A sequence of elements in a field K is a function s : N → K; s(n) = an
Denote the sequence by {an}. Let SKbe the set of all sequences of elements of K and define in SK the following operations:
{an} + {bn} = {an+ bn}, k{an} = {kan}; ∀k ∈ K Prove that endowed with these operation SK is a K-vector space.
I Associativity:
({an} + {bn}) + {cn} ={an+ bn} + {cn} = {(an+ bn) + cn} =
={an+ (bn+ cn)} = {an} + {bn+ cn} =
={an} + ({bn} + {cn})
I Commutativity:
{an} + {bn} = {an+ bn} = {bn+ an} = {bn} + {an}
I Neutral element for the sum: Let {0K} be the sequence of constant value 0K. Then
{an} + {0K} = {an+ 0K} = {an} so {0K} is the neutral element for the sum.
I Opposite: Given any {an},
{an} + {−an} = {an+ (−an)} = {0K} so {−an} is the opposite of {an}.
I
1K{an} = {1Kan} = {an}
I Distributivity:
I (k + l ){an} = {(k + l )an} = {kan+ lan} = {kan} + {lan} = k{an} + l {an}
I k({an} + {bn}) = k{an+ bn} = {k(an+ bn)} = {kan+ kbn} = {kan} + {kbn} = k{an} + k{bn}
I Homogeneity:
k(l {an}) = k{lan} = {k(lan)} = {(kl )an} = (kl ){an}
Consider the set C(R) of all continuous functions f : R → R endowed with the operations
(f + g )(x ) = f (x ) + g (x ); (af )(x ) = af (x ) ∀a ∈ R.
Check that C(R) is a real vector space. Answer the following questions:
(1) Is it true that the set consisting of the three functions 1 (constant function), sin2e cos2is linearly dependent?
(2) What about the functions 1, sin and cos?
(3) Consider the set {sin(nx ) | n ∈ N, n 6= 0} ∪ {cos(nx) | n ∈ N} and prove that it is a linearly independent set.
I Associativity:
[(f + g ) + h](x ) =(f + g )(x ) + h(x ) = [f (x ) + g (x )] + h(x ) =
=f (x ) + [g (x ) + h(x )] = f (x ) + (g + h)(x ) =
=[f + (g + h)](x )
I Commutativity:
(f + g )(x ) = f (x ) + g (x ) = g (x ) + f (x ) = (g + f )(x )
I Neutral element: Let Z : R → R be the constant function taking value 0 everywhere. Then Z ∈ C(R). Moreover, for every f ∈ C(R):
(f + Z )(x ) = f (x ) + Z (x ) = f (x ) so Z is neutral element for sum.
I Opposite: Given f ∈ C(R), let g : R → R be the function defined by
∀x ∈ R, g(x) = −f (x) Then
(f + g )(x ) = f (x ) + g (x ) = f (x ) + [−f (x )] = 0 = Z (x ) so g the opposite of f .
I
(1f )(x ) = 1f (x ) = f (x )
I Distributivity:
I [(a + b)f ](x ) = (a + b)f (x ) = af (x ) + bf (x ) = (af )(x ) + (bf )(x ) = (af + bf )(x )
I [a(f + g )](x ) = a(f + g )(x ) = a[f (x ) + g (x )] = af (x ) + ag (x ) = (af + ag )(x )
I Homogeneity:
[k(lf )](x ) = k[(lf )(x )] = k[lf (x )] = (kl )f (x ) = [(kl )f ](x )
(1) The linear combination of functions 1 − sin2− cos2
is the constant function Z taking value 0 everywhere, that is, the null vector in C(R). Thus the three functions 1, sin2, cos2are linearly dependent.
(2) Suppose that a1 + b sin +c cos is a null linear combination of the three functions, that is,
∀x ∈ R, a + b sin x + c cos x = 0 Choosing in particular for x the values 0,π2, π, one obtains
a + c = 0 a + b = 0 a − c = 0 whence a = b = 0.
