dim(u) = n 1 and {v j } j i

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SOLUTIONS Math B4900 Homework 1 2/7/2018 Unless otherwise specified, U, V, and W denote vector spaces over a common field F ; ϕ and ψ denote linear transformations; A, B, and C denote bases; A, B, and C denote matrices in M n (F ). 1. Vector spaces. (a) For a linear map ϕ : V V of vector spaces over a field F, show that ker(ϕ) and ϕ(v ) are subspaces of V and V, respectively. Proof. First let u, v ker(ϕ), so that ϕ(u) = 0 = ϕ(v). since ϕ is linear, we have ϕ(u + v) = ϕ(u) + f(v) = 0 + 0 = 0 and ϕ(αu) = αϕ(u) = α 0 = 0, for all α F. u + v and αu are in ker(ϕ). Finally, since 0 ker(ϕ), we have ker(ϕ) is a subspace of V. Similarly, for u, v ϕ(v ), so that there are u, v V such that ϕ(u) = u and ϕ(v) = v. since ϕ is linear, we have ϕ(v ) ϕ(u + v) = ϕ(u) + f(v) = u + v and ϕ(v ) ϕ(αu) = αϕ(u) = αu, for all α F. u + v and αu are in ϕ(v ). Finally, V, we have ϕ(v ), and is therefore a subspace of V. (b) Let V = R n and let a = (a 1,..., a n ) 0 be a fixed vector in V. Prove that U = {x = (x 1,..., x n ) V a 1 x 1 + + a n x n = 0} is a subspace of V. Determine its dimension and find a basis. Proof. We know that a i 0 for some i, and so if x U, we have x i = 1 a j x j. a i j i v j = e j a j e i, for j = 1,..., i 1, i + 1,..., n, a i where e j is the n-tuple with 0 s everywhere except for a 1 in the jthe factor. Now, if 0 = j i α j v j = j i α j e j 1 α j e i. a i j i since the e j s are linearly independent, we have that the v j s are linearly independent as well. But since F {v j } j i and n = dim(v ) > dim(u) dim(f {v j } j i ) = n 1, we have is a basis for U. dim(u) = n 1 and {v j } j i 1

2 (c) Show that Hom(V, W ) is a vector space under the binary operations (ϕ + ψ) : v ϕ(v) + ψ(v) and (αϕ) : v α(ϕ(v)), for all ϕ, ψ Hom(V, W ), v V, and α F ; and that if V = W, then it is also a ring under the addition above and the product (ϕψ) : v (ϕ ψ)(v) = ϕ(ψ(v)). Proof. By definition, for ϕ, ψ Hom(V, W ), since W is a vector space, we have ϕ + ψ and αϕ are both well-defined functions from V to W. Similarly, if V = W, ϕψ is also a well-defined function from V to itself. Also, because W is a vector space, by definition, the addition and scalar product defined on Hom(V, W ) satisfy the axioms of vector spaces so long as they are well-defined. it only remains to show that ϕ + ψ, αϕ, and ϕψ are linear, and the product distributes in the case when V = W, as follows. ϕ + ψ: For u, v V and α F, we have and (ϕ + ψ)(u + v) = ϕ(u + v) + ψ(u + v) = ϕ(u) + ϕ(v) + ψ(u) + ψ(v) = (ϕ(u) + ψ(u)) + (ϕ(v) + ψ(v)) = (ϕ + ψ)(u) + (ϕ + ψ)(v) (ϕ + ψ)(αu) = ϕ(αu) + ψ(αu) = αϕ(u) + αψ(u) = α(ϕ(u) + ψ(u)) = α(ϕ + ψ)(u). αϕ: For u, v V and β F, we have (αϕ)(u + v) = α(ϕ(u + v)) = α(ϕ(u) + ϕ(v)) = α(ϕ(u)) + α(ϕ(v)) = (αϕ)(u) + (αϕ)(v), and (αϕ)(βu) = α(ϕ(βu)) = αβ(ϕ(u)) = βα(ϕ(u)) = β(αϕ)(u). For ϕ, ψ, and φ End(V ), we have (ϕ(ψ + φ))(u) = ϕ((ψ + φ)(u)) = ϕ(ψ(u) + φ(u)) = ϕ(ψ(u)) + ϕ(φ(u)) since ϕ is linear; and ((ψ + φ)ϕ)(u) = (ψ + φ)(ϕ(u)) = ψ(ϕ(u)) + φ(ϕ(u)) = (ψϕ)(u) + (φϕ)(u) by definition of ψ + φ.

