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# Sect4 5

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### Sect4 5

1. 1. SECTION 4.5 GENERAL VECTOR SPACES In each of Problems 1-12, a certain subset of a vector space is described. This subset is a subspace of the vector space if and only if it is closed under the formation of linear combinations of its elements. Recall also that every subspace of a vector space must contain the zero vector. 1. It is a subspace of M33, because any linear combination of diagonal 3 × 3 matrices — with only zeros off the principal diagonal — obviously is again a diagonal matrix. 2. The square matrix A is symmetric if and only if AT = A. If A and B are symmetric 3 × 3 matrices, then (cA + dB)T = cAT + dBT = cA + dB, so the linear combination cA + dB is also symmetric. Thus the set of all such matrices is a subspace. 3. The set of all nonsingular 3 × 3 matrices does not contain the zero matrix, so it is not a subspace. 4. The set of all singular 3 × 3 matrices is not a subspace, because the sum 1 0 0  0 0 0  1 0 0  0 0 0  + 0 1 0  = 0 1 0        0 0 0    0 0 1    0 0 1    of singular matrices is not singular. 5. The set of all functions f : R → R with f (0) = 0 is a vector space, because if f (0) = g (0) = 0 then (a f + bg )(0) = a f (0) + bg(0) = a ⋅ 0 + b ⋅ 0 = 0. 6. The set of all functions f : R → R with f (0) ≠ 0 is not a vector space, because it does not contain the zero function f (0) ≡ 0. 7. The set of all functions f : R → R with f (0) = 0 and f (1) = 1 is not a vector space. For instance, if g = 2 f then g (1) = 2 f (1) = 2 ⋅1 = 2 ≠ 1, so g is not such a function. Also, this set does not contain the zero function. 8. A function f : R → R such that f (− x) = − f ( x) is called an odd function. Any linear combination af + bg of odd functions is again odd, because (a f + bg )(− x ) = a f (− x ) + bg (− x ) = − a f ( x ) − bg ( x ) = − (a f + bg )( x ). Thus the set of all odd functions is a vector space. Section 4.5 1
2. 2. For Problems 9-12, let us call a polynomial of the form a0 + a1 x + a2 x 2 + a3 x 3 a "degree at most 3" polynomial. 9. The set of all degree at most 3 polynomials with nonzero leading coefficient a3 ≠ 0 is not a vector space, because it does not contain the zero polynomial (with all coefficients zero). 10. The set of all degree at most 3 polynomials not containing x or x2 terms is a vector space, because any linear combination of such polynomials obviously is such a polynomial. 11. The set of all degree at most 3 polynomials with coefficient sum zero is a vector space, because any linear combination of such polynomials obviously is such a polynomial. 12. If the degree at most 3 polynomials f and g have all-integer coefficients, the linear combination a f + b g may have non-integer coefficient, because a and b need not be integers. Hence the set of all degree at most 3 polynomials having all-integer coefficients is not a vector space. 13. The functions sin x and cos x are linearly independent, because neither is a scalar multiple of the other. (This follows, for instance, from the facts that sin(0) = 0, cos(0) = 1 and sin(π / 2) = 1, cos(π / 2) = 0, noting that any scalar multiple of a function with a zero value must have the value 0 at the same point.) 14. The functions e x and xe x are linearly independent, since obviously neither is a scalar multiple of the other (their ratios xe x / e x = x and e x / xe x neither being constants). 