Problem 5

Question

Prove that the set $P_{n}$ of all polynomials of degree less than $n$ form a subspace of the vector space $F[x]$. Find a basis for $P_{n}$ and compute the dimension of $P_{n}$.

Step-by-Step Solution

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Answer

Based on the step-by-step solution, complete the following short answer. Question: Show that the set $P_n$ of all polynomials of degree less than $n$ is a subspace of the vector space $F[x]$, find a basis for $P_n$, and compute its dimension. Answer: To prove that $P_n$ is a subspace of $F[x]$, we showed that it is closed under vector addition and scalar multiplication. A basis for $P_n$ is given by the set $B = \{1, x, x^2, x^3, \cdots, x^{n-1}\}$, which is a linearly independent set and spans the space. The dimension of $P_n$ is equal to the number of elements in its basis, which is $n$.

1Step 1: Verify that $P_n$ is a subspace of $F[x]$

To prove that $P_n$ is a subspace of $F[x]$, we need to verify that $P_n$ is closed under vector addition and scalar multiplication. Let's consider two polynomials $p(x)$ and $q(x)$ in $P_n$, and a scalar $c\in F$. For vector addition, we have $(p(x) + q(x))$. Since $p(x)$ and $q(x)$ both have degree less than $n$, their sum will also have a degree less than $n$. Therefore, $p(x) + q(x) \in P_n$. For scalar multiplication, we have $(c\cdot p(x))$. Since $p(x)$ has a degree less than $n$, and multiplying by a constant doesn't change the degree of a polynomial, the result will still have a degree less than $n$. Therefore, $c\cdot p(x) \in P_n$. Since $P_n$ is closed under vector addition and scalar multiplication, it is a subspace of the vector space $F[x]$.

2Step 2: Find a basis for $P_n$

In order to find a basis for $P_n$, we need to find a set of linearly independent vectors that span the space. Recall that the standard basis for the vector space of all polynomials is given by $\{1, x, x^2, x^3, \cdots\}$. Since we are dealing with polynomials of degree less than $n$, our basis for $P_n$ will contain n elements. Therefore, a basis for $P_n$ is given by: $$B = \{1, x, x^2, x^3, \cdots, x^{n-1}\}$$ This basis is linearly independent because no element in the set can be expressed as a linear combination of the others, and it spans the space because any polynomial of degree less than $n$ can be written as a linear combination of these basis elements.

3Step 3: Compute the dimension of $P_n$

The dimension of a vector space is the number of elements in its basis. From Step 2, we know that the basis for $P_n$ has n elements. Therefore, the dimension of $P_n$ is: Dimension$(P_n) = n$.

Key Concepts

SubspaceVector SpaceBasisDimension

Subspace

A subspace is a special kind of subset within a vector space that must satisfy three conditions: it must be non-empty, closed under addition, and closed under scalar multiplication.
To understand this better, let's consider the set of all polynomials of degree less than $ n $, denoted by $ P_n $.

Non-empty: The zero polynomial, which is a polynomial of degree less than any number, must exist in $ P_n $.
Closed under addition: If you take any two polynomials in $ P_n $ and add them, the result should also be in $ P_n $. Since the degree of the sum is still less than $ n $, this condition is met.
Closed under scalar multiplication: Multiplying any polynomial in $ P_n $ by a scalar won't change its degree beyond the acceptable limit, keeping it in the set $ P_n $.

In the context of polynomial algebra, showing these properties confirms that $ P_n $ is indeed a subspace of $ F[x] $, the vector space of all polynomials.

Vector Space

A vector space is a fundamental concept in linear algebra encompassing a collection of objects, typically called vectors, which can be added together and multiplied by scalars to produce another object in the same space.
For polynomial algebra, a vector space like $ F[x] $ includes all polynomials with coefficients in field $ F $, and degree can vary infinitely.

Closure under addition: Any two polynomials can be added to generate another polynomial.
Closure under scalar multiplication: Multiplying any polynomial by a scalar results in another polynomial.
Contains additive identity: The zero polynomial acts as the additive identity.
Contains additive inverses: For every polynomial, there exists another polynomial such that their sum is the zero polynomial.

Understanding these properties helps clarify why $ P_n $, the set of polynomials with degree less than $ n $, inherits these characteristics as a subspace of $ F[x] $. This forms the basis for exploring its structure and properties further.

Basis

The concept of a basis is pivotal in vector spaces because it provides the minimum number of vectors needed to express every element of that space.
For $ P_n $, a basis can be thought of as the set $ \{1, x, x^2, \, ... \, , x^{n-1}\} $.
This set is significant because:

Linearly independent: None of these vectors, which are polynomials in this case, can be expressed as a combination of others within the set.
Spans the space: Any polynomial of degree less than $ n $ can be represented as a combination of elements from this basis.

Having a basis allows us to express any polynomial of interest using these fundamental building blocks. This approach makes calculations more manageable and gives a clearer understanding of the structure of $ P_n $.

Dimension

Dimension in vector spaces refers to the number of vectors in the basis of the space. It's a measure of how big or complex the space is.
For $ P_n $, the dimension is $ n $ because there are $ n $ elements in its basis: $ \{1, x, x^2, ..., x^{n-1}\} $.
Additionally:

Represents degrees of freedom: The number of ways to construct polynomials of degree less than $ n $.
Informative about structure: Knowing the dimension helps understand the nature and potential of the vector space.

Recognizing the dimension of a subspace like $ P_n $ is crucial for delving deeper into polynomial spaces and for applications like solving polynomial equations. This concept links the abstract structure to practical manipulations and outcomes.

Problem 4

Problem 6

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Problem 4

Prove that the complex numbers are a vector space of dimension 2 over $\mathbb{R}$.

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Problem 6

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Problem 7

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