Math 110: Worksheet 3

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1 Math 110: Worksheet 3 September 13 Thursday Sept. 7: Fix A M n n (F ) and define T : M n n (F ) M n n (F ) by T (B) = AB BA. (a) Show that T is a linear transformation. Let B, C M n n (F ) and a F. We then have We conclude that T is linear. (b) What does the kernel of T represent? T (ab + C) = A(aB + C) (ab + C)A = (aab + AC) (aba + CA) = a(ab BA) + (AC CA) = at (B) + T (C). Let B Ker(T ). We then have T (B) = 0 AB BA = 0. Thus, Ker(T ) consists exactly of those matrices that commute with A. 2. Let V be a vector space and let T : V V be linear. Prove that the following statements are equivalent: (i) Ker(T ) Im(T ) = {0}; (ii) if T (T (v)) = 0 for some v V, then T (v) = 0. (i) (ii) Assume Ker(T ) Im(T ) = {0} and suppose that T (T (v)) = 0 for some v V. Note then that T (v) is the image under T of some element in V so it belongs to Im(T ). Moreover, as T sends T (v) to 0, we have T (v) Ker(T ). Thus, T (v) Ker(T ) Im(T ) = {0} so T (v) = 0. (ii) (i) Conversely, assume that if T (T (v)) = 0 for some v V, then T (v) = 0. Let w Ker(T ) Im(T ). There then exists some u V such that w = T (u); we also have T (w) = 0 so we infer that T (T (u)) = 0. From the assumed hypothesis, this yields w = T (u) = 0. We conclude that Ker(T ) Im(T ) = {0}. 3. Let T : V W be a linear transformation between vector spaces. 1

2 (a) Prove that if dim(v ) < dim(w ) then T cannot be onto. From the dimension theorem, we have nullity(t ) + rank(t ) = dim(v ). Recall that T is onto if and only if rank(t ) = dim(w ); this would then yield nullity(t ) + dim(w ) = dim(v ) < dim(w ). As nullity(t ) 0, this is clearly false so we conclude that T cannot be onto. (b) Prove that if dim(v ) > dim(w ) then T cannot be one-to-one. vspace.3 cm From the dimension theorem, we have nullity(t ) + rank(t ) = dim(v ). Recall that T is one-to-one if and only if nullity(t ) = 0; this would then yield rank(t ) = dim(v ) > dim(w ). However, as rank(t ) dim(w ), this is clearly false so we conclude that T cannot be one-to-one. If V and W are R 2 and R 3 (not necessarily in that order), come up with a physical description of the implications of the statements above. Note that dim(r 2 ) = 2 < 3 = dim(r 3 ) so (a) implies that there cannot be a linear transformation from R 2 onto R 3. Similarly, (b) shows that there cannot be a one-to-one linear transformation from R 3 to R Let a, b R with a b and consider T : P n (R) P n+2 (R) defined by T (f)(x) = (x a)(x b)f(x). (a) Show that T is linear and find its nullity and rank. Let f, g P n (R) and c R. We then have T (cf + g)(x) = (x a)(x b)(cf(x) + g(x)) = c(x a)(x b)f(x) + (x a)(x b)g(x) = ct (f) + T (g). We conclude that T is linear. To find the nullity of T, let f Ker(T ); we must then have T (f)(x) = (x a)(x b)f(x) = 0 for all x R. This is only possible if f 0 so we infer that Ker(T ) = {0}. Thus, nullity(t ) = 0. The dimension theorem then gives rank(t ) = dim(p n (R)) nullity(t ) = n

3 (b) Describe Im(T ) and find a basis for it. (For instance, Im(T ) is the space of all polynomials of degree at most n + 2 that... ; for finding the basis, use the nullity of T and what it says about the image of linearly independent sets) Im(T ) is the space of all polynomials of degree at most n + 2 that are zero at x = a and x = b. As Ker(T ) = {0}, T is one-to-one. By exercise 14(a) of section 2.1, this implies that T carries linearly independent sets in P n (R) to linearly independent sets in P n+2 (R). Recall that S = {1, x, x 2,..., x n } is a linearly independent subset of P n (R) so that T (S) = {T (1), T (x), T (x 2 ),..., T (x n )} is a linearly independent subset of P n+2 (R). Moreover, there are precisely n + 1 vectors in T (S), the same as the dimension of Im(T ). We conclude that T (S) is a basis for Im(T ). 5. Suppose that V and W are finite dimensional spaces and that U is a subspace of V. Prove that there exists T L(V, W ) such that Ker(T ) = U if and only if dim(u) dim(v ) dim(w ). Suppose first that there exists T L(V, W ) such that Ker(T ) = U. Using the dimension theorem and the fact that rank(t ) dim(w ), we have dim(u) + rank(t ) = dim(v ) dim(u) dim(v ) dim(w ). Conversely, assume that dim(u) dim(v ) dim(w ). Setting k = dim(u), n = dim(v ) and m = dim(w ) gives k n m m n k. Let {v 1,..., v k } be a basis for U. By the replacement theorem, we can extend it to a basis for V : {v 1,..., v k, v k+1,..., v n }. As m n k, there exists a linearly independent subset of W containing n k vectors: {w 1,..., w n k }. Define the map T : V W by { 0W 1 i k T (v i ) = k + 1 i n w i k Recall that describing T on a basis of V is sufficient to define it uniquely. Note that we still need to show that Ker(T ) = U. First observe that U Ker(T ) as T sends all the basis vectors of U to 0 W. Next, let x Ker(T ). There then exist scalars a 1,..., a n F such that x = a 1 v 1 + a 2 v a n v n. Apply T on both sides: T (x) = T (a 1 v 1 + a 2 v a n v n ) 0 W = a 1 T (v 1 ) + a 2 T (v 2 )... + a n T (v n ) 0 W = a k+1 w a n w n k The linear independence of {w 1,..., w n k } implies then that a k+1 =... = a n = 0, showing that x is a linear combination of {v 1,..., v k } so that x U. This proves that Ker(T ) = U and completes the proof. 3