(3) Suppose, by contradiction, that there is a null linear combination whose coefficients are not all 0, and let k be minimal for which such a linear combination exists, say
a +
k
X
n=1
bnsin nx +
k
X
n=1
cncos nx = 0
Per item (2) above, k > 1.
Differentiating twice this equation:
k
X
n=1
nbncos nx −
k
X
n=1
ncnsin nx = 0
−
k
X
n=1
n2bnsin nx −
k
X
n=1
n2cncos nx = 0
Multiplying last equality by 1k2one gets
−
k−1
X
n=1
n2
k2bnsin nx − bksin kx −
k−1
X
n=1
n2
k2cncos nx − ckcos kx = 0 Summing this equality to the initial linear combination, it follows that
a +
k−1
X
n=1
1 − n2
k2
bnsin nx +
k−1
X
n=1
1 −n2
k2
cncos nx = 0
By the minimality of k, all coefficients of this linear combination must be null, whence
a = b0= . . . = bk−1= c0= . . . = ck−1= 0
Consequently
∀x ∈ R, bksin kx + ckcos kx = 0
In particular, choosing x being 0 and π2, it also follows bk = ck = 0, against the hypothesis that not all coefficients of the linear combination were 0.
Thus the set under consideration is linearly independent.
Check whether the following sets in R3 are linearly independent (1) {(2, 1, 1), (0, 1, 0), (1, 0, 1)}
(2) {(1, 1, 1), (0, 0, 1), (0, 1, 0), (3, 1, 1)}
(3) {(1, 5, 0), (0, 0, 4), (1, 0, 2)}
(4) 20
9, 4, 12 , 59, 1, 3 , 1013 , 0,118
(1) Let
a(2, 1, 1) + b(0, 1, 0) + c(1, 0, 1) = (0, 0, 0) This means (2a + c, a + b, a + c) = (0, 0, 0), equivalently
2a + c = 0 a + b = 0 a + c = 0
From the second and third equations, b = c = −a, and substituting in the first equation one obtains a = 0, and consequently
a = b = c = 0.
Thus the set is linearly independent.
(2) A four-element set in R3 is always linearly dependent.
(3) Let
a(1, 5, 0) + b(0, 0, 4) + c(1, 0, 2) = (0, 0, 0) This means
(a + c, 5a, 4b + 2c) = (0, 0, 0)
The second equation gives a = 0, then the first provides c = 0, and finally the third ones yields c = 0.
Consequently, the set is linearly independent.
(4) Observe that
20 9 , 4, 12
= 4 5 9, 1, 3
thus the three vectors are linearly dependent.
Determine which among the following subsets of R4are vector subspaces and, in the affirmative, find a basis:
(1) W1= {(a, b, 2a, −b) | a, b ∈ R}
(2) W2= {(a, 1 + a, a − b, 0) | a, b ∈ R}
(3) W3= {(a + b, b − a, 0, −b) | a, b ∈ R}
(1) Let (a, b, 2a, −b), (a0, b0, 2a0, −b0) ∈ W1. Then
(a, b, 2a, −b) + (a0, b0, 2a0, −b0) = (a + a0, b + b0, 2a + 2a0, −b − b0) =
= (a + a0, b + b0, 2(a + a0), −(b + b0)) ∈ W1
− (a, b, 2a, −b) = (−a, −b, −2a, −(−b)) = (−a, −b, 2(−a), −(−b)) ∈ W1
Therefore W1is a vector subspace of R4. A basis is obtained by letting successively a = 1, b = 0 and a = 0, b = 1:
((1, 0, 2, 0), (0, 1, 0, −1))
Indeed, these two vectors are linearly independent, since they are not proportional. Moreover, for all (a, b, 2a, −b) ∈ R4,
(a, b, 2a, −b) = a(1, 0, 2, 0) + b(0, 1, 0, −1)
(2) Notice that (0, 0, 0, 0) /∈ W2, since
(a, 1 + a, a − b, 0) = (0, 0, 0, 0) would imply a = a + 1 = 0, which is impossible.
