1. Vector spaces (1) Define vector space: V is a vector space if ... (2) Define vector subspace: W is a subspace of a vector space V if ... (3) Determine which of the following sets are vector subspaces. x (g) fp(t) 2 P : p(t) = p(−t)g (a) : x ≥ 0; y ≥ 0 2 y (h) fp(t) 2 P2 : − p(t) = p(−t)g x 8 2 3 9 (b) : xy ≥ 0 1 y < 3×3 = (i) A 2 R : 425 2 ker(A) 82 3 9 : 3 ; < 1 − a = (c) 4a − 6b5 : a; b 2 R (j) B 2 3×3 : B is in reduced row-echelon form : 2b + a ; R 1 1 2 0 82 3 9 (k) A 2 2×2 : A = A < 2b + 3c = R 1 1 0 0 (d) 4 b 5 : b; c 2 R : c ; (l) The set of infinite sequences of the form (a; a + k; a + 2k; a + 3k; : : :) where a and (e) fp(t) 2 P2 : p(2) = 0g,(P2 is the set k are constants. of polynomials of degree at most 2). 1 (m) The set of infinite sequences of the form R 2 3 (f) fp(t) 2 P2 : 0 p(t) dt = 0g (a; ak; ak ; ak ;:::) where a and k are constants. (4) Find a basis and determine the dimension of each of the following subspaces. R 1 (a) The space of all polynomials f(t) 2 P3 such that f(1) = 0 and −1 f(t) dt = 0. (b) The space of all lower triangular 2 × 2 matrices. 1 1 0 0 (c) The space of all 2 × 2 matrices A such that A = . 1 1 0 0 (5) Find all solutions of the differential equation f 00(x) + 8f 0(x) − 20f(x) = 0. (See Example 3 in the 4.2 notes.) selected answers: (3) Recall that a space W is a vector subspace if • W contains the zero element. • W is closed under addition: if f and g are in W , then f + g is in W . • W is closed under scalar multiplication: if f is in W , then kf is in W for every scalar k. x (a) W = : x ≥ 0; y ≥ 0 is not a subspace because it is not closed under scalar y 1 1 −1 multiplication: the vector is in W , but − = is not in W . 1 1 −1 0 Note that this is the only reason why W fails to be a vector space. W contains , 0 x x the zero element, and W is closed under addition: if 1 and 2 are in W , then y1 y2 x1 x2 x1 + x2 + = is in W since x1 + x2 ≥ 0 and y1 + y2 ≥ 0. y1 y2 y1 + y2 1 2 x (b) W = : xy ≥ 0 is not a vector subspace because it is not closed under addition: y 1 0 1 0 1 the vectors and are in W , but + = is not in W . 0 −1 0 −1 −1 0 This space does contain , the zero element, and this space is closed under scalar 0 x x kx multiplication: if is in W , then k = is in W since (kx)(ky) = k2xy ≥ 0. y y ky x (xy ≥ 0 since is in W ). y 82 3 9 < 1 − a = (c) W = 4a − 6b5 : a; b 2 R is not a vector subspace because, for example, it does not : 2b + a ; 2 1 − a 3 contain the zero element: there are no real numbers a and b such that 4a − 6b5 = 2b + a 0; 0; 0. R 1 (f) W = fp(t) 2 P2 : 0 p(t) dt = 0g is a vector subspace: 1) W contains the zero function 2 R 1 2 p0(t) = 0 + 0t + 0t since 0 0 + 0t + 0t dt = 0. 2) W is closed under addition: R 1 if p1(t) and p2(t) are in W , then p1(t) + p2(t) is in W since 0 p1(t) + p2(t) dt = R 1 R 1 0 p1(t) dt + 0 p2(t) dt = 0 + 0 = 0. 3) W is closed under scalar multiplication: if p(t) R 1 R 1 is in W , then kp(t) is in W since 0 kp(t) dt = k 0 p(t) dt = k · 0 = 0: 8 213 9 < 3×3 = (i) W = A 2 R : 425 2 ker(A) is a vector subspace: 1) W contains the zero matrix : 3 ; 20 0 03 213 20 0 03 40 0 05 since 425 is in the kernel of 40 0 05. 2) W is closed under addition: if 0 0 0 3 0 0 0 213 213 213 A1 and A2 are in W , then A1 + A2 is in W since (A1 + A2) 425 = A1 425 + A2 425 = 3 3 3 203 203 203 405 + 405 = 405. 