= A+A t +αb +αb t = T(A)+αT(B). Thus, T is a linear transformation.

Transcription

1 Math 56 - Homework 3 Solutions Prof Arturo Magidin Recall from Homework that V {r R r > 0} is a real vector space with addition v w vw product as real numbers and scalar multiplication α v v α exponentiation as real numbers Let T : V R be given by Tv lnv Show that T is a linear transformation where R is a vector space under the usual operations Proof We have Tv w lnv w lnvw lnv+lnw Tv+Tw; Tα v lnα v lnv α αlnv αtv Thus, T is a linear transformation, as claimed 2 Prove that there exists a linear transformation T: R 2 P 2 R such that T, + x 2 and T2,3 x+4x 2 What is T8, for that linear transformation? Proof Note that β [,,2,3] is a basis for R 2 Indeed, it is linearly independent: if ρ,+σ2,3 0,0, then ρ+2σ ρ+3σ 0; this means that ρ 2σ 3σ, hence σ 0, so ρ 0 Since β is linearly independent and has dimr 2 elements, it is a basis By a theorem proven in class, if β is a basis and we specify where we want the basis elements to map, then this determines a unique linear transformation Invoking that theorem, we have that there exists a unique linear transformation T with the desired properties To finish off, we just need to use the given information to figure out the value of T8, To do that, we need to express 8, in terms of β If ρ,+σ2,3 8,, then ρ+2σ 8 and ρ+3σ Subtracting the first equation from the second we have that σ 3, and hence ρ 2 We therefore have: T8, T 2,+32,3 2T,+3T2,3 2+x 2 +3 x+4x 2 5 3x+4x 2 3 Let T : M 2 2 R M 2 2 R be given by TA A+A t, where A t is the transpose of A Show that T is a linear transformation you may take the properties of the transpose for granted, and find bases for NT and RT; use the bases to find nullityt and rankt Answer Let A and B be 2 2 matrices, and α a scalar We have Thus, T is a linear transformation TA+αB A+αB+A+αB t A+αB +A t +αb t A+A t +αb +αb t TA+αTB Let us first find NT: if A NT, then A+A t 0; that is, A t A, so A is skew symmetric a b Thus, if A, then c d a c a b c d a b c d

2 Thus, a a, c b, b c, d d This means that a d 0, and b c { } 0 Thus, we have that is a basis for NT: the calculations above show that every 0 element of NT is a scalar multiple of this vector, and the set is clearly linearly independent In particular, nullityt At this point we know from the Rank-Nullity Theorem that the rank must be 3 To find RT, we may note that for all A, TA t A+A t t A t +A t t A t +A TA a b That is, TA is symmetric Is every symmetric matrix in RT? Suppose that C is a symmetric matrix Then T 2 a 2 b 2 a 2 b + 2 a 2 b a b 2 b 2 b 2 b Thus, every matrix in the image is symmetric, every symmetric matrix is in the image Thus, a basis for RT is given by [ ] ,,, and this matches our previous observation that the rank must be equal to three 4 Let V R be the vector space of real sequences Define T l,t r : V V, the left shift and right shift operators, by T l a,a 2,a 3, a 2,a 3,a 4,, T r a,a 2,a 3, 0,a,a 2,a 3, i Prove that T l and T r are linear transformations Proof Let a,a 2, and b,b 2, be elements of V, and α R We have: T l a,a 2,+αb,b 2, T l a +αb,a 2 +αb 2, a 2 +αb 2,a 3 +αb 3, a 2,a 3,+αb 2,b 3, T l a,a 2,+αT l b,b 2, T r a,a 2,+αb,b 2, T r a +αb,a 2 +αb 2, Thus, T l and T r are both linear transformations 0,a +αb,a 2 +αb 2, 0,a,a 2,+α0,b,b 2, Ta,a 2,+αTb,b 2, ii Show that T l is onto but not one-to-one Proof Let b,b 2, V Then T l 0,b,b 2, b,b 2,, so T l is onto Since T l,b,b 2, b,b 2, T l 0,b,b 2,, it follows that T l is not one-to-one iii Show that T r is one-to-one, but not onto Proof If T r a,a 2, T r b,b 2,, then 0,a,a 2, 0,b,b 2,, hence a i b i for all i, so a,a 2, b,b 2, Thus, T r is one-to-one 2

