Math 4242 Fall 2016 (Darij Grinberg): homework set 6 due: Mon, 21 Nov 2016 Let me first recall a definition.

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1 Math 4242 Fall 206 homework page Math 4242 Fall 206 Darij Grinberg: homework set 6 due: Mon, 2 Nov 206 Let me first recall a definition. Definition 0.. Let V and W be two vector spaces. Let v = v, v 2,..., v m be a basis of V. Let w = w, w 2,..., w n be a basis of W. Let L : V W be a linear map. The matri representing L with respect to v and w is the n m-matri M v,w,l defined as follows: For every j {, 2,..., m}, epand the vector L v j with respect to the basis w, say, as follows: L v j = α,j w + α 2,j w α n,j w n. Then, M v,w,l is the n m-matri whose i, j-th entry is α i,j. For eample, if n = 3 and m = 2, then where M v,w,l = α, α,2 α 2, α 2,2 α 3, α 3,2, L v = α, w + α 2, w 2 + α 3, w 3 ; L v 2 = α,2 w + α 2,2 w 2 + α 3,2 w 3. The purpose of this matri M v,w,l is to allow easily epanding L v in the basis w, w 2,..., w n of W if v is a vector in V whose epansion in the basis v, v 2,..., v m of V is known. For instance, if v is one of the basis vectors v j, then the epansion of L v j can be simply read off from the j-th column of Mv,w,L ; otherwise, it is an appropriate linear combination: L λ v + λ 2 v λ m v m = λ L v + λ 2 L v λ m L v m where the L v j can be computed by. You can abbreviate M v,w,l as M L, but it s your job to ensure that you know what v and w are and they aren t changing midway through your work. 5 7 Eample 0.2. Let A be the 2 2-matri. Let L : R R 2 be the linear map L A. Recall that this is the map R 2 R 2 that sends every vector v R 2 to Av. Consider the following basis v = v, v 2 of the vector space R 2 : v = 2, v 2 = 3.

2 Math 4242 Fall 206 homework page 2 Consider the following basis w = w, w 2 of the vector space R 2 : w = 2, w 2 = 2 What is the matri M v,w,l representing L with respect to these two bases v and w? First, let me notice that it is not A or at least it doesn t have to be A a priori, because our two bases v and w are not the standard basis of R 2. Only if we pick both v and w to be the standard bases of the respective spaces we can guarantee that M v,w,l will be A. Without having this shortcut, we must resort to the definition of M v,w,l. It tells us to epand L v and L v 2 in the basis w of R 2, and to place the resulting coefficients in a 2 2-matri. Let s do this. We begin with L v : L v = L A v = Av = = 2 How do we epand this in the basis w? This is a typical eercise in Gaussian elimination we just need to solve the equation L v = λ w + λ 2 w 2 in the two unknowns λ and λ 2, and the result is L v = 7 3 w w 2. Similarly, we take care of L v 2, obtaining L v 2 = 3w + 5w 2.. Thus, the required matri is M v,w,l = Eercise. Let A be the 3 2-matri. Let L : R R 2 be the linear map L A. Recall that this is the map R 3 R 2 that sends every vector v R 3 to Av. Consider the following basis v = v, v 2, v 3 of the vector space R 3 : v =, v 2 = 0, v 3 = 0 0

3 Math 4242 Fall 206 homework page 3 Consider the following basis w = w, w 2 of the vector space R 2 : w =, w 2 = 0 a Find the matri M v,w,l representing L with respect to these two bases v and w. [5 points] b Let v be the basis v 3, v 2, v of R 3. Let w be the basis w 2, w of R 2. Find the matri M v,w,l. [5 points] Solution. a We follow Definition 0.. Thus, we compute the three vectors L v, L v 2, L v 3, and epand them in the basis w: We have. L v = L A v = Av by the definition of L A =L A = = 9 2 = 2w + 3 w We found the coefficients 2 and 3 by solving the equation = λ 2 w + λ 2 w 2 in the unknowns λ and λ 2. This is a straightforward eercise in Gaussian elimination. Thus, we have epanded L v in the basis w. Similarly, we can epand L v 2 and L v 3 in w; the results are L v 2 = 0w + 2 w 2 ; 3 L v 3 = 6w + w 2. 4 The matri M v,w,l now can be built from the coefficients of the right hand sides of 2, 3 and 4: Namely, we get M v,w,l = b Let us introduce new names for the vectors v 3, v 2, v and w 2, w. Namely: Denote the list v 3, v 2, v by v, v 2, v 3 ; thus, v = v 3, v 2 = v 2 and v 3 = v. Thus, v = v 3, v 2, v = v, v 2, v 3. Denote the list w 2, w by w, w 2 ; thus, w = w 2 and w 2 = w. Thus, w = w 2, w = w, w 2.

