Chapter 4: Problem 14
The set \(\mathcal{B}=\left\\{1-t^{2}, t-t^{2}, 2-2 t+t^{2}\right\\}\) is a basis for \(\mathbb{P}_{2}\) . Find the coordinate vector of \(\mathbf{p}(t)=3+t-6 t^{2}\) relative to \(\mathcal{B} .\)
Short Answer
Expert verified
The coordinate vector is \([7, -3, -2]\).
Step by step solution
01
Understanding the Basis
The basis \( \mathcal{B} = \{ 1-t^{2}, t-t^{2}, 2-2t+t^{2} \} \) consists of three polynomials of degree at most 2, which form a basis for the vector space \( \mathbb{P}_2 \), the space of polynomials of degree at most 2.
02
Express Polynomial in Terms of Basis
We want to find coefficients \( c_1, c_2, c_3 \) such that \( \mathbf{p}(t) = 3 + t - 6t^2 = c_1(1-t^2) + c_2(t-t^2) + c_3(2-2t+t^2) \).
03
Set Up the Equation
Expand the right side of the equation: \( c_1(1-t^2) = c_1 - c_1t^2 \), \( c_2(t-t^2) = c_2t - c_2t^2 \), and \( c_3(2-2t+t^2) = 2c_3 - 2c_3t + c_3t^2 \). Combine them to get: \( (c_1 + 2c_3) + (c_2 - 2c_3)t - (c_1 + c_2 - c_3)t^2 \).
04
Match Coefficients
Match the corresponding coefficients of \( \, 3 + t - 6t^2 \) and \( (c_1 + 2c_3) + (c_2 - 2c_3)t - (c_1 + c_2 - c_3)t^2 \):1. \( c_1 + 2c_3 = 3 \)2. \( c_2 - 2c_3 = 1 \)3. \( -(c_1 + c_2 - c_3) = -6 \) or equivalently \( c_1 + c_2 - c_3 = 6 \).
05
Solve the System of Equations
Solve the system from Step 4:- From equation 1: \( c_1 + 2c_3 = 3 \).- From equation 2: \( c_2 = 1 + 2c_3 \).- From equation 3: \( c_1 + c_2 - c_3 = 6 \). Substitute \( c_2 = 1 + 2c_3 \) into the third equation:\( c_1 + (1 + 2c_3) - c_3 = 6 \).Simplifying gives \( c_1 + c_3 = 5 \).Now, solve these equations:- \( c_1 + 2c_3 = 3 \)- \( c_1 + c_3 = 5 \).Subtract the second from the first:\( c_3 = -2 \).Substitute \( c_3 = -2 \) back, find \( c_1 = 7 \). Finally, substitute \( c_3 = -2 \) into \( c_2 = 1 + 2(-2) = -3 \).
06
Form the Coordinate Vector
The coordinate vector of \( \mathbf{p}(t) \) relative to the basis \( \mathcal{B} \) is \( [c_1, c_2, c_3] = [7, -3, -2] \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Coordinate Vector
A coordinate vector refers to a representation of a polynomial in terms of the coefficients that express it as a linear combination of basis vectors. In our example, the polynomial \( \mathbf{p}(t) = 3 + t - 6t^2 \) is expressed in terms of the basis \( \mathcal{B} = \{1-t^2, t-t^2, 2-2t+t^2\} \). This means we are looking to find coefficients \( c_1, c_2, \) and \( c_3 \) such that the polynomial can be written as a combination of these bases.
- The coordinate vector of \( \mathbf{p}(t) \) relative to \( \mathcal{B} \) will be \([7, -3, -2]\).
- This vector uniquely represents \( \mathbf{p}(t) \) within the vector space \( \mathbb{P}_2 \).
Polynomial Expressions
Polynomial expressions are mathematical expressions consisting of variables (such as \( t \)), coefficients, and the operations of addition, subtraction, and multiplication. A polynomial of degree \( n \) has the general form \( a_n t^n + a_{n-1} t^{n-1} + \dots + a_1 t + a_0 \).
- In our exercise, we dealt with polynomials of degree at most 2, i.e., quadratic polynomials.
- The polynomial \( \mathbf{p}(t) = 3 + t - 6t^2 \) is a quadratic polynomial.
Basis for Vector Space
In linear algebra, a basis for a vector space is a set of vectors such that:- They span the vector space, meaning any vector in the space can be expressed as a linear combination of these basis vectors.- They are linearly independent, meaning no vector in the set can be represented as a combination of the others.In our exercise, the set \( \mathcal{B} = \{1-t^2, t-t^2, 2-2t+t^2\} \) forms a basis for \( \mathbb{P}_2 \), the vector space of polynomials with degree at most 2.
- Each polynomial in the basis is unique and cannot be formed by combining the others.
- Any polynomial in \( \mathbb{P}_2 \) can be expressed as a linear combination of the polynomials in \( \mathcal{B} \).