-dimensional, then because dim V dim V, we conclude that Φ is an isomorphism Problem 8 Suppose that V and W are two vector spaces a Suppose that T : V W is an isomorphism Prove that if v,, v m are a basis for V, then T v,, T v m are a basis for W I first claim that T v,, T v m are linearly independent Suppose that c T v + + c m T v m Then T (c v + + c m v m 6

7 Since T is an isomorphism, this means that c v + + c m v m Since the v i s are a basis, they are linearly independent, which means that each c i is Now, since T is an isomorphism, dim W dim V m Since the m vectors T v,, T v m are linearly independent, they are automatically a basis for W b Suppose that V is a vector space with basis v,, v n Prove that the n vectors v, v +v, v + v + v,, v + + v n is a basis for V Consider the map T : V V given with respect to v,, v m by the matrix M Then T v v, T v v + v, T v v + v + v, Since M is upper triangular and has no on the diagonal, it is invertible, so that T is an isomorphism The claim then follows from (a Problem 9 Suppose that V is a vector space and U and U are two subspaces a Define a direct sum, and give an example of V, U, and U for which U + U is not a direct sum Let V R, and U U V This is not a direct sum, since U U {} b Suppose that U U is a direct sum Prove that (U U /U is isomorphic to U (You do not need to prove your maps are linear, but you should prove they are well-defined Define a map T : U (U U /U by T (u u + U Define S : (U U /U U by T (v + U u, where v u + u is the (unique way to write v as the sum of elements of U and U For T we don t need to worry about well-definedness For the other one, we do If v v, then v v U This means v u + u and v u + u, with the two u s the same This proves that the map is well-defined From the construction, we have ST T S Id, which means the maps are inverses, and hence the spaces are isomorphic Problem Let P n (R denote the space of polynomials of degree less than or equal to n, and define an inner product on this space by setting f, g f(xg(x dx Let D : P(R P(R denote the differentiation map a Consider the restriction D P (R : P (R P (R Give bases for these two spaces, and compute the matrix M(D P (R with respect to your bases 7

8 Let s use the basis for these spaces given by, x, x, x and, x, x Then D( ( + (x + (x, D(x ( + (x + (x, D(x x ( + (x + (x, D(x x ( + (x + (x So the matrix is b Let V P n (R be the set of polynomials satisfying f( f( Prove that V is a subspace of P n (R First notice that V (here means the polynomial by definition Suppose that f, g V and c is a scalar Then This shows that f + g V Similarly, which means that cf V (f + g( f( + g( f( + g( (f + g( (cf( c f( c f( (cf(, c Prove that (D V D V The definition of (D V is that for any f, g V : What needs checked is that for any f, g, Using integration by parts, we have (D V f, g (D V f, g f, (D V g (D V f, g f, (D V g f (xg(x dx (f(xg(x (f(g( f(g( f, (D V g f(xg (x dx f(xg (x dx f(xg (x dx (Notice that it s important that f( f( and g( g(; that s why we have to restrict to V to make this work 8

9 Problem a Suppose that V is a vector space, and that T,, T n V are injective linear maps Prove that the composition T T T n is injective Let s prove this by induction on the number of linear maps The case n is trivial Now suppose that (T T T n (v This means T ((T T n (v Since T is injective, this means (T T n (v By induction, since there are only n injective maps here, we know that v This shows that the whole composition is injective b Suppose that V and W are finite dimensional Prove that there exists an injective linear map T : V W if and only if dim V dim W If dim V > dim W, we have already proved a theorem that there can t be an injective linear map So all we need to show is that if dim V < dim W, there is such a map Well, pick a basis v,, v m for V and w,, w n for W, so m n To define a linear map T : V W, I just have to tell you where the basis vectors go, and I declare that T (v i w i I claim that this is injective: if T (c v + + c m v m, then c w + + c m w m, which means the c s are all because the w i are linearly independent 9