4 Math 4242 Fall 206 homework page 4 You do not have to do this; but I am doing this in order not to get confused with subscripts playing unusual roles. The advantage of writing v as v, v 2, v 3 instead of v 3, v 2, v is that the -st entry has a subscript, etc. Now, we have a basis v = v, v 2, v 3 of V and a basis w = w, w 2 of W and 0 we know the entries of these bases eplicitly: for eample, v = v 3 = 0, and we want to find the matri M v,w,l. We can do this using the same method as in part a of this eercise; the final result is M v,w,l = Notice that this matri can be obtained from the matri M v,w,l shown in 5 by reversing the order of the rows and reversing the order of columns. This should not be a surprise, since the bases v and w are obtained from the bases v and w by reversing the order of the vectors. If you make this observation, you can of course find the matri M v,w,l right away without computing anything, provided you have solved part a. For every n N, we let P n denote the vector space of all polynomial functions with real coefficients of degree n in one variable. This vector space has dimension n +, and its simplest basis is,, 2,..., n. We call this basis the monomial basis of P n. If f is a polynomial in one variable, then I shall use the notation f [y] for substitute y for into f. For eample, if f = , then f [5] = = 62. This would normally be denoted by f y, but this is somewhat ambiguous, since the notation + could then stand for two different things namely, substitute + into the polynomial function or multiply by +, whereas the notation f [y] removes this ambiguity. Eample 0.3. a Define a map S a : P 2 R by S a f = f [2] + f [3]. Then, S a is linear, because:. If f and g are two elements of P 2, then S a f + g = f + g [2] = f [2]+g[2] = f [2] + f [3] =S a f 2. If f P 2 and λ R, then. + f + g [3] = f [2] + g [2] + f [3] + g [3] = f [3]+g[3] + g [2] + g [3] = S a f + S a g. =S a g S a λ f = λ f [2] + λ f [3] = λ f [2] + λ f [3] = λ f [2] + f [3] = λs a f. =S a f

5 Math 4242 Fall 206 homework page 5 Let v be the monomial basis,, 2 of P 2, and let w be the one-element basis of R. What is the matri M v,w,sa? Again, follow the definition of M v,w,sa. It tells us to epand S a, S a and S a 2 in the basis w of R, and to place the resulting coefficients in a 3- matri. Epanding things in the basis w is particularly simple, since w is a one-element list; specifically, we obtain the epansions S a = [2] = S a = [2] Thus, the required matri is =2 + [3] = + = 2 = 2 ; = = = 5 = 5 ; + [3] =3 S a 2 = 2 [2] + 2 [3] = = 3 = 3. =2 2 =3 2 M v,w,sa = b Define a map S b : P 2 P 4 by S b f = f [ 2]. Notice that we chose P 4 as the target space, because substituting 2 for will double the degree of a polynomial. The map S b is linear for reasons that are similar to the ones that convinced us that S a is linear. Let v be the monomial basis,, 2 of P 2, and let w be the monomial basis,, 2, 3, 4 of P 4. What is the matri M v,w,sb? Again, follow the definition of M v,w,sb. It tells us to epand S b, S b and S b 2 in the basis w of P 4, and to place the resulting coefficients in a 5 3- matri. The epansions are as follows: [ S b = 2] = = ; [ S b = 2] = 2 = ; S b 2 [ = 2 2] = 2 2 = 4 = Thus, the required matri is M v,w,sb =

