Vector spaces and Fourier theory — Exam (1) Let V and W be vector spaces over R. (a) Define what it means for a map α : V → W to be linear. (3 marks) (b) Define what it means for a subset U ⊆ V to be a subspace. (3 marks) (c) Suppose that α is linear. Define the kernel of α, and prove that it is a subspace of V . (5 marks) (d) Which of the following functions are linear? (You should justify your answers briefly.) (6 marks) (i) ρ : M2 (R) → R given by ρ(A) = trace(A2 ) h 3 2 2 i +y ) (ii) σ : R2 → R2 given by σ [ xy ] = xy 3 /(x /(x2 +y 2 ) (iii) τ : M2 (R) → M2 (R) given by τ (A) = [ 11 11 ] A [ 11 11 ] (e) Which of the following sets is a subspace? (You should justify your answers briefly.) (8 marks) (i) U0 = {f ∈ R[x] | f (1) ≤ f (2) ≤ f (3)} (ii) U1 = {A ∈ M2 (R) | det(A + I) = det(A − I)} (iii) U2 = {A ∈ M2 (R) | trace(A2 ) = 0} (Hint: which elements of the standard basis for M2 (R) lie in U2 ?) Solution: (a) Bookwork. α is linear if α(tv + t0 v 0 ) = tα(v) + t0 α(v 0 ) for all t, t0 ∈ R and v, v 0 ∈ V . [3] (b) Bookwork. U is a subspace if (i) 0 ∈ U and (ii) for all t, t0 ∈ R and all u, u0 ∈ U we have tu + t0 u0 ∈ U . [3] (c) Bookwork. The kernel of α is the set {u ∈ V | α(u) = 0}.[1]As α is linear we have α(0) = 0, so 0 ∈ ker(α). [1]Now suppose that t, t0 ∈ R and u, u0 ∈ ker(α). We then have α(u) = 0 = α(u0 ) [1]and also α(tu + t0 u0 ) = tα(u) + t0 α(u0 ) = t.0 + t0 .0 = 0, so tu + t0 u0 ∈ ker(α). [2] This shows that ker(α) is a subspace. (d) Similar to problem sheets (i) We have ρ(I) = 2 = ρ(−I), so ρ(−I) 6= −ρ(I), so ρ is not linear. [2] (ii) We have σ([1, 0]T ) = [03 /(12 + 02 ), 13 /(12 + 02 )]T = [0, 1]T σ([0, 1]T ) = [13 /(12 + 02 ), 03 /(12 + 02 )]T = [1, 0]T σ([1, 1]T ) = [13 /(12 + 12 ), 13 /(12 + 12 )]T = [1/2, 1/2]T so σ(e1 + e2 ) 6= σ(e1 ) + σ(e2 ). Thus, σ is not linear. [2]

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(iii) Put Q = [ 11 11 ] for brevity. We have τ (tA + t0 A0 ) = Q(tA + t0 A0 )Q = QtAQ + Qt0 A0 Q = tQAQ + t0 QA0 Q = tτ (A) + t0 τ (A0 ), so τ is linear. [2] (e) Similar to problem sheets (i) U0 is not a subspace, because the function f (x) = x lies in U0 , but −f does not. [2]   (ii) Given a matrix A = ac db we have   b det(A + I) = det a+1 c d+1 = ad + a + d + 1 − bc   b det(A − I) = det a−1 c d−1 = ad − a − d + 1 − bc det(A + I) − det(A − I) = 2(a + d).  b  This means that A lies in U1 iff a + d = 0 iff A has the form ac −a . Given this, it is clear that U1 is a subspace. [3] h 2 i   a +bc b(a+d) (iii) Given a matrix A = ac db we have A2 = c(a+d) d2 +bc , so trace(A2 ) = a2 + d2 + 2bc. Using this we see that the matrices [ 00 10 ] and [ 01 00 ] lie in U2 , but their sum does not. Thus, U2 is not a subspace. [3] (2) Consider the linear map α : R[x]≤2 → M2 (R) given by h i f (0) f (1) α(f ) = f 0 (0) f 0 (1) . (a) Write down a basis U for R[x]≤2 and a basis V for M2 (R). (3 marks) (b) Find the matrix of α with respect to the bases U and V. (5 marks) (c) Show that α is injective. (4 marks) (d) Give a basis for the image of α. (4 marks)   (e) Find a nonzero matrix X = ac db such that hX, α(xi )i = 0 for i = 0, 1, 2. (5 marks) (Here we use the standard inner product for square matrices.) (f) Show (by an explicit example) that α is not surjective. (4 marks) Solution: (a) Similar to problem sheets The obvious basis for R[x]≤2 consists of the polynomials p0 (x) = 1, p1 (x) = x and p2 (x) = x2 .[1]The obvious basis for M2 (R) consists of the matrices E1 = [ 10 00 ]

