Problem 21 - Differential Equations and Linear Algebra

Question

Find the change-of-basis matrix \(P_{C \leftarrow B}\) from the given ordered basis \(B\) to the given ordered basis \(C\) of the vector space \(V.\) $$ V=P_{1}(\mathbb{R}) ; B=\\{7-4 x, 5 x\\} ; C=\\{1-2 x, 2+x\\} $$

Step-by-Step Solution

Answer

The change-of-basis matrix \(P_{C \leftarrow B}\) from the given ordered basis \(B = \{7-4x, 5x\}\) to the ordered basis \(C = \{1-2x, 2+x\}\) of the vector space \(V = P_1(\mathbb{R})\) is the following matrix: $$ P_{C \leftarrow B} = \begin{bmatrix} 2 & -1 \\ 1 & 6 \end{bmatrix} $$

Step 1: Write the B and C basis

First, we'll write down the given bases \(B\) and \(C\). \(B\) consists of the vectors \(7-4x\) and \(5x\); \(C\) consists of the vectors \(1-2x\) and \(2+x\).

Step 2: Express each vector from B in terms of C

Now we will express each vector from the basis \(B\) as a linear combination of the vectors in basis \(C\). That is, we want to find constants \(a\) and \(b\) such that: \( 7 - 4x = a(1 - 2x) + b(2+x) \) Similarly, we need to find constants \(c\) and \(d\) such that: \( 5x = c(1 - 2x) + d(2+x) \)

Step 3: Solve for the coefficients a, b, c, d

Now we'll solve the system of equations to find \(a, b, c\) and \(d\). For the first equation, we can equate the coefficients of \(x\) and the constant terms: \( a + 2b = 7 \) \( -2a + b = -4 \) Solving this system of equations, we get \(a=2\) and \(b=1\). For the second equation, we can do the same: \( -2c + d = 0 \) \( c + d = 5 \) Solving this system of equations, we get \(c=-1\) and \(d=6\). So now we have: \(7 - 4x = 2(1 - 2x) + 1(2 + x) \) \( 5x = -1(1 - 2x) + 6(2 + x) \)

Step 4: Construct the change-of-basis matrix

Finally, we construct the change-of-basis matrix \(P_{C \leftarrow B}\) using the coefficients \(a, b, c\), and \(d\) as the columns of the matrix: $$ P_{C \leftarrow B} = \begin{bmatrix} 2 & -1 \\ 1 & 6 \end{bmatrix} $$ The matrix \(P_{C \leftarrow B}\) represents the change of basis from \(B\) to \(C\) for the given vector space \(V = P_1(\mathbb{R})\).

Key Concepts

Vector Space

In the realm of mathematics, particularly linear algebra, a vector space is a fundamental concept that establishes the playground for vectors. It's essentially a collection of objects called 'vectors' that can be scaled and added together in a way that results in another vector belonging to the same space. This setting is bound by a set of rules, known as 'axioms', ensuring operations like addition and scalar multiplication always lead to a consistent result.

Consider our example of the vector space denoted by \(V = P_{1}(\mathbb{R})\), which represents all the polynomials of degree less or equal to one with real coefficients. This includes any linear expression \(ax + b\), where \(a\) and \(b\) are real numbers. This vector space is equipped to contain all linear polynomials and allows us to investigate things like change-of-basis using ordered bases, such as \(B\) and \(C\) in the given problem. Creating new bases or transitioning from one to another needs to follow the axioms of the vector space, ensuring the original relationships between vectors are preserved.

Linear Combination

The idea of a linear combination is at the heart of much of linear algebra. Any time we combine vectors using scalar multiplication and addition, we've formed a linear combination. The goal is to create new vectors through this process of combination while remaining within our vector space.

To see a linear combination in action, let's refer to our exercise, where vectors from basis \(B\) are expressed as linear combinations of vectors from basis \(C\). This process is like a recipe; you're given certain ingredients (the vectors in \(C\)) and you need to mix them in the right amounts (the scalars \(a, b, c, d\)) to reproduce the flavors (vectors in \(B\)). In essence, finding these scalars is a way of translating one language (basis \(B\)) to another (basis \(C\)), much like finding the equivalent phrase in two different dialects of the same language.

Systems of Equations

When we reach the point of needing to find the specific scalars for our linear combination, we are faced with a system of equations. These are sets of equations where each one represents a relationship between the scalars we're looking for. Solving a system of equations means finding the values for our unknowns that satisfy all equations simultaneously.

In the solution for our exercise, systems of equations pop up as we balance the equation of polynomials. Equating coefficients on both sides of each equation is an approach that helps us break down the problem into smaller, solvable pieces. Once we solve these systems, it's like cracking the code – we are rewarded with the exact 'recipe' of scalars needed for our linear combination to accurately transition from one basis to another in our vector space.

The systems are solved using methods such as substitution, elimination, or even matrix operations like row reduction, which are beyond the scope of this article but are likewise important tools in the linear algebra toolkit.