6 Math 4242 Fall 206 homework page 6 Eercise 2. Which of the following maps are linear? For every one that is, represent it as a matri with respect to the monomial bases of its domain and its target. [6 points for each part, split into 2+4 if the map is linear] a The map T a : P 2 P 2 given by T a f = f [ + ]. Thus, T a is the map that substitutes + for into f. Thus, T a n = + n for every n {0,, 2}. b The map T b : P 2 P 3 given by T b f = f []. Notice that f [] is the same as f, because substituting for changes nothing. I am just writing f [] to stress that f is a function of. c The map T c : P 2 P 4 given by T c f = f [] f []. d The map T d : P 2 P 4 given by T d f = f [ 2 + ] [ ]. e The map T e : P 2 P 2 given by T e f = 2 f. g The map T g : P 3 P 3 given by T g f = f []. [There is no part f because I want to avoid calling a map T f while the letter f stands for a polynomial.] [Note: Proofs are not required.] Solution. a The map T a is linear, because:. If f and g are two elements of P 2, then T a f + g = f + g [ + ] = f [ + ] =T a f 2. If f P 2 and λ R, then + g [ + ] = T a f + T a g. =T a g T a λ f = λ f [ + ] = λ f [ + ] = λt a f. =T a f Let v be the monomial basis,, 2 of P 2. Let w be the monomial basis,, 2 of P 2. To find the matri M v,w,ta that represents T a with respect to the bases v and w, we proceed as in Eample 0.3. Thus, we epand T a, T a and T a 2 in the basis w of P 2, and to place the resulting coefficients in a 3 3-matri. The epansions are as follows: T a = [ + ] = = ; T a = [ + ] = + = ; T a 2 = 2 [ + ] = + 2 = = Thus, the required matri is M v,w,ta = b The map T b is linear, because:

7 Math 4242 Fall 206 homework page 7. If f and g are two elements of P 2, then 2. If f P 2 and λ R, then T b f + g = f + g [] = f [] + g [] = f []+g[] = f [] + g [] = T b f + T b g. =T b f =T b g T b λ f = λ f [] =λ f [] = λ f [] = λt b f. =T b f Let v be the monomial basis,, 2 of P 2.,, 2, 3 of P 3. Let w be the monomial basis To find the matri M v,w,tb that represents T b with respect to the bases v and w, we proceed as in Eample 0.3. Thus, we epand T b, T b and T b 2 in the basis w of P 3, and to place the resulting coefficients in a 4 3-matri. The epansions are as follows: T b = [] = = ; = T b = [] = = 2 = ; = T b 2 = 2 [] = 2 = 3 = = 2 Thus, the required matri is M v,w,tb = c The map T c is not linear. d The map T d is linear, because: If f and g are two elements of P 2, then [ ] [ ] T d f + g = f + g 2 + = f 2 + =T d f [ + g ] 2 + } {{ } =T d g = T d f + T d g. For eample, it fails to satisfy the if f P 2 and λ R, then T c λ f = λt c f condition if we pick f = and λ = 2, because in this case we have T c λ f = λ f [] λ f [] = 2 [] 2 [] = =2 =2 =2 =2 2 2 = 4 and λ =2 T c f = f [] f [] = 2 f = [] f = [] = 2 [] = [] = = 2.

8 Math 4242 Fall 206 homework page 8 2. If f P 2 and λ R, then T d λ f = λ f [ ] 2 + [ = λ f ] 2 + } {{ } =T d f = λt d f. Let v be the monomial basis,, 2 of P 2.,, 2, 3, 4 of P 4. Let w be the monomial basis To find the matri M v,w,td that represents T d with respect to the bases v and w, we proceed as in Eample 0.3. Thus, we epand T d, T d and T d 2 in the basis w of P 4, and to place the resulting coefficients in a 5 3-matri. The epansions are as follows: [ ] T d = 2 + = = ; [ ] T d = 2 + = 2 + = ; T d 2 [ ] 2 = = 2 + = = Thus, the required matri is M v,w,td = e The map T e is linear, because:. If f and g are two elements of P 2, then [ ] T e f + g = 2 f + g [ ] [ ] = f +g = 2 f [ ] + g [ ] [ ] [ ] = 2 f + 2 g = T e f + T e g. =T e f =T e g 2. If f P 2 and λ R, then [ ] [ ] T e λ f = 2 λ f = λ 2 f = λt e f. [ ] =T e f =λ f

