Problem 18 - 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 \(V.\) $$\begin{aligned}&V=\mathbb{R}^{2} ; B=\\{(-5,-3),(4,28)\\}\\\&C=\\{(6,2),(1,-1)\\}\end{aligned}$$

Step-by-Step Solution

Answer

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 \(V\) is \[ P_{C \leftarrow B}= \begin{bmatrix} \frac{1}{41} & -\frac{1}{41} \\ \frac{1}{41} & -\frac{7}{41} \\ \end{bmatrix}. \]

Step 1: Write the matrix with columns representing the vectors in basis B

We are given the basis \(B=\{(-5,-3),(4,28)\}\), so the matrix with columns representing the vectors in the basis \(B\) is \[ A_B= \begin{bmatrix} -5 & 4 \\ -3 & 28 \\ \end{bmatrix}. \]

Step 2: Express each vector in basis C as a linear combination of the vectors in basis B

The basis \(C=\{(6,2),(1,-1)\}\) represents the vectors in the new basis. We need to find the coefficients \(x_1, x_2, y_1, y_2\) such that: \[ \begin{cases} -5x_1+4x_2=6 \\ -3x_1+28x_2=2 \\ -5y_1+4y_2=1 \\ -3y_1+28y_2=-1 \end{cases} \] To solve for these coefficients, we can use the inverse of matrix \(A_B\).

Step 3: Find the inverse of matrix A_B

The inverse of matrix \(A_B\) is: \[ A_B^{-1}= \frac{1}{(-5)(28)-(-3)(4)} \begin{bmatrix} 28 & -4 \\ 3 & -5 \\ \end{bmatrix} = \frac{1}{164} \begin{bmatrix} 28 & -4 \\ 3 & -5 \\ \end{bmatrix}. \]

Step 4: Calculate the coefficients x1, x2, y1, y2

Now we can calculate the coefficients by multiplying \(A_B^{-1}\) with the vectors of the basis \(C\): \[ x= \frac{1}{164} \begin{bmatrix} 28 & -4 \\ 3 & -5 \\ \end{bmatrix} \begin{bmatrix} 6 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{41} \\ \frac{1}{41} \\ \end{bmatrix}, \] \[ y= \frac{1}{164} \begin{bmatrix} 28 & -4 \\ 3 & -5 \\ \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} = \begin{bmatrix} -\frac{1}{41} \\ -\frac{7}{41} \\ \end{bmatrix}. \]

Step 5: Write the change-of-basis matrix P_C←B

Place the coefficients found in step 4 as the columns of the change-of-basis matrix: \[ P_{C \leftarrow B}= \begin{bmatrix} \frac{1}{41} & -\frac{1}{41} \\ \frac{1}{41} & -\frac{7}{41} \\ \end{bmatrix}. \] So, 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 \(V\) is \[ P_{C \leftarrow B}= \begin{bmatrix} \frac{1}{41} & -\frac{1}{41} \\ \frac{1}{41} & -\frac{7}{41} \\ \end{bmatrix}. \]

Key Concepts

Matrix Inversion

Matrix inversion is a fundamental concept in linear algebra that allows us to find a matrix's inverse. This inverse is akin to a reciprocal number in basic arithmetic. In our exercise, we needed to find the inverse of the matrix \(A_B\) to transition between bases. To compute the inverse of a square matrix, it should be non-singular, meaning its determinant is not zero. In simpler terms, non-singular matrices are invertible.

For a 2x2 matrix, the formula to find its inverse is by swapping its diagonal elements, changing the signs of its off-diagonal elements, and dividing by its determinant. Specifically, if a matrix is \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), its inverse is given by:

• First, compute the determinant: \(ad - bc\).
• Then, form \(\frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\).

In this exercise, the inverse helps solve systems of equations required to express a new basis in terms of an old one.

Linear Combinations

A linear combination involves creating new vectors from existing ones using scalar multiplication and addition. In this exercise, we used linear combinations to express vectors from the basis \(C\) using vectors from the basis \(B\). Essentially, it means finding specific scalars such that when we multiply them to the vectors in one basis and add them up, we get the vectors in another basis.

For instance, given vectors \(b_1\) and \(b_2\) from basis \(B\), the challenge is to determine values \(x_1, x_2\) so that \(x_1b_1 + x_2b_2\) equals a vector from basis \(C\).

• This forms a system of linear equations that we solve using the inverse matrix.
• Each solution helps us build the change-of-basis matrix, vital in transforming vector space notations.

Linear combinations thus enable translation from one vector expression to another within different bases.

Vector Spaces

Vector spaces are abstract mathematical constructs that generalize our intuitive idea of vectors. They consist of objects known as vectors, which can be added together and multiplied by scalars to produce another vector. In our exercise, the vector space \(\mathbb{R}^2\) provided the setting where bases \(B\) and \(C\) live.

Within a vector space:

• Vectors follow specific rules like commutative and associative addition, and distributive multiplication.
• An ordered basis, like \(B\) or \(C\), allows us to define every vector uniquely through linear combinations of those basis vectors.

Understanding vector spaces helps in solving more complex algebraic problems and is foundational to many areas of mathematics and physics.

Ordered Bases

An ordered basis is a specific sequence of vectors used to define all possible vectors in a vector space. Every vector in the space can be uniquely represented as a linear combination of the basis vectors. The order of these vectors is crucial, as different orders can lead to different vector representations.

In our exercise, we dealt with two such bases, \(B\) and \(C\). To change a basis, as we did to find the change-of-basis matrix \(P_{C \leftarrow B}\), we must understand:

• Each basis is like a coordinate system to express vectors within the vector space.
• The change-of-basis matrix is calculated using linear combinations and matrix inversion, facilitating the conversion of vector descriptions from one basis to another.

By understanding ordered bases, one learns how to switch perspectives within vector spaces, a key skill in advanced mathematical reasoning.