E2 = [ 00 10 ]

E3 = [ 01 00 ]

E4 = [ 00 01 ] .[2]

(b) Similar to problem sheets We have α(p0 ) = [ 10 10 ] = 1.E1 + 1.E2 + 0.E3 + 0.E4 α(p1 ) = [ 01 11 ] = 0.E1 + 1.E2 + 1.E3 + 1.E4 α(p2 ) = [ 00 12 ] = 0.E1 + 1.E2 + 0.E3 + 2.E4 [3] so the matrix of α with respect to our bases is 1 0 0 A = 10 11 10 .[2] 012

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(c) Similar to problem sheets From the above we see that α(a + bx + cx2 ) = aα(p0 ) + bα(p1 ) + cα(p2 ) =

 a a+b+c  b b+2c

.[2]

Thus, if α(a + bx + cx2 ) = 0 we see that a = a + b + c = b = b + 2c = 0, which easily implies that a = b = c = 0. This shows that ker(α) = {0} and thus that α is injective. [2] (d) Unseen As α is injective, the matrices α(p0 ) = [ 10 10 ], α(p1 ) = [ 01 11 ] and α(p2 ) = [ 00 12 ] form a basis for the image. [4] (e) Similar to problem sheets We have   hX, α(1)i = h ac db , [ 10 10 ]i = a + b   hX, α(x)i = h ac db , [ 01 11 ]i = b + c + d   hX, α(x2 )i = h ac db , [ 00 12 ]i = b + 2d[2] We therefore must  have  a + b = b + c + d = b + 2d = 0, which gives a = 2d, b =  2 −2d  and −2 c = d, so X = d 21 −2 . [2] Here d is arbitrary so we can take d = 1 and so X = 1 1 1 . [1] (f) Unseen Observe that hX, Xi = 22 +(−2)2 +12 +12 = 10 6= 0, but hX, Ai = 0 for all A in the image of α (by part (d)). It follows that X 6∈ image(α), and thus that image(α) 6= M2 (R), so α is not surjective. [4] (3) (a) Define the standard inner product on the space M2 (R). (2 marks) (b) What is hA, Bi, where A = [ 10 11 ] and B = [ 11 01 ]? (2 marks) h i − sin(θ) (c) Let Rθ denote the rotation matrix cos(θ) sin(θ) cos(θ) . Simplify hRθ , Rφ i, and thus describe when hRθ , Rφ i = 0. (6 marks) (d) Put V = {X ∈ M2 (R) | X [ 11 ] = 0} and W = {Y ∈ M2 (R) | Y



1 −1



= 0}.

