Problem 3.1 - Functions as Vectors Consider the collection of all polynomials (with real coefficients) of degree less than or equal to N in the real variable x, which we may write as {È N Σ ατ' (1) PN = a; El j=0 {x> We can take as our starting point that the space of all polynomials (with no upper limit on the degree) is a vector space. To check whether a subset of a vector space is itself a vector space (in which case we call the subset a subspace) we only need to check that it is closed under vector addition and scalar multiplication that is, a linear combination of any elements of the subset is itself an element of the subset. (a) For each of the following subsets, state whether or not the set constitutes a vector space. If so, suggest a convenient basis and give the dimension of the space. If not, explain or show why not. You do not need to explicitly or rigorously prove that your suggested basis is a basis, but you should argue that your set is indeed linear independent and does indeed span the subset. • The full set PN given by Eq. 1. • The subset of PN such that the leading coefficient of the polynomial is 1 (that is, the set such that aN = = 1). • The subset of PN such that the polynomials have the value 0 at x = 1. • Supplementary Part (Not for Credit): The subset of PN such that the polynomials have the value 1 at x = 0. [Note: For a given vector space V, the full space V and the subset {0} just containing the zero vector are always subspaces. The zero subspace is zero-dimensional.]
Problem 3.1 - Functions as Vectors Consider the collection of all polynomials (with real coefficients) of degree less than or equal to N in the real variable x, which we may write as {È N Σ ατ' (1) PN = a; El j=0 {x> We can take as our starting point that the space of all polynomials (with no upper limit on the degree) is a vector space. To check whether a subset of a vector space is itself a vector space (in which case we call the subset a subspace) we only need to check that it is closed under vector addition and scalar multiplication that is, a linear combination of any elements of the subset is itself an element of the subset. (a) For each of the following subsets, state whether or not the set constitutes a vector space. If so, suggest a convenient basis and give the dimension of the space. If not, explain or show why not. You do not need to explicitly or rigorously prove that your suggested basis is a basis, but you should argue that your set is indeed linear independent and does indeed span the subset. • The full set PN given by Eq. 1. • The subset of PN such that the leading coefficient of the polynomial is 1 (that is, the set such that aN = = 1). • The subset of PN such that the polynomials have the value 0 at x = 1. • Supplementary Part (Not for Credit): The subset of PN such that the polynomials have the value 1 at x = 0. [Note: For a given vector space V, the full space V and the subset {0} just containing the zero vector are always subspaces. The zero subspace is zero-dimensional.]
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Problem 3.1 - Functions as Vectors
Consider the collection of all polynomials (with real coefficients) of degree less than or equal to N
in the real variable x, which we may write as
{È
N
Σ ατ'
(1)
PN =
a; El
j=0
{x>
We can take as our starting point that the space of all polynomials (with no upper limit on the
degree) is a vector space.
To check whether a subset of a vector space is itself a vector space (in which case we call the subset
a subspace) we only need to check that it is closed under vector addition and scalar multiplication
that is, a linear combination of any elements of the subset is itself an element of the subset.
(a) For each of the following subsets, state whether or not the set constitutes a vector space. If
so, suggest a convenient basis and give the dimension of the space. If not, explain or show why not.
You do not need to explicitly or rigorously prove that your suggested basis is a basis, but you should
argue that your set is indeed linear independent and does indeed span the subset.
• The full set PN given by Eq. 1.
• The subset of PN such that the leading coefficient of the polynomial is 1 (that is, the set such
that aN =
= 1).
• The subset of PN such that the polynomials have the value 0 at x = 1.
• Supplementary Part (Not for Credit): The subset of PN such that the polynomials have the
value 1 at x =
0.
[Note: For a given vector space V, the full space V and the subset {0} just containing the zero vector
are always subspaces. The zero subspace is zero-dimensional.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53e89d9d-bfd3-4a29-98a6-3c58c69fa30f%2F161e7c1d-d310-4a6f-a5e1-caa5e27792c7%2F76lvzej.png&w=3840&q=75)
Transcribed Image Text:Problem 3.1 - Functions as Vectors
Consider the collection of all polynomials (with real coefficients) of degree less than or equal to N
in the real variable x, which we may write as
{È
N
Σ ατ'
(1)
PN =
a; El
j=0
{x>
We can take as our starting point that the space of all polynomials (with no upper limit on the
degree) is a vector space.
To check whether a subset of a vector space is itself a vector space (in which case we call the subset
a subspace) we only need to check that it is closed under vector addition and scalar multiplication
that is, a linear combination of any elements of the subset is itself an element of the subset.
(a) For each of the following subsets, state whether or not the set constitutes a vector space. If
so, suggest a convenient basis and give the dimension of the space. If not, explain or show why not.
You do not need to explicitly or rigorously prove that your suggested basis is a basis, but you should
argue that your set is indeed linear independent and does indeed span the subset.
• The full set PN given by Eq. 1.
• The subset of PN such that the leading coefficient of the polynomial is 1 (that is, the set such
that aN =
= 1).
• The subset of PN such that the polynomials have the value 0 at x = 1.
• Supplementary Part (Not for Credit): The subset of PN such that the polynomials have the
value 1 at x =
0.
[Note: For a given vector space V, the full space V and the subset {0} just containing the zero vector
are always subspaces. The zero subspace is zero-dimensional.]
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