MTH Midterm 2 Review Questions - Solutions

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1 MTH Midterm 2 Review Questions - Solutions The following questions will be good practice for some of the questions you will see on the midterm. However, doing these problems alone is not nearly enough. You should also study HW4 HW7 and their posted solutions as well as Sections & in the text. In particular, you should be able to reproduce the proofs of all important theorems from these sections. Question. In this question you are expected to use the notation developed in the course and found in the text. Let V be a finite dimensional vector space, β = {v,..., v n } and β be ordered bases for V and T : V V be a linear map. (a Write down an expression for the jth column of the matrix [T ] β. Solution: [T (v j ] β (b Write down an expression for the matrix Q that changes β coordinates to β coordinates. Solution: Q = [I V ] β β (c Using part (b prove that [u] β = Q[u] β holds for all u V. Solution: Q[u] β = [I V ] β β [u] β = [I V (u] β = [u] β (using theorem 2.4 (d Write down an expression for [T ] β in terms of [T ] β and Q. No proof is required. Solution: [T ] β = Q [T ] β Q (see theorem 2.23 For the remainder of this question let V = R 2, β be the standard ordered basis and β = { ( ( 0, ( } and T given by T ( x ( y = y x (e Using part (d or otherwise find [T ] β, showing all your work. Solution: We see T ( ( ( ( ( x 0 x 0 y = [T ] β =. The change of coordinates 0 y 0 ( matrix from β to β coordinates is Q =. From part (d we have: 0 ( ( ( ( [T ] β = Q 0 2 [T ] β Q = = 0

2 (f Write down [T ] β. Solution: By theorem 2.8, [T ] β = ([T ] β = ( 2 Question 2. Let a linear map T : V W between finite dimensional vector spaces over the same field. Suppose that W is a subspace of V and N(T is the nullspace of T. Show that dim(n(t V = dim(v dim(t (V. Use the result to show that dim(t (V dim(v. Solution: Consider T V, the restriction of T to V. Recall N(T V = N(T V. Applying the Rank-Nullity (Dimension Theorem to T V : dim(v = nullity(t V +rank(t V = dim(n(t V +dim(t V (V = dim(n(t V +dim(t (V We see that dim(n(t V = dim(v dim(t (V. Since dim(n(t V 0 dim(v dim(t (V 0 dim(t (V dim(v. Question 3. Let A, B be n n matrices with coefficients in a field F. (a Define the map L A : F n F n. Solution: L A (x = Ax for all x F n. (b Write down an expression for L AB in terms of L A and L B. Solution: L AB = L A L B (since L AB (x = (ABx = A(Bx = L A (L B (x = (L A L B (x for all x F n (c Deduce from the above that if A is an invertible matrix, then L A is an invertible map. [Hint: Can you guess the inverse of L A?] Solution: If A exists, then L A is defined. We have: L A L A = L AA = L In = I F n and similarly L A L A = I F n. By definition of inverse transformation, the inverse of L A is given by L A = L A. Question 4. A be an invertible square matrix. Prove that (A t = (A t. Is the square matrix (A t invertible? If it is give its inverse, if it isn t explain why. Solution: A t (A t = (A A t = In t = I n and (A t A t = (AA t = In t = I n. By definition, (A t = (A t. The matrix (A t is invertible. We have just shown (A t is the inverse of A t, which implies ((A t = A t. 2

3 Question 5. You are given three pairs of finite dimensional vector spaces V, W. In each case decide whether V and W are isomorphic. If they are write down an isomorphism. If they are not clearly explain why. (a V = {p P 2 : p(tdt P 2 }, W = P 2. Solution: Not isomorphic. V is a subspace of P 2 hence these spaces are isomorphic only if V = P 2. Since t 2 dt = t3 3 P 2, it follows that t 2 V. So V P 2. (b V = {(x, y, z R 3 : 2x y + 5z = 0} and W = R 2. Solution: Isomorphic. Write x = y/2 5z/2 V = {(s/2 5t/2, s, t : s, t R} = {s(/2,, 0 + t( 5/2, 0, : s, t R} = span{(/2,, 0, ( 5/2, 0, }. We see that {(/2,, 0, ( 5/2, 0, } is a basis for V since these vectors are clearly linearly independent. V is then isomorphic to R 2 since both spaces have dimension 2. Define T : R 2 V as the unique linear map satisfying T (, 0 = (/2,, 0 and T (0, = ( 5/2, 0, (exists and is unique by theorem note that in particular, T (a, b = (a/2 5b/2, a, b. This is an isomorphism since it maps the basis of R 2 to the basis of the isomorphic space V (see proof of theorem 2.9. (c V = {M M 2 2 (R : M t = M} and W = R. ( ( ( a b a b a c Solution: Note that V if and only if =. Then, a = a, d = c d c d b d d, ( b = c, c ( = b a = d {( = 0 and} b = c. So every matrix in V takes the form 0 b 0 0 = b. We see is a basis for V, hence V has dimension one and b ( 0 a is isomorphic to R. In particular, T (a = provides an isomorphism from R to V. a 0 Question 6. True or false. Decide whether the following statements are true or false. If they are true prove them (briefly state clearly any results you use; if they are false demonstrate this by means of an example. (a If ABC = I n where A, B, and C are n n matrices, then B is invertible. Solution: True. There are several ways to prove this but the intuition is that since L In dimension preserving map (i.e. rank(i n = n, the maps L A, L B, L C from F n F n must all be dimension preserving as well (i.e. all have rank n. Here is a proof utilizing theorem 3.7: From theorem 3.7 we have n = rank(i n = rank(abc = rank((abc rank(ab rank(b. Since B is n n, we also have rank(b n. It follows that rank(b = n which implies B is invertible. 3 is a

