MATH1025 AF

FW00F

Anton and Rorres

Section 5.1

5. The set of vectors from R2 of the form (x, 0) is a vector space. Let

u = (x, 0) and v = (x´, 0). Then u + v = (x + x´, 0) which is of the required form. Similarly ku = (kx, 0) which is also of the required form. Moreover, the zero vector (0, 0) and -u = (-x, 0) are both of this form. The remaining axioms hold since they are valid for any vector in R2.

6. The set of vectors from R2 of the form u = (x, y) with x greater or equal to zero is not a vector space since ku with k negative and -u are not of the required form as the first element is negative in these cases.

9. The set of 2 by 2 matrices of the form M = is not a vector space since if

N = then M + N =which is not of the required form. Also the zero matix is not of this form. Moreover, -M = and kM =are not of the required form if k is different from one.

10. The set of 2 by 2 matrices of the form M = is a vector space since if

N = then M + N =which is of the required form. Also the zero matix is of this form. Moreover, -M = and kM =are also of this form. The remaining axioms hold for any 2 by 2 matrix.

15. Here the vectors are the positive real numbers but `addition' is defined as multiplication and `multiplication by a scalar' is defined as exponentiation. Thus x + y = xy which is positive if x and y are and kx = xk which is also a positive number if x is. Since multiplication of scalars is commutative and associative, axioms 2 and 3 hold. The 'zero' vector is 1 since 1+x = 1x = x and -x = 1/x since x - x = x(1/x) = 1 which is the 'zero' vector. Now k(x+y) = (xy)k = xkyk = kx + ky so axiom 7 holds. Also (k+j)x = xk+j = xkxj = kx + jx so axiom 8 holds. As well k(jx) = (xj)k = xkj = (kj)x and 1x = x1 = x which shows that axioms 9 and 10 hold as well.

Section 5.2

1. To prove that a set is a subspace we only need to prove the two closure axioms.

(a) If u = (a, 0, 0) and v = (b, 0, 0) then u + v = (a+b, 0, 0) and ku = (ka, 0, 0) both of which are of the required form. Thus it is a subspace of R3.

(b) If u = (a, 1, 1) and v = (b, 1, 1) then u + v = (a+b, 2, 2) and ku = (ka, k, k) neither of which are of the required form. Thus it is not a subspace of R3.

(c) If u = (a, b, c) with b = a + c and v = (p, q, r) with q = p + r then u + v = (a+p, b+q, c+r) and b+q = (a + c) + (p+r) = (a+p) + (c+r). Futhermore. ku = (ka, kb, kc) and kb = k(a+c) = ka + kc. Thus both are of the required form and it is a subspace of R3.

(d) If u = (a, b, c) with b = a + c + 1 and v = (p, q, r) with q = p + r + 1 then u + v = (a+p, b+q, c+r) and b+q = (a+c+1) + (p+r+1) = (a+p) + (c+r) + 2. Futhermore. ku = (ka, kb, kc) and kb = k(a+c+1) = ka + kc + k. Thus neither are of the required form and it is not a subspace of R3.

3. (a) If u = a0 + a1x + a2x2 + a3x3 with a0 = 0 and v = b0 + b1x + b2x2 + b3x3 with b0 = 0 then u + v = a0 + a1x + a2x2 + a3x3 + b0 + b1x + b2x2 + b3x3 = a0 + b0 + (a1 + b1)x +

(a2 + b2)x2 + (a3 + b3)x3 with a0 + b0 = 0. Also ku = ka0 + ka1x + ka2x2 + ka3x3 with

ka0 = 0. Thus they both have the required form and it is a subspace of P3.

(b) If u = a0 + a1x + a2x2 + a3x3 with a0 + a1 + a2 + a3 = 0 and v = b0 + b1x + b2x2 + b3x3 with b0 + b1 + b2 + b3 = 0 then u + v = a0 + a1x + a2x2 + a3x3 + b0 + b1x + b2x2 + b3x3 = a0 + b0 + (a1 + b1)x + (a2 + b2)x2 + (a3 + b3)x3 with a0 + a1 + a2 + a3 + b0 + b1 + b2 + b3 = 0. Also ku = ka0 + ka1x + ka2x2 + ka3x3 with ka0 + ka1 + ka2 + ka3 = 0. Thus they both have the required form and it is a subspace of P3.