(3) Let (a + b, b − a, 0, −b), (a0+ b0, b0− a0, 0, −b0) ∈ W3. Then (a + b, b − a, 0, −b) + (a0+ b0, b0− a0, 0, −b0) =
= (a + b + a0+ b0, b − a + b0− a0, 0, −b − b0) =
= (a + a0+ b + b0, b + b0− (a + a0), 0, −(b + b0)) ∈ W3
− (a + b, b − a, 0, −b) = (−(a + b), −(b − a), −0, −(−b)) =
= (−a + (−b), −b − (−a), 0, −(−b)) ∈ W3
Therefore W3is a vector subspace of R4. A basis is obtained by letting successively a = 1, b = 0 and a = 0, b = 1:
((1, −1, 0, 0), (1, 1, 0, −1))
Indeed these two vectors are linearly independent, as they are not proportional. Moreover, for any element (a + b, b − a, 0, −b) ∈ W3,
(a + b, b − a, 0, −b) = a(1, −1, 0, 0) + b(1, 1, 0, −1)
Let R[X ]≤n= {p ∈ R[X ] | deg(p) ≤ n}. This is a finite-dimensional vector space (check!). Consider the following subsets:
(1) W1= {p ∈ R[X ]≤3| p(1) = 0};
(2) W2= {p ∈ R[X ]≤3| p(1) = 0 = p(0)};
Check that they are vector subspaces of R[X ]≤3, and find a basis for each of them.
To check that R[X ]≤n is a subspace of R[X ], notice that, for all p, q ∈ R[X ]≤n:
deg(p + q) ≤ max[deg(p), deg(q)] ≤ n ⇒p + q ∈ R[X ]≤n
deg(−p) = deg(p) ≤ n ⇒ − p ∈ R[X ]≤n The space R[X ]≤n is finite-dimensional, since it is generated by the monomials
1, X , X2, . . . , Xn Indeed, every
p = a0+ a1X + a2X2+ . . . + anXn∈ R[X ]≤n
is a linear combination of such monomials.
(1) For any p, q ∈ W1:
(p + q)(1) = p(1) + q(1) = 0 + 0 = 0 ⇒ p + q ∈ R[X ] (−p)(1) = −p(1) = −0 = 0 ⇒ −p ∈ W1
Therefore W1is a subspace of R[X ]≤3. To find a basis for W1, notice that
a + bX + cX2+ dX3∈ Wi⇔ a + b + c + d = 0 ⇔ d = −a − b − c Consequently, the generic element of W1has the form
a + bX + cX2− (a + b + c)X3
Choosing successively the values
a = 1, b = 0, c = 0 a = 0, b = 1, c = 0 a = 0, b = 0, c = 1 yields the three polynomials
p1=1 − X3 p2=X − X3 p3=X2− X3 They form a basis of W1, as every
p = a + bX + cX2− (a + b + c)X3∈ W1can be expresses in a unique way as a linear combination
p = ap1+ bp2+ cp3
(2) Notice that
W2= {p ∈ W1| p(0) = 0}
so to show that W2is a subspace of R[X ]≤3 it is enough to establish that W2is a subspace of W1.
Given any p, q ∈ W2:
(p + q)(0) = p(0) + q(0) = 0 + 0 = 0 ⇒ p + q ∈ W2
(−p)(0) = −p(0) = −0 = 0 ⇒ −p ∈ W2
A generic p = a + bX + cX2+ dX3∈ R[X ]≤3belongs to W2if and only if
p(1) = a + b + c + d = 0 p(0) = a = 0 if and only if
a = 0
d = −b − c
bX + cX2− (b + c)X3 for some b, c ∈ R.
Setting successively
b = 1, c = 0 b = 0, c = 1 yields the polynomials
q1=X − X3 q2=X2− X3
They form a basis of W2, since they are linearly independent (as they are not proportional) and they generate W2: any
p = bX + cX2− (b + c)X3∈ W2 is of the form p = bq1+ q2
