Chapter 9: Problem 19
Here are some vectors. $$ \left\\{x^{3}-2 x^{2}-x+2,3 x^{3}-x^{2}+2 x+2\right\\} $$ If these are linearly independent, extend to a basis for all of \(\mathbb{P}_{3}\).
Short Answer
Expert verified
Basis: \( \left\{ x^3 - 2x^2 - x + 2, 3x^3 - x^2 + 2x + 2, 1, x \right\} \).
Step by step solution
01
- Understanding Linear Independence
Check if the given vectors \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \) are linearly independent. For vectors to be linearly independent, the equation \( c_1 \text{{vec1}} + c_2 \text{{vec2}} = 0 \) should only have the trivial solution \( c_1 = 0 \) and \( c_2 = 0 \).
02
- Setting Up the Equation
Set up the equation with coefficients \(c_1\) and \(c_2\): \(c_1 (x^3 - 2x^2 - x + 2) + c_2 (3x^3 - x^2 + 2x + 2) = 0 \).
03
- Combine Like Terms
Expand the equation and combine like terms: \( c_1 x^3 - 2c_1 x^2 - c_1 x + 2c_1 + 3c_2 x^3 - c_2 x^2 + 2c_2 x + 2c_2 = 0 \). Grouping like terms gives: \( (c_1 + 3c_2)x^3 + (-2c_1 - c_2)x^2 + (-c_1 + 2c_2)x + (2c_1 + 2c_2) = 0 \).
04
- Solving for Coefficients
Set the coefficients of like terms equal to zero: \( \begin{cases} c_1 + 3c_2 = 0 \ -2c_1 - c_2 = 0 \ -c_1 + 2c_2 = 0 \ 2c_1 + 2c_2 = 0 \end{cases} \).
05
- Determining Solutions
Solve the system of linear equations. Notice from the first and second equations: - From \( c_1 + 3c_2 = 0 \), we get \( c_1 = -3c_2 \). - Substituting into \( -2c_1 - c_2 = 0 \), we get \( -2(-3c_2) - c_2 = 0 \), which simplifies to \( 6c_2 - c_2 = 0 \), implying \( 5c_2 = 0 \), so \( c_2 = 0 \) and \( c_1 = 0 \). This implies the only solution is the trivial one, confirming linear independence.
06
- Extending to a Basis
Since \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \) are linearly independent, extend to a basis for \( \mathbb{P}_{3} \) by adding other linearly independent vectors. Typically, choose the standard basis vectors for polynomials up to degree 3: \( 1, x, x^2 \). Combined, the basis for \( \mathbb{P}_{3} \) is: \( \left\{ x^3 - 2x^2 - x + 2, 3x^3 - x^2 + 2x + 2, 1, x, x^2 \right\} \).
07
- Final Basis
Recognize that \( \mathbb{P}_{3} \) has dimension 4. Our previous set had 5 elements (one extra). Check for linear independence among all 5 vectors and eliminate the dependent ones. Therefore, the valid basis set extended by the given vectors is: \( \left\{ x^3 - 2x^2 - x + 2, 3x^3 - x^2 + 2x + 2, 1, x \right\} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Basis
In mathematics, a polynomial basis is a set of polynomials that forms a basis for a polynomial space. This means it's a set of polynomials such that any polynomial in the space can be expressed as a linear combination of these basis polynomials. For polynomials of degree at most 3, such as in \( \mathbb{P}_{3} \), we typically choose the standard basis:
- 1
- x
- x^2
- x^3
Vectors in Polynomial Space
Polynomials can be thought of as vectors where each coefficient of the polynomial corresponds to a component in a vector. For example, consider a polynomial \(a_0 + a_1 x + a_2 x^2 + a_3 x^3\). This polynomial can be represented as the vector \([a_0, a_1, a_2, a_3]\). When we talk about vectors in polynomial space, we utilize similar concepts for operations as we do in regular vector spaces.
In our exercise, we have the polynomials \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \). These can be represented as vectors in polynomial space as follows:
In our exercise, we have the polynomials \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \). These can be represented as vectors in polynomial space as follows:
- \( x^3 - 2x^2 - x + 2 \) becomes \([2, -1, -2, 1] \)
- \( 3x^3 - x^2 + 2x + 2 \) turns into \([2, 2, -1, 3] \)
Basis Extension
After verifying the linear independence of our given vectors in the polynomial space, the next step is to extend these polynomials to a full basis for \( \mathbb{P}_{3} \). Since our space has a dimension of 4, we need exactly four linearly independent vectors to form a basis.
Our given vectors \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \) count as 2 out of the 4 vectors we need. To complete the basis, we can choose two additional polynomials that are guaranteed to be linearly independent from the ones we have. A typical choice are the lowest degree polynomials: \(1\), \(x\), and \(x^2\).
Thus, we can propose:
Our given vectors \( x^3 - 2x^2 - x + 2 \) and \( 3x^3 - x^2 + 2x + 2 \) count as 2 out of the 4 vectors we need. To complete the basis, we can choose two additional polynomials that are guaranteed to be linearly independent from the ones we have. A typical choice are the lowest degree polynomials: \(1\), \(x\), and \(x^2\).
Thus, we can propose:
- \(1\)
- \(x\)