(c) If u = a0 + a1x + a2x2 + a3x3 with a0, a1, a2 and a3 integers then ku = ka0 + ka1x + ka2x2 + ka3x3 and if k is not an integer the coefficients are not integers. Thus it is not a subspace of P3.

(d) If u = a0 + a1x and v = b0 + b1x then u + v = a0 + a1x + b0 + b1x = a0 + b0 +

(a1 + b1)x. Also ku = ka0 + ka1x. Thus they both have the required form and it is a subspace of P3.

5. (a) If A and B are n by n matrices and tr(A) = 0 and tr(B) = 0 then tr(A+B) = tr(A) + tr(B) = 0. Also tr(kA) = k tr(A) = 0. Thus these matrices are a subspace of Mnn.

(b) If A and B are n by n matrices and AT= -A and BT = -B then (A+B)T = AT + BT = -A + -B = -(A + B). Also (kA)T = k AT = k(-A) = -kA . Thus these matrices are a subspace of Mnn.

(c) If the n by n linear system Ax = 0 has only the trivial system then det(A) is not zero. If det(B) is not zero this does not mean that det(A+B) is not zero (if B = -A for example) so these matrices are not a subspace of Mnn.

(d) If A is an n by n matrix such that AB = BA for a fixed n by n matrix B and C is an n by n matrix such that CB = BC then (A+C)B = AB + AC = BA + BC = B(A+C). Also (kA)B = k(AB) = k(BA) = B(kA). Thus these matrices are a subspace of Mnn.

7. We must find scalars a and b such that au + bv is the given vector.

(a) We must have b = 2, -2a + 3b = 2 and 2a - b = 2. This linear system has the solution a = 2 and b = 2 so the vector is a linear combination of u and v.

(b) We must have b = 3, -2a + 3b = 1 and 2a - b = 5. This linear system has the solution a = 4 and b = 3 so the vector is a linear combination of u and v.

(c) We must have b = 0, -2a + 3b = 4 and 2a - b = 5. This linear system does not have a solution so the vector is not a linear combination of u and v.

(d) We must have b = 0, -2a + 3b = 0 and 2a - b = 0. This linear system has the solution a = 0 and b = 0 so the vector is a linear combination of u and v.

8. We must find scalars a, b and c such that au + bv + cw is the given vector.

(a) The coefficients are solutions of the linear system 2a + b + 3c = -9, a - b + 2c = -7 and 4a + 3b + 5c = -15. This has solution a = -2, b = 1 and c = -2 so that (-9, -7, -15) = -2u + v - 2w.

(b) The coefficients are solutions of the linear system 2a + b + 3c = 6, a - b + 2c = 11 and 4a + 3b + 5c = 6. This has solution a = 4, b = -5 and c = 1 so that (6, 11, 6) = 4u - 5v + w.

(c) The coefficients are solutions of the linear system 2a + b + 3c = 0, a - b + 2c = 0 and 4a + 3b + 5c = 0. This has solution a = 0, b = 0 and c = 0 so that (0, 0, 0) = 0u + 0v + 0w.

(d) The coefficients are solutions of the linear system 2a + b + 3c = 7, a - b + 2c = 8 and 4a + 3b + 5c = 9. This has solution a = 0, b = -2 and c = 3 so that (7, 8, 9) = 0u - 2v + 3w.

9. We must find scalars a, b and c such that ap1 + bp2 + cp3 is the given polynomial.

(a) The coefficients are solutions of the linear system 2a + b + 3c = -9, a - b + 3c = -7 and 4a + 3b + 5c = -15. This has solution a = -2, b = 1 and c = -2 (see question 8(a)) so that -9 - 7x - 15x2 = -2p1 + p2 - 2p3.