3 2. Linear maps and matrices. (a) Give a function f : R 2 R 2 that is not R-linear (and justify). (b) Consider the three bases of V = R 3 given by A : a 1 = (1, 0, 0), a 2 = (0, 1, 0), a 3 = (0, 0, 1); B : b 1 = (1, 0, 0), b 2 = (0, 0, 1), b 3 = (0, 1, 0); and C : c 1 = (1, 0, 0), c 2 = (1, 1, 0), c 3 = (1, 1, 1). Denoting e 1 = (1, 0, 0), e 2 = (0, 1, 0), and e 3 = (0, 0, 1), we have a 1 = b 1 = c 1 = e 1, a 2 = b 3 = e 3, a 3 = b 2 = e 3, c 2 = e 1 + e 2, and c 3 = e 1 + e 2 + e 3. (i) Compute the matrices MA A(ϕ), M A C (ϕ), M C B(ϕ), and M C C (ϕ) for the linear map ϕ : V V defined by (x, y, z) ( y, x, x + y 2z). Answer. We have ϕ(e 1 ) = (0, 1, 1) = e 2 + e 3 = c 1 + c 3, ϕ(e 2 ) = ( 1, 0, 1) = e 1 + e 3 = c 1 c 2 + c 3, ϕ(e 3 ) = (0, 0, 2) = 2e 3 = 2c 2 2c 3, ϕ(c 2 ) = ( 1, 1, 2) = e 1 + e 2 + 2e 3 = 2c 1 c 2 + 2c 3, ϕ(c 3 ) = ( 1, 1, 0) = e 1 + e 2 = 2c 1 + c 2. M A A (ϕ) = M B C (ϕ) = 0 1 1, MA(ϕ) C = 0 1 2, 1 1 2 1 1 2 0 1 1 1 2 2 1 2 0, MC C (ϕ) = 0 1 1. 1 1 1 1 2 0 (ii) Compute M A A (id) and M A C (id). Answer. MA A (id) = 1 0, MC A (id) = 1 1 1 0 1 1. 1 1

4 (iii) Compute the change of basis matrix P from basis A to basis C, and use it to verify that your matrices MA A(ϕ) and M C C (ϕ) from part (i) match appropriately. Answer. We have P = MA(id) C = 1 0 1 1 and P 1 = MC A (id) = 1 1 1 0 1 1. 1 1 P 1 MC C (ϕ)p = 1 1 1 1 2 2 0 1 1 0 1 1 1 0 1 1 = 0 1 = MA A (ϕ). 1 1 2 0 1 1 1 2 (iv) Compute the dual bases A and C of V = R 3, representing elements as row vectors. [Hint: Set up a system of equations and solve.] Answer. You can check that A : a 1 = (1, 0, 0), a 2 = (0, 1, 0), a 3 = (0, 0, 1). To compute C, every linear function ϕ : V F comes in the form ϕ : (x 1, x 2, x 3 ) α 1 x 1 + α 2 x 2 + α 3 x 3 for some α 1, α 2, α 3 F. Now, since c i specifically sends c i 1 and c j 0 for i j, this gives the following equations: c 1 : c 1 1 = α 1, c 2 0 = α 1 + α 2, c 3 0 = α 1 + α 2 + α 3, so α 1 = 1, α 2 = 1, and α 3 = 0, i.e. c 1 = (1, 1, 0) ; c 2 : c 1 0 = α 1, c 2 1 = α 1 + α 2, c 3 0 = α 1 + α 2 + α 3, so α 1 = 0, α 2 = 1, and α 3 = 1, i.e. c 2 = (0, 1, 1) ; c 3 : c 1 0 = α 1, c 2 0 = α 1 + α 2, c 3 1 = α 1 + α 2 + α 3, so α 1 = 1, α 2 = 1, and α 3 = 0, i.e. c 3 = (0, 0, 1).