3 Alternatively, if 0,0, T r a,a 2, 0,a,a 2,, then a i 0 for all i Thus, NT r {0}, so T r is one-to-one To show T r is not onto, note that if b,b 2, T r a,a 2,, then b 0 In particular,,,, / RT r, showing that T r is not onto Note: One can also prove that T r is one-to-one and T l is onto by noting that T l T r I V Since the composition is bijective, it follows from set-theoretic properties that the first function, T r is one-to-one, and the second function, T l is onto iv Find nullityt r, rankt r, nullityt l, and rankt l Proof Since T r is one-to-one, nullityt r 0 By the Rank-Nullity Theorem, it follows that rankt r dimv 2 ℵ0 Now, T l is onto, so rankt l dimv 2 ℵ0 However, the Rank-Nullity Theorem does not now help us find nullityt l What is NT l? If a,a 2, NT l, then Thus, a i 0 for all i > Hence 0,0,0, T l a,a 2, a 2,a 3, NT l {a,0,0, V a R} A basis for NT l is then given by the single vector v,0,0,0,, hence nullityt l dimnt 5 Let β [,x,x 2 ] be the standard ordered basis for P 2 R, and let [ γ,,, be an ordered basis for M 2 2 R Let T: P 2 R M 2 2 R be given by p p Tpx 0 p2 p0 i Prove that T is a linear transformation Proof Note that the map e 3 : P 3 R R given by e 3 px p3, evaluation at 3, is linear: e 3 p + αq p + αq3 p3 + αq3 e 3 p + αe 3 q Similarly, e 2 and e 0, evaluation at 2 and at 0, are linear Finally, the map P 3 R P 3 R given by p p is linear, and composing with evaluation at 2, we get a linear map Thus, we have: p+αq p+αq Tpx+αqx 0 p+αq2 p+αq0 p+αq p 0+αq 0 Thus, T is linear p2+αq2 p0+αq0 p p 0 +α p2 p0 Tpx+αTqx ii Find the coordinate matrix of T relative to β and γ, [T] γ β Proof Evaluating, we have: 0 T ] q q 0 q2 q0 3

4 Tx 2 0 Tx Thus, we have [T] γ β Let V be a vector space and let W be a subspace of V Define a function T: V V/W by Tv v+w for all v V i Prove that T is a linear transformation Proof Let v,v 2 V, α F We have: Tv +αv 2 v +αv 2 +W v +W+αv 2 +W Tv +αtv 2, hence T is linear ii Prove that T is onto and that NT W Proof Let v+w V/W Then Tv v+w, so T is onto Now, letv V Thenv NTifandonlyif0 V/W Tv v+w Since0 V/W 0+W, v NT if and only if v+w 0+W From Homework, we know that his holds if and only if v 0 v W Thus, v NT if and only if v W, proving that NT W iii Find a formula that relates dimv, dimw, and dimv/w Answer From the Dimension Theorem we have dimv nullityt+rankt dimw+dimv/w Thus, dimv dimw + dimv/w If V is finite dimensional, then this gives dimv/w dimv dimw, but in generaly we cannot write it this way, because there is no subtraction of cardinal numbers iv Let δ be a basis for W, and extend δ to a basis β δ γ with δ γ for V Prove that {v+w v γ} is a basis for V/W Answer Suppose that v,,v n γ are pairwise distinct, α,,α n F, and 0+W α v +W+ +α n v n +W α v + +α n v n +W We want to show that α α n 0 We have that α v + +α n v n NT W, so this vector is a linear combination of elements of δ: let w,,w m δ be pairwise distinct, and ρ,,ρ m be scalars such that α v + +α n v n ρ w + +ρ m w m Then α v + + α n v n ρ w ρ m w m 0 V Since this is al inear combination of pairwise distinct elements of β equal to 0 V, it follows that α α n 0 and ρ ρ m 0 Thus, {v+w v γ} is linearly independent 4

5 In the finite dimensional case we would be done, because this set has the correct number of elements But since we are not assuming V is finite dimensional, we must also prove that this set spans Let v + W V/W Then v V, so there exist w,w m δ, v,,v n γ pairwise distinct, and scalars ρ,,ρ m,σ,,σ n F, such that Then v ρ w + +ρ m w m +σ v + +σ n v n v+w Tv Tρ w + +ρ m w m +σ v + +σ n v n ρ Tw + +ρ m Tw m +σ Tv + +σ n Tv n ρ w +W+ +ρ m w m +W+σ v +W+ +σ n v n +W ρ 0 V/W + +ρ m 0 V/W +σ v +W+ +σ n v n +W σ v +W+ +σ n v n +W Thus, v+w span{v+w v γ}, proving that {v+w v γ} is a spanning set Since we have already proven the set is linearly independent, it follows that it is basis, as claimed 7 If V is a vector space, and W, W 2 are subspaces, then V is the direct sum of W and W 2, written V W W 2, if and only if V W +W 2 and W W 2 {0} Prove that M n n R W W 2, where W is the set of all n n symmetric matrices matrices A such that A t A, and W 2 is the set of all skew-symmetric matrices matrices B such that B t B Proof First, we note that W and W 2 are subspaces If A,B W 2, then A t A, B t B, and so A+αB t A t +αb t A+α B A+αB, hence A+αB W 2 In addition, the zero matrix is in W 2 Thus, W 2 is a subspace Similarly, if A,B W, then A t A, B t B, so A+αB t A t +αb t A+αB And since the zero matrix is symmetric as well as skew-symmetric, W is a subspace as well Second, note that if A W W 2, then A A t A, so A A; therefore, every entry of A is equal to 0, so W W 2 {0} Finally, we need to show that W + W 2 M n n R There are several ways of doing it, but here is an elegant one: Let A be an n n matrix Let B 2 A+At Then B t 2 At +A B, so B is symmetric Let C 2 A At ; then C t 2 At A C, so C is skew-symmetric Thus B W and C W 2 Now note that B +C 2 A+At + 2 A At A, soa W +W 2 Thus, weconcludethatm n n R W +W 2, andthisprovesthatm n n R W W 2 5