4 (b If A is an invertible square matrix, then A 2 is invertible. Solution: True. If A exists we note that A 2 (A 2 = AAA A = AI n A = AA = I n. Similarly, (A 2 A 2 = I n. So A 2 is invertible (and in fact, (A 2 = (A 2. (c If A is an invertible square matrix, then A + A is invertible. Solution: True. A + A = 2A. If A exists, we note that (2A( 2 A = AA = I n. Similarly, ( 2 A (2A = I n. So 2A is invertible (and in fact, (2A = 2 A. (d If A, B are m n matrices and Ax = Bx for every x R n, then A = B. Solution: True. The jth column of A is given by Ae j and similar for B. If Ae j = Be j for each j, it follows that A and B have the same columns, and hence are the same matrix. (e Let P be the space of polynomials (of any degree with real coefficients. If T : P P is linear and onto, then T is one-to-one. Solution: False. Consider the map D(p = p where p is the derivative of p P. This map is linear and onto, since for any q P, there exists p P such that D(p = q (in particular, p is the anti-derivative of q, which is also a polynomial. However, this map is not one-to-one, which can be seen since D(x = = D(x + (for example. (f Let P be the space of polynomials (of any degree with real coefficients. If T : P P is linear and one-to-one, then T is onto. Solution: False. Let T (p = x 0 p(tdt. This is a linear, one-to-one map from P P. This can ( x p(tdt ( = d x q(tdt 0 dx 0 be seen since if T (p = T (q, then x p(tdt = x q(tdt d 0 0 dx p(x = q(x for all x. However, this map is not onto since there does not exist p P such that T (p = (for example. (g If T : R 2 R 2 is linear and onto, then T is one-to-one. Solution: True. This follows directly from theorem 2.5. (h If T : R 2 R 3 is linear and one-to-one, then T is onto. Solution: False. It is impossible for any linear map T : R 2 R 3 to be onto, since as a consequence of the Rank-Nullity theorem rank(t 2 in this case. So if one wants to construct a counterexample, any linear one-to-one map from R 2 to R 3 will work. For example, T (x, y = (x, y, 0 is clearly not onto, but is one-to-one since N(T = {(0, 0}. (i There exists a linear map T : R 3 R 3 such that R(T = N(T. Solution: False. Let T : R 3 R 3 be linear. By the Rank-Nullity theorem, 3 = dim(r(t + dim(n(t. Since 3 is odd and dimensions are positive integers or 0, it follows that dim(r(t 4

5 dim(n(t here which implies R(T N(T. Question 7. Let V be a finite dimensional vector space and U, T : V V be linear maps. Prove that rank(u T min{rank(u, rank(t }. Deduce that rank(ab min{rank(a, rank(b} for n n matrices A and B. You may use any result (or definition you require provided it is stated clearly. Solution: rank(ut rank(u follows directly from the observation that R(UT = U(T (V U(W = R(U, since T (V W. For rank(ut rank(t, we apply the result of question 2 above (here the subspace is T (V and the map is U to obtain dim(u(t (V = dim(t (V. Then, rank(ut rank(t directly follows. This proves rank(ut min{rank(u, rank(t }. Using these results applied to L A and L B, we have rank(ab = rank(l AB = rank(l A L B rank(l A = rank(a and rank(ab = rank(l AB = rank(l A L B rank(l B = rank(b. This proves rank(ab min{rank(a, rank(b}. 2 Question 8. Express the matrix A = as a product of elementary matrices showing all your work Solution: Note that A can be transformed into the identity by the following sequence of row operations: ( Swap rows 2 and 3 (2 Multiply row 2 by /3 (3 Add - times row 2 to row (4 Add -2 times row 3 to row 0 It follows that E 4 E 3 E 2 E A = I n, where E =, E 2 = 0 /3 0, E 3 = 0 0, E 4 = 0 0. Note that E 4 E 3 E 2 E A = I n A = E E2 E3 E4. The inverses of these matrices are the elementary matrices corresponding to the inverse row operations of (-(4 above. We have: A =

