Linear Transformations And Change Of Basis In R^2 A Comprehensive Guide

Jul 26, 2025 by Admin 72 views

Understanding Linear Transformations and Change of Basis in R^2

Hey guys! Let's dive into a fascinating topic in linear algebra: linear transformations and change of basis. In this article, we're going to explore how to represent a linear transformation with respect to different bases and how to switch between these representations. We'll tackle a specific problem involving bases in R^2 and a linear transformation, breaking it down step by step.

Problem Setup: Bases, Linear Transformations, and Matrices

We're given two bases for R^2:

B = { (1, 3), (-2, -2) }
B' = { (-12, 0), (-4, 4) }

Think of a basis as a fundamental set of vectors that can be used to describe any other vector in the space. R^2, the familiar 2D plane, can be spanned by different sets of basis vectors. B and B' are two such sets.

We also have a linear transformation T: R^2 → R^2. This transformation takes vectors in R^2 and maps them to other vectors in R^2, while preserving certain properties like vector addition and scalar multiplication. The matrix A represents this transformation relative to the basis B:

A =  
\begin{bmatrix}
0 & 3 \\
2 & 4
\end{bmatrix}

What does this matrix A actually mean? Well, it tells us how T transforms the basis vectors in B. The first column of A represents the image of the first basis vector (1, 3) in B, expressed in terms of the basis B. Similarly, the second column represents the image of the second basis vector (-2, -2) in B, also expressed in terms of B. Let's make this super clear: T(1,3) when expressed in B has coordinates (0,2) and T(-2,-2) when expressed in B has coordinates (3,4).

Our goal is to understand how to move between different representations of this linear transformation. We'll start by finding the transition matrix between the bases B and B'.

Finding the Transition Matrix P from B' to B

The transition matrix P allows us to convert coordinate vectors from one basis to another. Specifically, the transition matrix from B' to B, denoted as P, will help us express vectors given in the B' basis in terms of the B basis. In other words, if we have a vector's coordinates in B', multiplying it by P will give us the same vector's coordinates in B. The transition matrix is a cornerstone concept when working with multiple bases in linear algebra.

To find P, we need to express the basis vectors of B' as linear combinations of the basis vectors of B. Let's break it down:

Express (-12, 0) as a linear combination of (1, 3) and (-2, -2): We want to find scalars c1 and c2 such that: (-12, 0) = c1(1, 3) + c2(-2, -2) This gives us a system of equations:
- c1 - 2c2 = -12
- 3c1 - 2c2 = 0 Solving this system (you can use substitution, elimination, or any method you prefer), we get c1 = 6 and c2 = 9. Therefore, (-12, 0) = 6(1, 3) + 9(-2, -2). The coordinates of (-12,0) relative to B are (6,9).
Express (-4, 4) as a linear combination of (1, 3) and (-2, -2): Similarly, we want to find scalars d1 and d2 such that: (-4, 4) = d1(1, 3) + d2(-2, -2) This gives us another system of equations:
- d1 - 2d2 = -4
- 3d1 - 2d2 = 4 Solving this system, we get d1 = 4 and d2 = 4. Therefore, (-4, 4) = 4(1, 3) + 4(-2, -2). The coordinates of (-4,4) relative to B are (4,4).

Now, we construct the transition matrix P. The coordinates we found become the columns of P:

P = 
\begin{bmatrix}
6 & 4 \\
9 & 4
\end{bmatrix}

This matrix P is the transition matrix from B' to B. Remember, it transforms coordinate vectors in B' to coordinate vectors in B.

Finding the Transition Matrix from B to B'

Now, let's find the transition matrix from B to B'. We'll call this matrix P^(-1) because it is the inverse of the transition matrix P from B' to B. Intuitively, it does the opposite: it converts coordinate vectors in B to coordinate vectors in B'.