3) W is closed under scalar multiplication: if A is in W , then kA 0 0 0 213 213 203 203 is in W since (kA) 425 = kA 425 = k 405 = 405. 3 3 0 0 (l) W = f(a; a + k; a + 2k; a + 3k; : : :)g is a vector space: 1) W contains the sequence (0; 0; 0;:::)(a = k = 0). 2) W is closed under addition: if (a1; a1 + k1; a1 + 2k1;:::) and (a2; a2 + k2; a2 + 2k2;:::) are in W , then the sum of the sequences is in W since (a1; a1 + k1; a1 + 2k1;:::) + (a2; a2 + k2; a2 + 2k2;:::) = (a1 + a2; a1 + a2 + k1 + k2; a1 + a2 + 2k1 + 2k2;:::) = (a; a + k; a + 2k; : : :) where a = a1 + a2 and k = k1 + k2. (4) To find a basis, we need to write down a general element of the vector space and determine the number of free variables. 3 (a) A general polynomial of degree 3 has the form f(t) = a + bt + ct2 + dt3. Additionally, R 1 if f(1) = 0, then a + b + c + d = 0, and if −1 f(t) dt = 0, then Z 1 bt2 ct3 dt4 1 2 3 0 = a + bt + ct + dt dt = at + + + −1 2 3 4 −1 b c d b c d 2c = a + + + − −a + − + = 2a + : 2 3 4 2 3 4 3 We must now solve the system of equations 2c 2a + = 0 3 a + b + c + d = 0: Putting these in an augmented matrix, we have 2 0 2=3 0 0 1 0 1=3 0 0 1 0 1=3 0 0 ∼ ∼ ; 1 1 1 1 0 1 1 1 1 0 0 1 2=3 1 0 c 2c hence a = − 3 and b = − 3 − d. Plugging these back into the equation for f(t), we see that c 2c 1 2 f(t) = − − + d t + ct2 + dt3 = c − − t + t2 + d −t + t3 : 3 3 3 3 3 2 2 1 Thus f is a linear combination of the polynomials t − t and t − 3 t − 3 . Moreover, these polynomials form a linearly independent set, and thus give a basis for this space: 3 2 2 1 B = t − t; t − 3 t − 3 . There are two elements in the basis, so the dimension of this vector space is 2. a b (c) A general 2 × 2 matrix is A = . For A to satisfy the condition to be in the set, c d we have 0 0 a b 1 1 a + b a + b = = 0 0 c d 1 1 c + d c + d Hence a = −b and c = −d, and −b b −1 1 0 0 A = = b + d : −d d 0 0 −1 1 −1 1 0 0 Thus a basis for this space is B = ; , and the dimension of this 0 0 −1 1 space is 2. 2. Linear transformations (1) Define linear transformation: a function T from a vector space V to a vector space W is a linear transformation if ... (2) Which of the transformations are linear? For those that are linear, determine whether they are isomophisms. 2×2 2×2 (a) T (M) = M + I2 from R to R . 2×2 2×2 (b) T (M) = 7M from R to R . 2 2×2 2×2 (c) T (M) = M from R to R . 1 2 (d) T (M) = M from 2×2 to 2×2. 3 4 R R 4 1 2 (e) T (M) = M M from 2×2 to 2×2. 3 4 R R (f) T (a + bi) = a from C to C.(C is the set of complex numbers.) (g) T (a + bi) = b + ai from C to C. (h) T (f(t)) = f(7) from P2 to R. 00 (i) T (f(t)) = f (t)f(t) from P2 to P2. (j) T (f(t)) = f(−t) from P2 to P2. (k) T (f(t)) = f(2t) − f(t) from P2 to P2. (3) Define image: the image of a linear transformation T is ... (4) Define rank: the rank of a linear transformation T is ... (5) Define kernel: the kernel of a linear transformation T is ... (6) Define nullity: the nullity of a linear transformation T is ... (7) Find the image, rank, kernel, and nullity of the transformation T from P2 to P2 defined by T (f(t)) = f 00(t) + 4f 0(t). (8) Find the image, rank, kernel, and nullity of the transformation T from P2 to R defined by R 3 T (f(t)) = −2 f(t) dt. 2×2 2×2 (9) Find the image, rank, kernel, and nullity of the transformation T from R to R defined 1 2 1 2 by T (M) = M − M.