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.
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.
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.
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.
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.
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.
(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
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