6 Question 9. Find the rank of the following matrices briefly justifying your answers. ( 2 0 (a A = Solution: rank(a = 2 since the dimension of the subspace spanned by the column vectors of A is 2 (note that the first two column vectors are linearly independent. 2 (b B = Solution: Recall that elementary column operations are rank preserving. By adding 2 times 0 column to column 2, we obtain Since the subspace spanned by the column vectors 3 0 of this matrix has dimension, it follows that this matrix has rank, and rank(b =. 0 0 (c C = Solution: This matrix can be transformed into I 4 by elementary row operations. Since these are rank preserving operations and rank(i 4 = 4, it follows that rank(c = 4. Question 0. Let V, W, Z be a finite dimensional vector spaces over the same field and T : V W and U : W Z be linear maps. Suppose that N(U = {0}. Then nullspace(ut = nullspace(t. Solution: N(UT = {x V : (UT (x = 0} = {x V : U(T (x = 0} = {x V : T (x = 0}. The last equality holds since N(U = {0}. Since N(T = {x V : T (x = 0} the result follows. Question. Let V be a finite dimensional vector space and T, U : V V be linear maps. Recall that T 0 : V V is the zero map. (a Prove that UT = T 0 if and only if R(T N(U. Solution: = : UT = T 0 implies that UT (x = 0 for all x V. So T (x N(U for all x V and consequently R(T = {T (x : x V } N(U. 6

7 = : UT (V = U(R(T = {0} as R(T N(U and so UT (x = 0 for all x V. (b Let T : R 2 R 2 be a linear map of rank. Prove that there exists a non-zero linear map U such that UT = T 0. Solution: We know that the range of T is a one dimensional subspace of R 2. So it is spanned by a vector v. We extend the singleton {v} to a basis {v, u} of R 2. We then define U on the basis by U(v = 0 and U(u = u and extend linearly. Formally we define T (λv + µu = µu. This is a non-zero linear map and UT = T 0 as T (V = span{v} and UT (V = U(span{v} = span{u(v} = {0}. Question 2. Let T : P 2 P 2 be the linear map defined by T (p(x = xp (x + p(x and β = { + x, x}, β = {x, } be ordered bases for P 2. Find [T ] β β and [T (2x + 2] β showing all your work. Solution: The first column of [T ] β β is [T ( + x] β = [ + 2x] β = ( 2 as + 2x = 2 x +. Similarly [T ( x] β = [] β = ( ( 0 and so [T ] β β = 2 0. [T (2x + 2] β = [4x + 2] β = ( 4 2 as 4x + 2 = 4 x + 2. Question 3. Let T : P 2 P 2 be dened by T (p = 4p + p. (a Let β be the standard ordered basis for P 2. Find [T ] β showing all your work. Solution: T ( = =, T (x = 4 + x = 4 + x, T (x 2 = 8x + x 2 = 8 x + x 2. So 4 0 [T ] β = 0 8. (b Let γ = { + x, x, x 2 } be an ordered basis for P 2. Find [T ] γ showing all your work. Solution: The change of coordinates matrix from γ coordinates to β coordinates is Q = [I P2 ] β γ = 0 0. The change of coordinates matrix from β coordinates to γ coordinates is Q = [I P2 ] γ β = 2 2 0, as for example = ( + x + ( x x2 and so the first column of the matrix is (,, 2 2 0t. 7

8 3 0 4 We have [T ] γ = Q [T ] β Q = Question 4. Let A = (a Find a basis for R(L A and give a brief justification why the set is a basis. 0 2 Solution: R(L A = L A (R 3 = span 2,, 4. The first and third vectors are multiples of one another; the second is not a multiple of either. So 2, is a 3 linearly independent spanning set and hence a basis for R(L A. (b Compute nullity(l A. Solution: From the Rank-Nullity Theorem we have nullity(l A = dim(r 3 rank(l A = 3 2 = as rank(l A = dim(r(l A. 2 (c Find a basis for L A (W, where W = span,, Solution: L A (W = span L A, L A, L A 0 = span,, 0 = span. So a basis is. 4 4 Question 5. State and prove the Rank-Nullity (or Dimension theorem for a linear map T : V W. Solution: Read the book pp