There are a couple of ways to find the inverse of a matrix. One common method is to use the formula for the inverse of a 2x2 matrix:

If A =

\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}

then A^(-1) = (1/(ad - bc))

\begin{bmatrix}
d & -b \\
-c & a
\end{bmatrix}

where (ad - bc) is the determinant of A.

Applying this to our matrix P:

P =

\begin{bmatrix}
6 & 4 \\
9 & 4
\end{bmatrix}

The determinant of P is (6 * 4) - (4 * 9) = 24 - 36 = -12.

Therefore, P^(-1) = (1/-12)

\begin{bmatrix}
4 & -4 \\
-9 & 6
\end{bmatrix}

Simplifying, we get:

P^(-1) =

\begin{bmatrix}
-1/3 & 1/3 \\
3/4 & -1/2
\end{bmatrix}

So, this P^(-1) is the transition matrix from B to B'. If you have the coordinates of a vector in the B basis, multiplying it by P^(-1) will give you the coordinates of the same vector in the B' basis. Understanding the inverse transition matrix is crucial for moving back and forth between different coordinate systems.

Finding the Matrix for T Relative to B'

Okay, we've found the transition matrices. Now, let's tackle the main challenge: finding the matrix for the linear transformation T relative to the basis B'. We'll call this matrix A'.

We know that A represents T relative to B, and we have the transition matrices P and P^(-1). How can we connect these pieces to find A'? Here's the key relationship:

A' = P^(-1) * A * P

This formula might look intimidating at first, but it's a powerful statement. It says that to find the matrix representation of T in the B' basis (A'), we need to perform a change of basis. Let's break down what this formula actually does:

P: Takes a coordinate vector in B' and converts it to B.
A: Applies the linear transformation T to the vector in B (since A is the matrix representation of T in B).
P^(-1): Takes the result (which is in B) and converts it back to B'.

In essence, we're going from B' to B, applying the transformation, and then going back to B'. This gives us the transformation in terms of B'.

Let's perform the matrix multiplication:

A =

\begin{bmatrix}
0 & 3 \\
2 & 4
\end{bmatrix}

P =

\begin{bmatrix}
6 & 4 \\
9 & 4
\end{bmatrix}

P^(-1) =

\begin{bmatrix}
-1/3 & 1/3 \\
3/4 & -1/2
\end{bmatrix}

First, let's calculate A * P:

\begin{bmatrix}
0 & 3 \\
2 & 4
\end{bmatrix} * \begin{bmatrix}
6 & 4 \\
9 & 4
\end{bmatrix} = \begin{bmatrix}
27 & 12 \\
48 & 24
\end{bmatrix}

Now, let's multiply P^(-1) by the result:

\begin{bmatrix}
-1/3 & 1/3 \\
3/4 & -1/2
\end{bmatrix} * \begin{bmatrix}
27 & 12 \\
48 & 24
\end{bmatrix} = \begin{bmatrix}
7 & 4 \\
-1/2 & 0
\end{bmatrix}

Therefore, the matrix A' representing T relative to the basis B' is:

A' =

\begin{bmatrix}
7 & 4 \\
-1/2 & 0
\end{bmatrix}

This matrix A' tells us how the linear transformation T acts on vectors expressed in the B' basis. The columns of A' represent the images of the basis vectors in B' under the transformation T, expressed in terms of B'.

Conclusion: Putting It All Together

We've successfully navigated the world of linear transformations and change of basis! We started with two different bases for R^2 and a linear transformation T represented by a matrix A relative to one basis (B). We then found the transition matrices P and P^(-1), which allow us to switch between coordinate representations in the two bases. Finally, we used these transition matrices to find the matrix A' that represents the same linear transformation T, but now relative to the other basis (B').

This process highlights a fundamental concept in linear algebra: linear transformations can be represented differently depending on the choice of basis. Understanding how to change between these representations is crucial for solving a wide range of problems in mathematics, physics, computer graphics, and many other fields.

I hope you guys found this explanation helpful! Keep exploring the fascinating world of linear algebra!