MATH 110: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 6 SOLUTIONS

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1 MATH 0: LINEAR ALGEBRA FALL 2007/08 PROBLEM SET 6 SOLUTIONS You may use without proof any results from the lectures as well as any basic properties of matrix operations and calculus operations. P will denote the vector space of polynomials in x with coefficients in R.. Let V be a vector space over R. Show that the following maps T i : V V, i = a, b, c, d, are linear operators. (a) V = R m m, A R m m is a fixed matrix, and T a : R m m R m m is defined by T a (X) = AX XA for all X R m m. Solution. Let α, β R and X, X 2 R m m. Then T a (αx + βx 2 ) = A(αX + βx 2 ) (αx + βx 2 )A = αax + βax 2 αx A βx 2 A = α(ax X A) + β(ax 2 X 2 A) = αt a (X ) + βt b (X 2 ). (b) V = R m n, A R m m and B R n n are fixed matrices, and T b : R m n R m n is defined by T b (X) = AXB for all X R m n. Solution. Let α, β R and X, X 2 R m m. Then (c) V = P, and T c : P P is defined by T b (αx + βx 2 ) = A(αX + βx 2 )B = αax B + βax 2 B = αt b (X ) + βt b (X 2 ). T c (p(x)) = p (x) for all p(x) P. Solution. Let α, β R and p(x), q(x) P. Then by the basic properties of differentiation that you have learned in high-school calculus, T c (αp(x) + βq(x)) = d (αp(x) + βq(x)) dx = α d dx p(x) + β d dx q(x) = αp (x) + βq (x) = αt c (p(x)) + βt c (q(x)). Date: November 5, 2007 (Version.).

2 (d) V = P, and T d : P P is defined by T d (p(x)) = xp(x) for all p(x) P. Solution. Let α, β R and p(x), q(x) P. Then by the basic properties of polynomial multiplication that you have learned in high-school algebra, T d (αp(x) + βq(x)) = x(αp(x) + βq(x)) = αxp(x) + βxq(x) = αt d (p(x)) + βt d (q(x)). 2. Let F : R 3 R 2, G : R 3 R 2, H : R 2 R 2 be defined by F (x, y, z) = (y, x + z), G(x, y, z) = (2z, x y), H(x, y) = (y, 2x) respectively. (a) Find the formulas that define 3F 5G, H F, H G, H (F + G), and state the respective domains and codomains of these maps. Are these linear transformations? Solution. 3F 5G : R 3 R 2 and its formula is given by (3F 5G)(x, y, z) = 3F (x, y, z) 5G(x, y, z) H F : R 3 R 2 and its formula is given by = 3(y, x + z) 5(2z, x y) = (3y 0z, 2x + 3z + 5y). H F (x, y, z) = H(F (x, y, z)) = H(y, x + z) = (x + z, 2y). H G : R 3 R 2 and its formula is given by H G(x, y, z) = H(G(x, y, z)) = H(2z, x y) = (x y, 4z). H (F + G) : R 3 R 2 and its formula is given by H (F + G)(x, y, z) = H((F + G)(x, y, z)) = H(y + 2z, 2x y + z) = (2x y + z, 2y + 4z). Alternatively, we may use Problem 3(c) below: H (F + G) = H F + H G. All four maps are linear transformations. (b) We define H 2 = H H, H 3 = H H H, and so on. Find the formulas that define H 2, H 3, and H 2 (F + G), and state the respective domains and codomains of these maps. Are these linear transformations? Solution. H 2 : R 2 R 2 and its formula is given by H 2 (x, y) = H(H(x, y)) = H(y, 2x) = (2x, 2y). H 2 : R 2 R 2 and its formula is given by H 3 (x, y) = H(H 2 (x, y)) = H(2x, 2y) = (2y, 4x). H 2 (F + G) : R 3 R 2 and its formula is given by H 2 (F + G)(x, y) = H 2 ((F + G)(x, y)) All three maps are linear transformations. = H 2 (y + 2z, 2x y + z) = (2y + 4z, 4x 2y + 2z). 3. Let U, W, V be finite-dimensional vector spaces over F. Let α, β F. 2

3 (a) Let T : V W be a linear transformation. Show that rank(t ) dim(v ). Solution. Let dim(v ) = n. Suppose dim(im(t )) = rank(t ) > n. Then there must be at least n + vectors w,..., w n+ Im(T ) that are linearly independent (eg. choose the first n + vectors in a basis of Im(T )). Since w,..., w n+ are in Im(T ), there exist v,..., v n+ V such that T (v i ) = w i, i =,..., n +. Since dim(v ) = n, v,..., v n+ must be linearly dependent, so there exist α,..., α n+ F, not all zero, such that α v + + α n+ v n+ = 0 V. Hence we have found α,..., α n+ F, not all zero, such that T (α v + + α n+ v n+ ) = T (0 V ), α T (v ) + + α n+ T (v n+ ) = 0 W, α w + + α n+ w n+ = 0 W, which implies that w,..., w n+ are linearly dependent a contradiction. original assumption must have been false, we must have dim(im(t )) n. (b) Let S : U V and T : V W be linear transformations. Show that and rank(t S) rank(t ) rank(t S) rank(s). Hence our Solution. Recall the following elementary fact from set theory (Math 74): A B implies f(a) f(b) for any function f. In particular, since S(U) V, we must have T (S(U)) T (V ), T S(U) T (V ), im(t S) im(t ) (recall that im(ϕ) is just another way of writing ϕ(v ) both denote the range of ϕ). Now both of these are subspaces of W and so dim(im(t S)) dim(im(t )), rank(t S) rank(t ). For the second part, we define the function ϕ : im(s) W by ϕ(v) = T (v) for all v im(s). Note that ϕ is a linear transformation since for all v, v 2 im(s) and α, β F, we have ϕ(αv +βv 2 ) = T (αv +βv 2 ) = αt (v )+βt (v 2 ) = αϕ(v )+βϕ(v 2 ). So applying part (a) to ϕ, we get rank(ϕ) dim(im(s)) = rank(s). (3.) But note that im(t S) im(ϕ) since if w im(t S), then w = T (S(u)) for some u U and therefore w = T (v) = ϕ(v) where v = S(u) im(s); hence w im(ϕ). So Combining (3.) and (3.2) then yields dim(im(t S)) dim(im(ϕ)), rank(t S) rank(ϕ). (3.2) rank(t S) rank(s). 3