(b) The coefficients are solutions of the linear system 2a + b + 3c = 6, a - b + 2c = 11 and 4a + 3b + 5c = 6. This has solution a = 4, b = -5 and c = 1 (see question 8(b)) so that 6 + 11x + 6x2 = 4p1 - 5p2 + p3.

(c) The coefficients are solutions of the linear system 2a + b + 3c = 0, a - b + 2c = 0 and 4a + 3b + 5c = 0. This has solution a = 0, b = 0 and c = 0 (see question 8(c)) so that 0 = 0p1 + 0p2 + 0p3.

(d) The coefficients are solutions of the linear system 2a + b + 3c = 7, a - b + 2c = 8 and 4a + 3b + 5c = 9. This has solution a = 0, b = -2 and c = 3 (see question 8(b)) so that 7 + 8x + 9x2 = 0p1 - 2p2 + 3p3.

11. Given any vector x = (x, y, z) we must be able to find coefficients a, b and c such that x = av1 + bv2 + cv3.if the vectors span R3.

(a) In this case the linear system 2a = x, 2a + c = y, 2a + 3b + c = z must have a unique solution. This is true if det(A) is not zero where A =. Since the determinant equals -6 the vectors span R3.

(b) In this case the linear system 2a + 4b + 8c = x, -a + b - c = y, 3a + 2b + 8c = z must have a unique solution. This is true if det(A) is not zero where A =. Since the determinant equals 0 the vectors do not span R3.

15. Any vector x = (x, y, z) in the plane can be written as a linear combination au + bv, i.e. x = -a + 3b, y = a + 4b, z = a + 4b. If x lies in the plane then its components must be such that the linear system is consistent when solving for a and b. The augmented matrix for this system iswhich has row echelon form . For consistency, the last row mus be zero, i.e. z-y = 0 gives the equation of the plane spanned by u and v.

Section 5.3

2. Since the two vectors in parts (a) and (c) are not multiples of each other,they are linearly independent. In part (d) there are 4 vectors in R3 so they must be linearly dependent. To check the vectors in part (c) we must solve the system

a(-3, 0, 4) + b(5, -1, 2) + c(1, 1, 3) = (0, 0, 0) for a, b and c. The coefficient matrix for this system is which has determinant 9, the system has only the trivial solution and the vectors are linearly independent.

5. The vectors lie in a plane only if they are linearly dependent. We proceed as in question 2(c) to check this.

(a) The coefficient matrix is which has determinant -72 so the vectors are linearly independent and do not lie in a plane.

(b) The coefficient matrix is which has determinant 0 so the vectors are linearly dependent and do lie in a plane.

7. (a) We must find scalars a, b and c such that a(0, 3, 1, -1) + b(6, 0, 5, 1) + c(4, -7, 1, 3) = (0, 0, 0, 0). One solution is a = 7, b = -2, c = 3. Thus the vectors are linearly dependent.

(b) From part a) we can write v1 = 2/7 v2 - 3/7 v3, v2 = 7/2 v1 - 3/2 v3, v3 = -7/3 v1 + 2/3 v2.

8. Using s instead of lambda, we must find values for s such that the equation

a(s, -1/2, -1/2) + b(-1/2, s, -1/2) + c(-1/2, -1/2, s) = (0, 0, 0) has a non-trivial solution.

The coefficient matrix is with determinant s3 -3/4 s - 1/4 which must be zero for a non-trivial solution. The roots are s = -1/2 (twice) and 1. The first value should be obvious since it makes all three vectors equal and hence linearly dependent.

14. If v1 and v2 are linearly independent then there are no non-=zero values for a and b such that av1 + bv2 = 0. Now assume that v1, v2 and v3 are linearly dependent, that is there exist non-zero values for a, b and c such that av1 + bv2 + cv3 = 0. This means that c must be non-zero since otherwise v1 and v2 would be linearly dependent. Thus we can write v3 = a/c v1 + b/c v2. But this means that v3 is in the span of v1 and v2 which leads to a contradiction. Thus the three vectors must be linearly indeopendent.

15. It is obvious that (u - v) + (v - w) + (w - u) = 0 and hence the vectors are linearly dependent.