Problem 27 - Differential Equations and Linear Algebra

Question

Determine a basis \(S\) for \(P_{3}(\mathbb{R})\) whose elements all have the same degree. Be sure to prove that \(S\) is a basis.

Step-by-Step Solution

Answer

A basis \(S\) for \(P_{3}(\mathbb{R})\) with same-degree elements is \(S = \{x^3, x^3 + x, x^3 + x^2, x^3 + x^2 + x\}\). We have proved that \(S\) is a basis because it is linearly independent and spans the vector space.

Step 1: Determine the dimension of the vector space

The dimension of \(P_{3}(\mathbb{R})\) is 4, as any polynomial of degree at most 3 can be written as \(a_3x^3 + a_2x^2 + a_1x + a_0\) for some real numbers \(a_0, a_1, a_2,\) and \(a_3\).

Step 2: Find a set of same-degree polynomials

Since the dimension of the vector space is 4, our basis must contain exactly 4 elements. Let's try to form a set of same-degree polynomials. A cubic polynomial has a degree of 3, so we consider a set of four cubic polynomials: \(S = \{p_1(x), p_2(x), p_3(x), p_4(x) \}\), where \(p_1(x) = x^3\), \(p_2(x) = x^3 + x\), \(p_3(x) = x^3 + x^2\), and \(p_4(x) = x^3 + x^2 + x\).

Step 3: Check linear independence

We need to prove the set \(S\) is linearly independent. Suppose a linear combination of the polynomials is equal to the zero polynomial: \(\alpha_1(x^3) + \alpha_2(x^3 + x) + \alpha_3(x^3 + x^2) + \alpha_4(x^3 + x^2 + x) = 0\). We can rewrite it as: \((\alpha_1+\alpha_2+\alpha_3+\alpha_4)x^3 + (0 + \alpha_2 + \alpha_3 + \alpha_4)x^2 + (0 + \alpha_2 + 0 + \alpha_4)x = 0\). Comparing the coefficients of the polynomial on the left and the zero polynomial, we have the system of linear equations: \[ \begin{cases} \alpha_1+\alpha_2+\alpha_3+\alpha_4 = 0 \\ \alpha_2+\alpha_3+\alpha_4 = 0 \\ \alpha_2+\alpha_4 = 0 \end{cases} \] Solving these equations, we get the unique solution: \(\alpha_1 = \alpha_2 = \alpha_3 = \alpha_4 = 0\). Thus, the set \(S\) is linearly independent.

Step 4: Check if S spans the vector space

Since our vector space \(P_{3}(\mathbb{R})\) has dimension 4 and the set \(S\) consists of 4 linearly independent elements, it forms a basis for \(P_{3}(\mathbb{R})\). We can show that any polynomial of degree at most 3 can be written as a linear combination of the elements in \(S\). Let \(p(x) = a_3x^3 + a_2x^2 + a_1x + a_0\) be any polynomial in \(P_{3}(\mathbb{R})\). We can rewrite \(p(x)\) as: \(p(x) = a_3 p_1(x) + (a_1 + a_3) p_2(x) - a_1 p_3(x) - (a_0 - a_1 - a_2) p_4(x)\). Therefore, any polynomial in \(P_{3}(\mathbb{R})\) is a linear combination of elements from \(S\), which means that \(S\) spans the vector space.

Step 5: Conclusion

We found the basis \(S\) for \(P_{3}(\mathbb{R})\) consisting of same-degree polynomials: \(S = \{x^3, x^3 + x, x^3 + x^2, x^3 + x^2 + x\}\). We proved that \(S\) is a basis because it is linearly independent and spans the vector space.

Key Concepts

Basis of a Vector Space

In linear algebra, the basis of a vector space is a set of vectors that are not only linearly independent but also span the entire vector space. This means that every vector in the space can be expressed uniquely as a linear combination of these basis vectors.
For a vector space to have a basis, the basis must:

• Contain a specific number of vectors equal to the dimension of the vector space.
• Be linearly independent, meaning no vector in the basis can be written as a linear combination of the others.
• Span the vector space, indicating that every vector in the space can be formed as a combination of the basis vectors.

Understanding the basis is key in simplifying problems in linear algebra, as it provides a foundation for working within the vector space.

Polynomials

Polynomials are mathematical expressions involving variables and coefficients. They are made up of terms, each consisting of a variable raised to a non-negative integer power times a coefficient. For example, a cubic polynomial of the form \(a_3x^3 + a_2x^2 + a_1x + a_0\) has four terms and is of degree 3 because the highest exponent of the variable \(x\) is 3.
In vector spaces of polynomials, the set \(P_3(\mathbb{R})\) represents all polynomials with real coefficients up to degree 3. Understanding polynomials in vector spaces is crucial as it extends the concept of vector spaces from finite-dimensional to function spaces, allowing for broader applications in calculus and differential equations.

Linear Independence

Linear independence is a fundamental concept in determining the validity of a basis in any vector space. A set of vectors or polynomials is said to be linearly independent if no vector in the set can be expressed as a linear combination of the others. This means that the only solution to the equation \(c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0\) is that all coefficients \(c_1, c_2, \ldots, c_n\) must be zero.
Understanding linear independence helps us confirm whether a candidate set of vectors forms a basis. If linearly independent, the set does not error in creating dependencies which compromise the basis structure.

Spanning Set

A spanning set in a vector space is a collection of vectors that can be linearly combined to form every vector in that space. For a set of vectors to span a vector space, you must be able to construct any vector in the space by choosing appropriate coefficients in the linear combinations of the set.
If a set of vectors spans a vector space and is linearly independent, it forms a basis for that space. In practice, checking the span involves expressing an arbitrary vector as a linear combination of the vectors in your candidate set and ensuring this is possible for any vector in the space. For a polynomial vector space, this translates to verifying that any polynomial of degree at most 3 can be decomposed into a linear combination of polynomials from your candidate basis.