(i) Find the general form for elements of V , and thus the dimension of V .(3 marks) (ii) Find the general form for elements of W , and thus the dimension of W .(2 marks) (iii) Show that every element of V is orthogonal to every element of W .(1 marks) (iv) By comparing dimensions, prove that V ⊥ = W .(4 marks) (e) Find orthonormal bases for V and W . (5 marks) Solution: (a) Bookwork. The standard inner product on M2 (R) is hA, Bi = trace(AB T ). [2] (b) In particular, if A = [ 10 11 ] and B = [ 11 01 ] we have B T = A so AB T = A2 = [ 10 21 ] [1]so hA, Bi = trace [ 10 21 ] = 2 [1]. (c) Unseen. We have h ih i h i − sin(θ) cos(φ) sin(φ) cos(θ) cos(φ)+sin(θ) sin(φ) cos(θ) sin(φ)−sin(θ) cos(φ) Rθ RφT = cos(θ) = .[2] sin(θ) cos(θ) − sin(φ) cos(φ) sin(θ) cos(φ)−cos(θ) sin(φ) cos(θ) cos(φ)+sin(θ) sin(φ) Taking traces, we get hRθ , Rφ i = 2(cos(θ) cos(φ) + sin(θ) sin(φ)) = 2 cos(θ − φ).[2] In particular, this means that hRθ , Rφ i = 0 iff θ − φ has the form (n + 21 )π for some integer n. [2] 3

(d) Similar to problem sheets.   (i) For X = [ wy xz ] ∈ M2 (R) we have X [ 11 ] = w+x y+z , so X ∈ V iff x = −w and z = −y. [1]This means that    1 −1   0  X = wy −w + x 01 −1 .[1] −y = w 0 0  1 −1   0 0  , 1 −1 . [1] 0 0    p−q  1 (ii) Similarly, for a matrix Y = [ pr qs ] ∈ M2 (R) we have Y −1 = r−s , so Y ∈ W iff p = q and r = s. This means that It follows that V has dimension 2, with basis

Y = [ pr pr ] = p [ 10 10 ] + q [ 01 01 ] .[1] It follows that W also has dimension 2. [1] (iii) We have hX, Y i = trace(XY T ) = trace

 w −w  y −y


[ pp rr ] = trace [ 00 00 ] = 0.[1]

(iv) Part (iii) tells us that W ≤ V ⊥ . [1]However, we also have dim(W ) = 2 and dim(V ⊥ ) = dim(M2 (R)) − dim(V ) = 4 − 2 = 2 = dim(W ), so we must have V ⊥ = W . [3]   0 0  for V [1], and (e) Similar to problem sheets. We have seen that 10 −1 , 1 −1 is a basis 0   1 −1 1 1 0 0 √ √ these two matrices are orthogonal [1]. It follows that 2 0 0 , 2 1 −1 is an orthonormal basis for V [1]. Similarly, the list √12 [ 10 10 ] , √12 [ 01 01 ] is an orthonormal basis for W . [2] (4) (a) State and prove the Cauchy-Schwartz inequality. (10 marks) R1 (b) Find constants α and β such that −1 f (x) dx = αf (−1) + βf (0) + αf (1) for all f ∈ R[x]≤2 . (6 marks) √ R1 (c) Deduce that if f ∈ R[x]≤2 and −1 f (t)2 dt = 1, then |f (−1) + 4f (0) + f (1)| ≤ 3 2. (9 marks) R1 (You may assume that the rule hf, gi = −1 f (t)g(t) dt gives an inner product on R[x]≤2 .) Solution: (a) Bookwork. Let V be a vector space with an inner product, and let u and v be elements of V . Then |hu, vi| ≤ kukkvk.[2] Proof: For any s and t we have 0 ≤ ksu−tvk2 [1] = hsu−tv, su−tvi = s2 hu, ui−2sthu, vi+t2 hv, vi = s2 kuk2 +t2 kvk2 −2sthu, vi.[1] Now take s = kvk2 and t = hu, vi [2] to get 0 ≤ kuk2 kvk4 + hu, vi2 kvk2 − 2kvk2 hu, vi2 = kvk2 (kuk2 kvk2 − hu, vi2 ).[1] If v = 0 then we have |hu, vi| = 0 = kukkvk so the claim holds. [1]If v 6= 0 then kvk2 > 0 so the above inequality will remain valid after dividing by kvk2 , giving hu, vi2 ≤ kuk2 kvk2 . √ 2 [1]We now take square roots (and note that a = |a|) to get |hu, vi| ≤ kukkvk, as claimed.[1]

