Problem 8 - Differential Equations and Linear Algebra

Question

Determine the component vector of the given vector in the vector space \(V\) relative to the given ordered basis \(B\). $$\begin{aligned} &V=P_{2}(\mathbb{R}) ; B=\left\\{x^{2}+x, 2+2 x, 1\right\\}\\\ &p(x)=-4 x^{2}+2 x+6 \end{aligned}$$

Step-by-Step Solution

Answer

The component vector of \(p(x) = -4x^2 + 2x + 6\) in the vector space \(V = P_2(\mathbb{R})\) relative to the ordered basis \(B = \{x^2 +x, 2+2x, 1\}\) is \(\begin{bmatrix} -4 \\ 3 \\ 0 \end{bmatrix}\).

Step 1: Express p(x) as a linear combination of basis vectors

We start by expressing \(p(x)\) as a linear combination of the basis vectors: \[p(x) = c_1(x^2 +x) + c_2(2+2x) + c_3(1)\]

Step 2: Equate the coefficients of corresponding terms

We equate the coefficients of corresponding terms for both sides of the equation in step 1: For \(x^2\) terms: \(-4 = c_1\) For \(x\) terms: \(2 = c_1 + 2c_2\) For constant terms: \(6 = 2c_2 + c_3\)

Step 3: Solve for the coefficients c_1, c_2, and c_3

Solve the system of linear equations obtained in step 2: Since we already have \(c_1 = -4\), we can substitute it into the equation for the x terms: \(2 = -4 + 2c_2\) Adding 4 to both sides gives: \(6 = 2c_2\) Divide by 2: \(c_2 = 3\) Now substitute \(c_2=3\) into the equation for constant terms: \(6 = 2(3) + c_3\) Which simplifies to: \(6 = 6 + c_3\) Subtracting 6 from both sides gives: \(c_3 = 0\)

Step 4: Write the component vector

Finally, express the coefficients obtained in step 3 as a vector: \[\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} -4 \\ 3 \\ 0 \end{bmatrix}\] Thus, the component vector of \(p(x) = -4x^2 + 2x + 6\) in the vector space \(V = P_2(\mathbb{R})\) relative to the ordered basis \(B = \{x^2 +x, 2+2x, 1\}\) is \(\begin{bmatrix} -4 \\ 3 \\ 0 \end{bmatrix}\).

Key Concepts

Basis Vectors

A set of basis vectors serves as the foundation for constructing a vector space. Imagine building a house where you first establish a sturdy base; similarly, in linear algebra, the basis is a set of vectors that are both linearly independent and span the vector space. This means that every other vector in that space can be represented as a combination of these basis vectors.

For example, in the case of a two-dimensional plane, two non-parallel vectors can serve as a basis. If we move into higher dimensions, we continue to add more vectors, keeping in mind they must remain linearly independent from each other. The usefulness of basis vectors becomes apparent when we perform various operations such as finding the component vector of a given vector, which essentially means representing a vector in terms of the basis of the space it inhabits.

In our exercise, the basis is \(B=\{x^2 + x, 2 + 2x, 1\}\), a set of polynomials that define a vector space for second-degree polynomials. Any second-degree polynomial can be expressed using these basis vectors.

Component Vector

The component vector of a vector in a vector space is analogous to the coordinates of a point on a map, pinpointing a specific location within a system. In linear algebra, this concept refers to expressing a vector through the quantities that correspond to each of the basis vectors of the space.

To find the component vector, we align our vector with the basis vectors and determine the coefficients (or 'weights') required to construct the original vector as a linear combination of the basis. These coefficients essentially tell us how much of each basis vector we need to add together to get back to our original vector.

This process was demonstrated in our exercise. The polynomial \(p(x) = -4x^2 + 2x + 6\) was represented by the component vector \(\begin{bmatrix} -4 \ 3 \ 0 \end{bmatrix}\), which are the coefficients for the basis vectors in \(B\), signifying that \(p(x)\) can be written as \( -4\) times the first basis vector, \(3\) times the second, and \(0\) times the third.

Linear Combination

The principle of linear combination is akin to a chef combining basic ingredients to create diverse dishes. In linear algebra, a linear combination involves combining basis vectors with assigned coefficients to form any vector within the same vector space.

The crucial aspect is the idea of scalars multiplying vectors and their subsequent addition. When we say a vector is a linear combination of basis vectors, we are specifying that we can move flexibly within the vector space using only these basic constituents. It's the algebraic equivalent of having a diverse cooking pantry: with a few select ingredients (basis vectors) and the varying amounts of each (scalars), you can make any recipe (vector) in the cookbook (vector space).

In the context of our example, finding the coefficients \(c_1\), \(c_2\), and \(c_3\) was essential as they defined how to 'mix' our basis vectors to end up with the polynomial \(p(x)\). The use of these coefficients in this linear combination gives a clear approach to writing any vector (polynomial) in the vector space \(V\) relative to the given basis \(B\).