15. If c1 (1 + x) + c2 (1 − x) + c3 (1 − x 2 ) = (c1 + c2 + c3 ) + (c1 − c2 ) x − c3 x 2 = 0, then c1 + c2 + c3 = c1 − c2 = c3 = 0. It follows easily that c1 = c2 = c3 = 0, so we conclude that the functions (1 + x ), (1 − x ), and (1 − x 2 ) are linearly independent. 16. (−1) ⋅ (1 + x) + (1) ⋅ ( x + x 2 ) + (1) ⋅ (1 − x 2 ) = 0, so the three given polynomials are linearly dependent. 17. cos 2 x = cos 2 x − sin 2 x according to a well-known trigonometric identity. Thus these three trigonometric functions are linearly dependent. 18. If c1 (2 cos x + 3sin x) + c2 (4 cos x + 5sin x) = (2c1 + 4c2 ) cos x + (3c1 + 5c2 )sin x = 0
3. 3. then the fact that sin x and cos x are linearly independent (Problem 13) implies that 2c1 + 4c2 = 3c1 + 5c2 = 0. It follows readily that c1 = c2 = 0, so we conclude that the two original linear combinations of sin x and cos x are linearly independent. 19. Multiplication by ( x − 2)( x − 3) yields x − 5 = A( x − 3) + B( x − 2) = ( A + B) x − (3 A + 2 B). Hence A + B = 1 and 3 A + 2 B = 5, and it follows readily that A = 3 and B = –2. 20. Multiplication by x ( x 2 − 1) yields 2 = A( x 2 − 1) + Bx( x + 1) + Cx( x − 1) = − A + ( B − C ) x + ( A + B + C ) x 2 . Hence − A = 2, B − C = 0 and A + B + C = 0. It follows readily that A = –2 and B = C = 1. 21. Multiplication by x ( x 2 + 4) yields 8 = A( x 2 + 4) + Bx 2 + Cx = 4 A + Cx + ( A + B) x 2 . Hence 4 A = 8, C = 0 and A + B = 0. It follows readily that A = 2 and B = –2. 22. Multiplication by ( x + 1) ( x + 2)( x + 3) yields 2 x = A ( x + 2)( x + 3) + B( x + 1)( x + 3) + C ( x + 1) ( x + 2) = ( A + B + C ) x 2 + (5 A + 4 B + 3C ) x + (6 A + 3B + 2C ). Hence A+ B+ C = 0 5 A + 4 B + 3C = 2 6 A + 3B + 2C = 0, and we solve these three equations for A = –1, B = 4, and C = –3. 23. If y′′′( x) = 0 then y′′( x) = ∫ y′′′( x) dx = ∫ (0) dx = A, y′( x) = ∫ y′′( x) dx = ∫ A dx = Ax + B, and y ( x) = ∫ y ′( x ) dx = ∫ ( Ax + B ) dx = Ax + Bx + C , 1 2 2 Section 4.5 3
4. 4. where A, B, and C are arbitrary constants of integration. It follows that the function y ( x) is a solution of the differential equation y′′′( x) = 0 if and only if it is a quadratic (at most 2nd degree) polynomial. Thus the solution space is 3-dimensional with basis { } 1, x, x 2 . 24. If y (4) ( x ) = 0 then y′′′( x ) =∫ y ( x) dx = ∫ (0) dx = A, (4) y′′( x ) = ∫ y′′′( x) dx = ∫ A dx = Ax + B, y′( x ) = ∫ y′′( x) dx = ∫ ( Ax + B) dx = Ax + Bx + C , and 1 2 2 y ( x ) = ∫ y′( x ) dx = ∫ ( Ax + Bx + C ) dx = Ax + Bx + Cx + D. 1 2 2 1 6 2 1 2 where A, B, C, and D are arbitrary constants of integration. It follows that the function y ( x) is a solution of the differential equation y (4) ( x ) = 0 if and only if it is a cubic (at most 3rd degree) polynomial. Thus the solution space is 4-dimensional with basis { } 1, x, x 2 , x3 . 25. If y ( x) is any solution of the second-order differential equation y′′ − 5 y′ = 0 and v( x) = y′( x), then v( x) is a solution of the first-order differential equation v′( x) = 5v( x) with the familiar exponential solution v( x) = Ce5 x . Therefore y( x) = ∫ y′( x) dx = ∫ v( x) dx = ∫ Ce dx = Ce5 x + D. 5x 1 5 We therefore see that the solution space of the equation y′′ − 5 y′ = 0 is 2-dimensional with { basis 1, e5 x . } 26. If y ( x) is any solution of the second-order differential equation y′′ + 10 y′ = 0 and v( x) = y′( x), then v( x) is a solution of the first-order differential equation v′( x) = −10v( x) with the familiar exponential solution v( x) = Ce −10 x . Therefore ∫ y′( x) dx ∫ v( x) dx ∫ Ce −10 x y( x) = = = dx = − 10 Ce10 x + D. 1 We therefore see that the solution space of the equation y′′ + 10 y′ = 0 is 2-dimensional with { basis 1, e −10 x . }
