Question

a. Let \(p(x)\) and \(q(x)\) lie in \(\mathbf{P}_{1}\) and suppose that \(p(1) \neq 0, q(2) \neq 0,\) and \(p(2)=0=q(1) .\) Show that \(\\{p(x), q(x)\\}\) is a basis of \(\mathbf{P}_{1}\). [Hint: If \(r p(x)+s q(x)=0,\) evaluate at \(x=1, x=2 .]\) b. Let \(B=\left\\{p_{0}(x), p_{1}(x), \ldots, p_{n}(x)\right\\}\) be a set of polynomials in \(\mathbf{P}_{n}\). Assume that there exist numbers \(a_{0}, a_{1}, \ldots, a_{n}\) such that \(p_{i}\left(a_{i}\right) \neq 0\) for each \(i\) but \(p_{i}\left(a_{j}\right)=0\) if \(i\) is different from \(j .\) Show that \(B\) is a basis of \(\mathbf{P}_{n}\).

Short answer

In both parts (a) and (b), the given polynomials form a basis because they are linearly independent and span their respective polynomial spaces.

Step-by-step solution

Understand given conditions for part (a)

We're given two polynomials, \(p(x)\) and \(q(x)\), in \(\mathbf{P}_{1}\), with specific properties: \(p(1) eq 0\), \(q(2) eq 0\), \(p(2) = 0\), and \(q(1) = 0\). Our task is to show that these two form a basis for the polynomial space \(\mathbf{P}_{1}\).

Use linear dependence condition (Part a)

Assume a linear combination \(r p(x) + s q(x) = 0\). Evaluate this at the unique points where each polynomial is non-zero based on given properties to determine the coefficients \(r\) and \(s\).

Evaluate at x=1 and x=2 (Part a)

Evaluate the linear combination at \(x=1\): \(r p(1) + s q(1) = 0\). Since \(q(1) = 0\), this reduces to \(r p(1) = 0\). Given \(p(1) eq 0\), it follows that \(r = 0\). Then evaluate at \(x=2\): \(r p(2) + s q(2) = 0\). Since \(p(2) = 0\), this reduces to \(s q(2) = 0\). Given \(q(2) eq 0\), it follows that \(s = 0\).

Conclude basis for part (a)

The only solution to the equation \(r p(x) + s q(x) = 0\) is \(r = 0\) and \(s = 0\). Therefore, \(\{p(x), q(x)\}\) are linearly independent and span \(\mathbf{P}_{1}\), making them a basis.

Understand given conditions for part (b)

We are given a set of polynomials \(B = \{p_0(x), p_1(x), \ldots, p_n(x)\}\) in \(\mathbf{P}_{n}\) with distinct root properties for polynomials to evaluate and show that this set is a basis for \(\mathbf{P}_{n}\).

Use distinct evaluation points (Part b)

The evaluation points condition implies that each polynomial is non-zero at its specific \(a_i\) and zero at \(a_j\) for \(i eq j\). This implies each polynomial contributes uniquely to the basis.

Conclude basis for part (b)

Since each polynomial \(p_i(x)\) in \(B\) contributes solely its own non-zero term at its evaluation point, and no other polynomial in the set contributes at that point, \(B\) is linearly independent. Together they span the space \(\mathbf{P}_{n}\). Thus, \(B\) is a basis.

Key concepts

Linear Independence

In mathematics, linear independence is a fundamental concept that helps determine whether a set of vectors (or functions) can be used as a basis for a vector space. For a set of functions or polynomials to be linearly independent, the only way a linear combination of them can equal zero is if all coefficients in the combination are zero.

For example, suppose we have two polynomials in the space oindent\[\{p(x), q(x)\}\]. If \(r p(x) + s q(x) = 0\), we must find that \(r = 0\) and \(s = 0\).

In general:

• If given more than one linear combination that adds to zero (other than the trivial one where all coefficients are zero), the functions or polynomials are linearly dependent.
• If the only combination that satisfies \(r p(x) + s q(x) = 0\) is the trivial one (when \(r = 0\) and \(s = 0\)), the set is linearly independent.

Linear independence is crucial for identifying basis sets in vector spaces, ensuring each vector (or function) in the set is necessary and contributes uniquely.

Polynomial Space

Polynomial space, often denoted as \(\mathbf{P}_n\), refers to the vector space composed of polynomials. Polynomials within this space have a maximum degree \(n\). This means any polynomial of degree \(n\) or less is considered part of \(\mathbf{P}_n\).

Key points about polynomial space:

• Each polynomial can be expressed generally as \(a_0 + a_1x + a_2x^2 + \cdots + a_nx^n\).
• The dimension of \(\mathbf{P}_n\) is \(n+1\), because polynomials include terms from \(x^0\) to \(x^n\).

Understanding this space is fundamental when working with polynomial expressions, as it helps determine the form and behavior of these mathematical objects, especially when identifying basis polynomials that describe this space entirely.

Linearly Independent Set

A linearly independent set is a collection of vectors or functions (like polynomials) in which no element can be written as a combination of the others. This characteristic is crucial for establishing a basis for a space, as any basis must be both linearly independent and span the space entirely.

To determine if a set of polynomials forms a linearly independent set, consider:

• If the only solution to their linear combination yielding zero is when all coefficients are zero, the set is linearly independent.
• This ensures each member of the set contributes something unique, preventing redundancy.

In the case of polynomials, like those in \(\mathbf{P}_1\) or \(\mathbf{P}_n\), linearly independent sets provide powerful tools for constructing the entirety of polynomial space efficiently and effectively.

Span of Polynomials

The span of polynomials refers to all possible polynomials that can be formed through combinations of a given set. A set of polynomials spans a space if every polynomial within that space can be expressed as a linear combination of the set's polynomials.

Key characteristics of span:

• If a set spans a polynomial space, it means the space is fully described by the polynomials present in the set.
• A basis of a space, combining the ideas of span and linear independence, guarantees the ability to uniquely describe every polynomial in the space.

When combined with linear independence, the span enables the identification of a basis for a polynomial space, ensuring all elements are essential and that no unnecessary elements are included.