4 Tuesday Sept. 12: Let T : R 3 R 3 be a linear transformation defined by T (x 1, x 2, x 3 ) = (3x 1 + x 3, 2x 1 + x 2, x 1 + 2x 2 + 4x 3 ) (1) (a) Find the matrix of T in the standard ordered basis for R 3. Observe that T (1, 0, 0) = (3, 2, 1), T (0, 1, 0) = (0, 1, 2) and T (0, 0, 1) = (1, 0, 4) so, denoting by α the standard basis, we have [T ] α = (b) Find the matrix of T in the ordered basis β = {v 1, v 2, v 3 } where v 1 = (1, 0, 1), v 2 = ( 1, 2, 1), v 3 = (2, 1, 1). Note that T (v 1 ) = (4, 2, 3), T (v 2 ) = ( 2, 4, 9), T (v 3 ) = (7, 3, 4). We next need to express these as linear combinations of {v 1, v 2, v 3 }. express this as the problem of finding a ij R such that We can T (v j ) = a 1j v 1 + a 2j v 2 + a 3j v 3 for j = 1, 2, 3. Then, [T ] β = (a ij ). We have a 11 a 12 a a 21 a 22 a 23 = a 31 a 32 a Writing this problem in the augmented form and performing row operations yields (R 3 R 1 R 3 ) (R 3 R 2 R 3 ) (2R 2 + R 3 R 2 ) (R 1 + R 3 R 1 ) (4R 1 + R 2 R 1 )

5 so that 17/4 35/4 11/2 [T ] β = 3/4 15/4 3/2. 1/2 7/ Let T : V V be a linear transformation and let β = {v 1,..., v n } be an ordered basis for V. Suppose the matrix A := [T ] β is strictly upper triangular, i.e., A ij = 0 for i j. (a) Show that for 1 i n, T (v i ) span({v 1,..., v i 1 }). (If the last vector in the set has a smaller index than the first, we can take it to be the empty set) We have, for 1 i n, [T (v i )] β = A 1i A 2i. A ni T (v i) = n A ji v j. j=1 As A ji = 0 for j i, we get T (v i ) = i 1 j=1 A jiv j T (v i ) span({v 1,..., v i 1 }). (b) Use (a) to show that, for any k 1, for 1 i n. T k (v i ) span({v 1,..., v i k }) We prove this by induction on k. The statement holds for k = 1 from (a). Assume that it holds for some k, that is, for 1 i n, T k (v i ) span({v 1,..., v i k }). Note then that for any 1 i n, T k+1 (v i ) = T (T k (v i )) span({t (v 1 ),..., T (v i k )}). From (a), we can further infer that for 1 j i k, T (v j ) span({v 1,..., v j 1 }) so that span({t (v 1 ),..., T (v i k )}) span({v 1,..., v i k 1 }). We deduce that the statement holds for k + 1 as well and conclude that the result holds for all k 1. (c) What can we conclude about T n and hence A n? Plugging in k = n in (b), we have for 1 i n, T n (v i ) span({v 1,..., v i n }). Note however that i n 0 for all i so this is in fact just the span of the empty set. Thus, T n (v i ) = 0 for all i, that is, T n is the zero map. As A n is the matrix representation of T n, we infer that A n must be the zero matrix. This shows that the nth power of an n n strictly upper triangular matrix is the zero matrix (the same procedure also works for strictly lower triangular matrices as well). Try this out by hand for n = 2 and n = 3. 5

6 8. Let T : R 2 R 2 be the linear transformation that takes a point (x, y) and projects it orthogonally onto the line y = x (that is, T (v) is the point on the line y = x closest to v). (a) If β is the standard basis for R 2, find [T ] β. It can be easily shown that T (1, 0) = (1/ 2, 1 2) and T (0, 1) = ( 1/ 2, 1 2) so that ( ) 1/ 2 1/ 2 [T ] β = 1/ 2 1/. 2 (b) Find a basis α of R 2 for which [T ] α is diagonal. Let α = {v 1, v 2 }. If we require [T ] α to be diagonal, then T (v i ) must be a multiple of v i for i = 1, 2. This can be achieved by choosing v 1 such that T leaves it unchanged; this will happen if the point corresponding to v 1 lies on the line y = x, say v 1 = (1, 1). We also need to choose v 2 so that it points in a different direction from v 1 and T (v 2 ) is a multiple of v 2. This can only be achieved if T (v 2 ) = 0; choosing v 2 = (1, 1) does the job. Thus, we have ( ) 1 0 [T ] α =