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(b) Similar to problem sheets. Consider a polynomial f (x) = ax2 + bx + c. We then have Z 1  1 f (x) dx = ax3 /3 + bx2 /2 + cx −1 = 2a/3 + 2c.[2] −1

On the other hand, we have αf (−1) + βf (0) + αf (1) = α(a − b + c) + βc + α(a + b + c) = 2αa + (2α + β)c.[2] For these to match up for all a, b and c we must have 2/3 = 2α and 2 = 2α + β, which gives α = 1/3 and β = 4/3. [2] (c) Unseen. Part (b) tells us that hf, 1i = (f (−1)+4f (0)+f (1))/3, so |f (−1)+4f (0)+f (1)| = hf, 3i [2]. The Cauchy-Schwartz inequality [1]tells us that this is at most kf kk3k [1]. Here R1 kf k2 = −1 f (t)2 dt, and we are given that this is equal to one, so kf k = 1 [1]. We also √ √ R1 have k3k2 = −1 32 dt = 18 [1]and so k3k = 18 = 3 2 [1]. Putting this together, we get √ |f (−1) + 4f (0) + f (1)| ≤ 3 2 [2] as claimed. (5) Let V be a vector space over R. (a) Define the map µV : Rn → V (where V = v1 , . . . , vn is a list of elements of V ). (2 marks) (b) Show that any linear map φ : Rn → V has the form φ = µV for some list V. (5 marks) (c) Some of the following situations are possible, and some are not. For each situation that is possible, give an example. For each situation that is impossible, give a brief argument to show that it is impossible. (8 marks) (i) A space V with a spanning list A of length 4 and a linearly independent list B of length 3. (ii) A space V with a spanning list A of length 3 and a linearly independent list B of length 4. (iii) A 3-dimensional space V with a list V of length 3 that is linearly independent but does not span. (iv) A 3-dimensional space V with a list V of length 3 that is linearly dependent and does not span. (d) Define what is meant by a jump in a sequence V. (3 marks) (e) Find the jumps in the following sequence: h0i h1i h2i v2 = 1 v3 = 2 v1 = 0 0

1

2

v4 =

h

1 0 −1

i

v5 =

h1i 2 3

v6 =

h3i 2 1

(7 marks) Solution: (a) Bookwork. The map µV is just given by µV ([λ1 , . . . , λn ]T ) =

P

i

λi vi . [2]

(b) Bookwork. Let φ : Rn → V be a linear map. Let e1 , . . . , en be the standard basis of Rn , P so x = i xi ei for all x ∈ Rn [1]. Put vi = φ(ei ) ∈ V [1]and V = v1 , . . . , vn . We claim that φ = µV [1]. Indeed, for any x ∈ Rn we have X X X φ(x) = φ( xi ei ) = xi φ(ei ) = xi vi = µV (x)[2] i

i

i

as claimed. 5

(c) Unseen. (i) An example is V = R3 with A = e1 , e2 , e3 , 0 and B = e1 , e2 , e3 . [2] (ii) This is impossible [1]by Steinitz’s Lemma: any spanning list must be at least as long as any linearly independent list. [1] (iii) This is impossible [1]: in a 3-dimensional space, a list of length three is independent iff it spans [1]. (iv) An example is V = R3 with V = 0, 0, 0.[2] (d) Bookwork. Put Vi = span(v1 , . . . , vi ) (with V0 = 0). We then say that i is a jump if vi 6∈ Vi−1 . [3] (e) Similar to problem sheets. As v1 = 0 ∈ V0 and v3 = 2v2 ∈ V2 and v5 = v3 − v4 ∈ V4 and v6 = v3 + v4 ∈ V5 , we see that 1, 3, 5 and 6 are not jumps [3]. As V1 = 0 and v2 6= 0, we see that 2 is a jump [2]. It is also clear that V3 is the set of vectors of the form [t, t, t]T , and v4 does not lie in that set, so 4 is a jump [2]. Thus, the set of jumps is precisely {2, 4}.

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