5. 5. 27. If we take the positive sign in Eq. (20) of the text, then we have v 2 = y 2 + a 2 where v( x) = y′( x). Then 2  dy  dx 1   = y + a , so = 2 2  dx  dy y + a2 2 (taking the positive square root as in the text). Then ⌠ dy ⌠ a du x =  =  ( y = a u) ⌡ y +a 2 2 ⌡ a u + a2 2 2 ⌠ du y =  = sinh −1 u + b = sinh −1 + b. ⌡ u +1 2 a It follows that y ( x ) = a sinh( x − b) = a (sinh x cosh b − cosh x sinh b ) = A cosh x + B sinh x. 28. We start with the second-order differential equation y′′ + y = 0 and substitute v( x) = y′( x), so dv dv dy dv y′′ = = = v = −y dx dy dx dy as in Example 9 of the text. Then v dv = − y dy, and integration gives 1 2 v2 = − 1 y 2 + C, 2 so v2 = a2 − y 2 (taking for illustration a positive value for the arbitrary constant C). Then 2  dy  dx 1   = a − y , so = 2 2  dx  dy a2 − y2 (taking the positive square root). Then ⌠ dy ⌠ a du x =  =  ( y = a u) ⌡ a2 − y2 ⌡ a 2 − a 2u 2 ⌠ du y =  = sin −1 u + b = sin −1 + b. ⌡ 1− u2 a It follows that Section 4.5 5
6. 6. y ( x ) = a sin( x − b) = a (sin x cos b − cos x sin b ) = A cos x + B sin x. Thus the general solution of the 2nd-order differential equation y′′ + y = 0 is a linear combination of cos x and sin x. It follows that the solution space is 2-dimensional with basis {cos x,sin x}. 29. (a) The verification in a component-wise manner that V is a vector space is the same as the verification that Rn is a vector space, except with vectors having infinitely many components rather than finitely many components. It boils down to the fact that a linear combination of infinite sequences of real numbers is itself such a sequence, a ⋅ {xn }1 + b ⋅ { yn }1 = {axn + byn }1 . ∞ ∞ ∞ (b) If e n = {0, , 0, 1, 0, 0, } is the indicated infinite sequence with 1 in the nth position, then the fact that c1e1 + c2e 2 + + ck ek = {c1 , c2 , , ck , 0, 0, } evidently implies that any finite set e1 , e 2 , , e k of these vectors is linearly independent. Thus V contains arbitrarily large sets of linearly independent vectors, and therefore is infinite-dimensional. 30. (a) If xn = xn −1 + xn −2 , yn = yn −1 + yn − 2 and zn = axn + byn for each n, then zn = a( xn −1 + xn − 2 ) + b( yn −1 + yn − 2 ) = (axn −1 + byn −1 ) + (axn − 2 + byn − 2 ) = zn −1 + zn − 2 . Thus W is a subspace of V. (b) Let v1 = {1, 0,1,1, 2,3,5, } be the element with x1 = 1 and x2 = 0, and let v 2 = {0,1,1, 2,3, 5, } be the element with x1 = 0 and x2 = 1. Then v1 and v2 form a basis for W. 31. (a) If z1 = a1 + i b1 and z2 = a2 + i b2 , then direct computation shows that c1a1 + c2 a2 −c1b1 − b2 a2  T (c1 z1 + c2 z2 ) = c1T ( z1 ) + c2T ( z2 ) =  c1a1 + c2 a2  .  c1b1 + b2 a2 
7. 7. (b) If z1 = a1 + i b1 and z2 = a2 + i b2 , then z1 z2 = (a1a2 − b1b2 ) + i (a1b2 + a2b1 ) and direct computation shows that  a1a2 − b1b2 −a1b2 − a2b1  T ( z1 z2 ) = T ( z1 )T ( z2 ) =  a1a2 − b1b2  .  a1b2 + a2b1  (b) If z = a + i b then 1 1 a − bi a −bi = ⋅ = 2 . z a + bi a −bi a + b2 Therefore −1 −1 1  a b  a −b  T (z ) = 2  −b a  =  b a  = T ( z ) −1 . a + b2     Section 4.5 7