4 (c) Let S : U V, S 2 : U V, and T : V W be linear transformations. Show that Solution. Let u U. Then T (αs + βs 2 ) = αt S + βt S 2. T (αs + βs 2 )(u) = T ((αs + βs 2 )(u)) Since this is true for any u U, we have that = T (αs (u) + βs 2 (u)) = αt (S (u)) + βt (S 2 (u)) = αt S (u) + βt S 2 (u). T (αs + βs 2 ) = αt S + βt S 2. (d) Let S : U V, T : V W, and T 2 : V W be linear transformations. Show that Solution. Let u U. Then (αt + βt 2 ) S = αt S + βt 2 S. (αt + βt 2 ) S(u) = (αt + βt 2 )(S(u)) Since this is true for any u U, we have that = αt (S(u)) + βt 2 (S(u)) = αt S(u) + βt 2 S(u) (αt + βt 2 ) S = αt S + βt 2 S. 4. Let T : R 3 R 3 be a linear transformation such that T 0 = 2 3, T = 3 0, 2 Find im(t ), ker(t ), and the formula that defines T. Solution. Let u = 0, u 2 =, u 3 = 2 7. T 2 7 = 2 3. Now if αu + βu 2 + γu 3 = 0, then solving the system gives α = β = γ = 0. So u, u 2, u 3 are linearly independent and since dim(r 3 ) = 3, {u, u 2, u 3 } is a basis for R 3. Hence by a result in the lectures, we have im(t ) = span{t (u ), T (u 2 ), T (u 3 )}. We will use the Throwing-Out Algorithm to obtain a basis for im(t ). Since T (u ) = T (u 3 ), we may throw out T (u 3 ). The remain set {T (u ), T (u 2 )} is easily seen to be linearly independent: λ µ 3 0 = implies λ = 0 (just examine the second coordinate) and so µ = 0. Hence {T (u ), T (u 2 )} is a basis and so im(t ) = span 2 3,

5 To find the formula defines T we will first need to determine how an arbitrary u = [x, y, z] R 3 may be expressed in terms of the basis {u, u 2, u 3 }. So we let x y = α 0 + β + γ 2 7 z and solve the system to get α = 6x + y 5z, β = 7x y + 7z, γ = x + z. (4.3) Since T is a linear operator, T x y = αt 0 + βt + γt 2 7 z = α β γ (α + γ) + 3β = 3(α + γ) (α + γ) 2β x y + 3z = 5x + 3y 2z 9x + y 0z upon substituting (4.3). To find ker(t ), we set T (u) = 0 and using the next to last expression in (4.3), we get 2(α + γ) + 3β 3(α + γ) = 0 0 (α + γ) 2β 0 which gives us α = γ and β = 0; 6x + y 5z = x z and 7x y + 7z = 0, x = 3 2 z and y = 7 2z. Hence ker(t ) = span Let T : V W be a linear transformation. Let V be finite-dimensional and dim(v ) = n. If w,..., w m W form a basis for im(t ) and if v,..., v m V are such that for i =,..., m, show that T (v i ) = w i V = span{v,..., v m } ker(t ). Solution. Let v V. Then T (v) Im(T ). Since w,..., w m form a basis for im(t ), there exist α,..., α m F such that Since T (v i ) = w i, we get T (v) = α w + + α m w m. T (v) = α T (v ) + + α m T (v m ), T (v) α T (v ) α m T (v m ) = 0 W, T (v α v α m v m ) = 0 W 5

6 which implies that v α v α m v m ker(t ). So given any v V, we could write v = (α v + + α m v m ) + (v α v α m v m ) span{v,..., v m } + ker(t ), which means that V = span{v,..., v m } + ker(t ). We will now show that this sum is in fact a direct sum. Let u,..., u k be a basis for ker(t ) where k = nullity(t ). We claim that {v,..., v m, u,..., u k } is linearly independent and by Homework 3, Problem 4(b), V = span{v,..., v m } + ker(t ) = span{v,..., v m } + span{u,..., u k } = span{v,..., v m, u,..., u k } = span{v,..., v m } span{u,..., u k } = span{v,..., v m } ker(t ). Suppose there exists λ,..., λ m, µ,..., µ k F such that λ v + + λ m v m + µ u + + µ k u k = 0 V. (5.4) Then applying T to both sides of the equation gives λ T (v ) + + λ m T (v m ) + µ T (u ) + + µ k T (u k ) = 0 W. Now since T (u i ) = 0 W for all i =,..., k, and T (v j ) = w j for all j =,..., m, we get λ w + + λ m w m = 0 W. As w,..., w m are linearly independent, we get λ = = λ m = 0. So we are left with µ u + + µ k u k = 0 V in (5.4). But since u,..., u k are also linearly independent, we then get µ = = µ k = 0. So v,..., v m, u,..., u k are linearly independent. 6