Download 3.1 15. Let S denote the set of all the infinite sequences

§3.1 15. Let S denote the set of all the infinite sequences of real numbers
with scalar multiplication and addition defined by
α{an } = {αan }
{an } + {bn } = {an + bn }
Show that S is a vector space.
Solution:(Joe)
Note: Let {0} denote an infinite sequence of zeros.
A1 : {an } + {bn } = {an + bn } = {bn + an } = {bn } + {an }
A2 : {an }+({bn }+{cn }) = {an }+{bn +cn } = {an +bn +cn } = {an +bn }+{cn } =
({an } + {bn }) + {cn }
A3 : {an } + {0} = {an + 0} = {an }
A4 : {an } + {−an } = {an + (−an )} = {0}
A5 : α({an } + {bn }) = α{an + bn } = {α(an + bn )} = {αan + αbn } =
{αan } + {αbn } = α{an } + α{bn }
A6 : (α + β){an } = {(α + β)an } = {αan + βan } = {αan } + {βan } =
α{an } + β{bn }
A7 : (αβ){an } = {(αβ)an } = {α(βan )} = α{βan } = α(β{an })
A8 : 1 ∗ {an } = {1 ∗ an } = {an }
§3.2 5. Determine whether the following are subspaces of P4 .
a) The set of polynomials in P4 of even degree
b) The set of all polynomials of degree 3
c) The set of all polynomials p(x) in P4 such that p(0) = 0
d) The set of all polynomials in P4 having at least one real root.
Solution:(Jeff)
a) The set of polynomials in P4 of even degree is not a subspace of P4 becuase the second condition of a subspace is not satisfied.
To see this consider the following elmements of the set given in (a):
x2 + x and −x2 + x.
When these two elements are added together we obtain:
(x2 + x) + (−x2 + x) = 2x.
Clearly 2x is a polynomial of odd degree and not in the set described by (a).
Thus, the second condition of a subspace is not satisfied.
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b) The set of all polynomials of degree 3 is not a subspace of P4 becuase the
first and second conditions of a subspace are not satisfied.
To see this consider the following elmements of the set given in (b):
x3 + x2 + x and −x3 + x2 + x.
We can see the first condition is not satsified because if we take the scalar
zero and the first element given above we obtain:
0(x3 + x2 + x) = 0.
Clearly zero is not a polynomial of degree 3 and the first condition of a subspace
is not satisified.
When the two elements above are added together we obtain:
(x3 + x2 + x) + (−x3 + x2 + x) = 2x2 + 2x.
Clearly 2x2 + 2x is not a polnomial of degree 3 and the second condition of
a subspace is not satisfied.
c) The set of all polynomials p(x) in P4 such that p(0) = 0 is a subspace of
P4 becuase it satisfies both conditions of a subspace.
To see this first note that all elements of the set described by (c) can be written
in the form p(x) = ax3 + bx2 + cx where a, b, c are real numbers.
The first condition is satisfied becuase if we take any element in the set described by (c) which I will represent by p(x) = ax3 + bx2 + cx and multiply it
by any scalar α we obtain:
αp(x) = α(ax3 + bx2 + cx) = αax3 + αbx2 + αcx.
Clearly αp(x) is still in the set described by (c) since αp(0) = αa03 +αb02 +αc0 =
0. Thus, the first condition of a subspace is satisfied.
The set described in (c) also satisfies the second condition of a subspace. To see
this consider any two elements of the set described by (c). I will call these two
elements p(x) = ax3 + bx2 + cx and P (x) = dx3 + ex2 + f x where a, b, c, d, e, f
are real numbers.
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If we take the sum of these two elements we obtain:
p(x)+P (x) = (ax3 +bx2 +cx)+(dx3 +ex2 +f x) = (a+d)x3 +(b+e)x2 +(c+f )x
Clearly p(x) + P (x) is still in the set described by (c) since p(0) + P (0) =
(a + d)03 + (b + e)02 + (c + f )0 = 0. Thus, the second condition of a subspace
is satisfied.
d) The set of all polynomials in P4 having at least one real root is not a subspace
of P4 becuase the second condition of a subspace is not satisfied.
To see this consider the following elmements of the set given in (d) which have
the real zeros of −1 and 1:
x + 1 and −x + 1.
When these two elements are added together we obtain:
(x + 1) + (−x + 1) = 2.
Clearly 2 is not a polynomial that has a real zero. Thus, the second condition of a subspace is not satisfied.
§3.2 18. Let U and V be subspaces of a vector space W . Prove that their
intersection U ∩ V is also a subspace of W .
Solution:(Joe)
Addition: Let u and v be elements of U ∩ V . Then u and v are elements of
both U and V . Because U is a subspace, u + v is an element of U . Similarly,
since V is a subspace, u + v is an element of V . Thus, u + v is an element of
U ∩V.
Scalar Multiplication: Let w be an element of U ∩ V . Let γ be a scalar.
Then w is an element of both U and V . Since U and V are subspaces, γw is an
element of both U and V . Thus, γw is an element of U ∩ V .
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