9 Math 4242 Fall 206 homework page 9 Let v be the monomial basis,, 2 of P 2. Let w be the monomial basis,, 2 of P 2. To find the matri M v,w,te that represents T e with respect to the bases v and w, we proceed as in Eample 0.3. Thus, we epand T e, T e and T e 2 in the basis w of P 2, and to place the resulting coefficients in a 3 3-matri. The epansions are as follows: [ ] T e = 2 = 2 = ; = [ ] T e = 2 = 2 = = ; = T e 2 [ ] = = = = = Thus, the required matri is M v,w,te = g The map T g is linear, because:. If f and g are two elements of P 3, then 2. If f P 3 and λ R, then T g f + g = f + g [] = f [] + g [] = f []+g [] = f [] + g [] = T g f + T g g. =T g f =T g g T g λ f = λ f [] =λ f [] = λ f [] = λt g f. =T g f Let v be the monomial basis,, 2, 3 of P 3. Let w be the monomial basis,, 2, 3 of P 3. To find the matri M v,w,tg that represents T g with respect to the bases v and w, we proceed as in Eample 0.3. Thus, we epand T g, T g, T g 2 and T g 3

10 Math 4242 Fall 206 homework page 0 in the basis w of P 3, and to place the resulting coefficients in a 4 4matri. The epansions are as follows: T g = [] = 0 = ; =0 T g = [] = = ; = T g 2 = 2 [] = 2 = 2 2 = ; =2 T g 3 = 3 [] = 3 2 = 3 3 = =3 2 Thus, the required matri is M v,w,tg = See the beginning of 3.2 of my notes, the Wikipedia, or various other sources, for eamples of injective, surjective and bijective maps. Eercise 3. a Which of the si maps in Eercise 2 are injective? [2 points per map] b Which of them are surjective? [2 points per map] [Note: Proofs are not required.] Solution. As we have seen in the solution to Eercise 2, five of our si maps more precisely, all the maps apart from T c are linear. Hence, it will be useful to know how to determine whether a linear map is injective. There is a method for that; to find it, we combine two results. First here is a result from homework set 4: Proposition 0.4. Let n N and m N. Let A be an n m-matri. a The matri A is right-invertible if and only if rank A = n. b The matri A is left-invertible if and only if rank A = m. c The matri A is invertible if and only if rank A = n = m. In particular, only square matrices can be invertible! Net, we recall a result from class:

11 Math 4242 Fall 206 homework page Proposition 0.5. Let V and W be two vector spaces with bases v and w, respectively. Let L : V W be a linear map. a The map L is injective if and only if the matri M v,w,l is left-invertible. b The map L is surjective if and only if the matri M v,w,l is right-invertible. c The map L is bijective if and only if the matri M v,w,l is invertible. Combining these two propositions, we obtain the following: Corollary 0.6. Let V and W be two vector spaces with bases v and w, respectively. Let m = dim V and n = dim W. Let L : V W be a linear map. a The map L is injective if and only if rank M v,w,l = m. b The map L is surjective if and only if rank M v,w,l = n. c The map L is bijective if and only if rank M v,w,l = n = m. Proof of Corollary 0.6. The basis v of V has size dim V = m, and the basis w of W has size dim W = n. Thus, M v,w,l is an n m-matri. a Proposition 0.5 a shows that the map L is injective if and only if the matri M v,w,l is left-invertible. But Proposition 0.4 b applied to A = M v,w,l shows that the matri M v,w,l is left-invertible if and only if rank M v,w,l = m. Combining these two if and only if statements, we conclude that the map L is injective if and only if rank M v,w,l = m. This proves Corollary 0.6 a. We have thus derived Corollary 0.6 a from Proposition 0.5 a and Proposition 0.4 b. Similarly, one can derive Corollary 0.6 b from Proposition 0.5 b and Proposition 0.4 a, and derive Corollary 0.6 c from Proposition 0.5 c and Proposition 0.4 c. Now, let us solve the actual eercise. a Let us find out whether T a is injective. Let v be the monomial basis,, 2 of P 2. Let w be the monomial basis,, 2 of P 2. From our solution of Eercise 2 a, we remember that the map T a : P 2 P 2 is linear, and is represented by the matri M v,w,ta = 0 2. Clearly, dim P 2 = and dim P 2 = 3. Hence, Corollary 0.6 a applied to V = P 2, W = P 2, m = 3, n = 3 and L = T a shows that the map T a is injective if and only if rank M v,w,ta = 3. Since rank M v,w,ta = 3 holds indeed, this can be checked straightforwardly, since we know the matri M v,w,ta, we thus conclude that the map T a is injective. By the same method, we can conclude that the maps T b, T d and T e are injective, but the map T g is not. Each time, we need to apply Corollary 0.6 a, but of course the vector spaces V and W and their dimensions m and n change from map to map thus, the rank of the matri isn t always being compared to 3. For eample, to tell whether T g is injective, we have to determine whether rank M v,w,tg = 4; but this is not satisfied because rank M v,w,tg = 3 = 4.

12 Math 4242 Fall 206 homework page 2 It remains to check whether the map T c is injective. The method we have showed for T a does not apply to T c because T c is not linear. Instead, we use the definition of the word injective. Recall that a map F : X Y from a set X to a set Y is injective if and only if it satisfies the following statement: If u and u 2 are two elements of X satisfying F u = F u 2, then u = u 2. Applying this to the map T c : P 2 P 4, we see that the map T c : P 2 P 4 is injective if and only if it satisfies the following statement: If u and u 2 are two elements of P 2 satisfying T c u = T c u 2, then u = u 2. But this statement is not true: For eample, the two elements u = and u 2 = 0 of P 2 satisfy T c u = T c u 2 because T c u = u [] u [] = 0 and = =0 T c u 2 = u 2 [] u 2 [] = 0 but not u = u 2. =0 Thus, the map T c is not injective. b The surjectivity of a linear map can be checked eactly in the same way as its injectivity, ecept that we have to use Corollary 0.6 b instead of Corollary 0.6 a. Thus, let me only summarize the results: The maps T a and T e are surjective, whereas the maps T b, T d and T g are not. Again, the map T c needs to be treated separately, since it is not linear. For this, we use the definition of the word surjective. Recall that a map F : X Y from a set X to a set Y is surjective if and only if it satisfies the following statement: For each v Y, there eists some u X such that v = F u. Applying this to the map T c : P 2 P 4, we see that the map T c : P 2 P 4 is surjective if and only if it satisfies the following statement: For each v P 4, there eists some u P 2 such that v = T c u. But this statement is not true: For eample, for v = 4 P 4, there eists no u P 2 such that v = T c u because T c u = u [] u [] would have to be a constant a polynomial of degree 2 a polynomial of degree 2, but v = 4 is not such a polynomial. Hence, the map T c is not surjective. [Remark: We might wonder whether the map T c becomes surjective if we restrict its target to P 2. In other words, is the map T c2 : P 2 P 2 given by T c f = f [] f [] surjective? We can no longer rule this out using v = 4, because 4 does not belong to P 2.

13 Math 4242 Fall 206 homework page 3 However, this new map T c2 is still not surjective. Indeed, for v = P 2, there eist no u P 2 such that v = T c2 u. In order to prove this, we assume the contrary. Thus, there eists some u P 2 such that v = T c2 u. Consider this u. We have = v = T c2 [u] = u [] u []. Substituting for in this equality, we obtain = u [] u [] = u [] 2. In other words, 0 = u [] 2. Hence, u [] = 0. Now, v = T c2 [u] = u [] u [] = 0. But this contradicts v =. This contradiction =0 shows that our assumption was wrong, qed.]