5 3. Forms and traces. (a) Let B be the standard basis of V = F n, and identify V with M n,1 (F ). As in class, let J M n (F ), and define, J : V V F by (u, v) u t Jv. (i) Show that if J is symmetric, then so is, J. See part (ii). (ii) Show that if J is skew-symmetric, then so is, J. Proof. Let ɛ = ±1, with ɛ = 1 if J is symmetric, and 1 if J is skew symmetric, so that J t = ɛj. Note that for any α F, thinking of α as a 1 1 matrix, we have α t = α. u, v J = u t Jv = (v t J t u) t = v t J t u since v t J t u F, = v t (ɛj)u = ɛv t Ju since v t is linear, = ɛ v, u. if J is symmetric, then so is, J ; and if J is skew symmetric, then so is, J. (b) Let V = R 3, and let A and C be the bases in the previous exercise. (i) Compute the images of c 1, c 2, and c 3 (in C) under the isomorphism f : V V defined by f(v) : u v, u, where v, u = v t u = v u, the usual dot product. Answer. Writing u = (α 1, α 2, α 3 ), we have f(c 1 ) : u α 1 so f(c 1 ) = (1, 0, 0); f(c 2 ) : u α 1 + α 2 so f(c 2 ) = (1, 1, 0); f(c 3 ) : u α 1 + α 2 + α 3 so f(c 3 ) = (1, 1, 1). Notice that these are exactly c 1, c 2, and c 3 back again. (ii) Compute the images of a 1, a 2, and a 3 under the isomorphism f : V V defined by f(v) : u v, u J, where J = 1 0. 2 Answer. Writing u = (α 1, α 2, α 3 ), we have Ju = 1 0 α 1 α 2 = α 1 + α 2 α 2. 2 α 3 2α 3

6 f(a 1 ) : u a 1 Ju = α 1 + α 2 so f(a 1 ) = (1, 1, 0); f(a 2 ) : u α 2 so f(a 2 ) = (0, 1, 0); f(a 3 ) : u 2α 3 so f(a 3 ) = (0, 0, 1). Notice that these are exactly the column vectors of J. (c) Show that trace is independent of basis, i.e. show that if A and B are both bases for a vector space V, and ϕ End(V ), then tr(m A A (ϕ)) = tr(m B B (ϕ)). [Hint: Use the fact that similar is the same thing as conjugate.] Proof. Since there is some P GL n (F ) such that P 1 MA A (ϕ)p = MB B (ϕ), we have tr(mb B (ϕ)) = tr(p 1 MA A (ϕ)p ) = tr(ma A (ϕ)p P 1 ) = tr(ma A (ϕ)). (d) Give examples showing that the following equations do not hold in general: tr(abc) = tr(acb), tr(ab) = tr(a)tr(b), tr(ma A (ϕ)) = tr(ma(ϕ)). B Answer. For the first two, let 0 1 A =, B = AB =, ABC =, and C =, and ACB =.. tr(abc) = 1 0 = tr(acb), and tr(ab) = 1 0 = 0 0 = tr(a)tr(b). For the last identity, see 2(b)(i) and compare tr(ma A(ϕ)) = 2 and tr(m A C (ϕ)) = 4. (e) Show that, for A, B M n (F ), if A is symmetric and B is antisymmetric, then tr(ab) = 0. Proof. Recall that tr(x t ) = tr(x). since A t = A and B t = B, we have tr(ab) = tr((ab) t ) = tr(b t A t ) = tr( BA) = tr(ba) = tr(ab). tr(ab) = 0.

7 (f) Show that the following is a bilinear form (on the vector space End(V )):, : End(V ) End(V ) F defined by ϕ, ψ = tr(ϕψ) for ϕ, ψ End(V ). Proof. For ϕ, ψ, φ End(V ), and α F, since trace is linear, we have ϕ + ψ, φ = tr((ϕ + ψ)φ) = tr(ϕφ + ψφ) = tr(ϕφ) + tr(ψφ) = ϕ, φ + ψ, φ ; φ, ϕ + ψ = tr(φ(ϕ + ψ)) = tr(φϕ + φψ) = tr(φϕ) + tr(φψ) = φ, ϕ + φ, ψ ; αϕ, ψ = tr(αϕψ) = αtr(ϕψ) = α ϕ, ψ ; and ϕ, αψ = tr(ϕαψ) = tr(αϕψ) = αtr(ϕψ) = α ϕ, ψ., is bilinear.

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