Question

Using a particular coordinate system, a linear transformation in an abstract vector space is represented by the matrix $$ \left(\begin{array}{lll} 2 & 1 & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 5 \end{array}\right) $$ and a particular (abstract) vector by the column vector $$ \left(\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right) $$ Give the matrix and column vector in a new coordinate system, in terms of which the old base vectors are represented by $$ \mathbf{e}_{\mathbf{1}}=\left(\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right) \quad \mathbf{e}_{2}=\left(\begin{array}{r} 1 \\ -1 \\ 0 \end{array}\right) \quad \mathbf{e}_{3}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right) $$

Short answer

In the new coordinate system, the matrix is \[ \begin{pmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 5 \end{pmatrix} \] and the vector is \( \begin{pmatrix} 1.5 \ -0.5 \ 3 \end{pmatrix} \).

Step-by-step solution

- Determine the change of basis matrix

The new coordinate system requires a change of basis. The base vectors in the new system are \[ \mathbf{e}_1=\begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}, \mathbf{e}_2=\begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix}, \mathbf{e}_3=\begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \]. Represent these as columns in a matrix: \[ P = \begin{pmatrix} 1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \].

- Find the inverse of the change of basis matrix

Compute the inverse of the matrix \( P \). Recall that if \( P \) is the change of basis matrix, then \( P^{-1} \) will transform vectors from the new basis back to the original basis. The inverse can be calculated as follows: \[ P^{-1} = \left( \begin{matrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{matrix} \right) \]

- Transform the vector to the new basis

To express the column vector \( \mathbf{v}=\begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} \) in the new basis, multiply it by the inverse of the change of basis matrix: \( \mathbf{v}' = P^{-1} \mathbf{v} \). Perform the matrix multiplication: \[ \mathbf{v}' = \left( \begin{matrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{matrix} \right) \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = \begin{pmatrix} 1.5 \ -0.5 \ 3 \end{pmatrix} \].

- Transform the matrix to the new basis

To find the representation of the linear transformation in the new basis, use the similarity transformation: \[ A' = P^{-1} A P \]. Perform the matrix multiplication step-by-step. First calculate \( P^{-1} A \): \[ P^{-1} A = \begin{pmatrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \end{pmatrix} = \begin{pmatrix} 1.5 & 1.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 5 \end{pmatrix} \]. Next, calculate \( A' = P^{-1} A P \): \[ A' = \begin{pmatrix} 1.5 & 1.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 5 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 5 \end{pmatrix} \].

Key concepts

change of basis

When we talk about 'change of basis' in linear algebra, we refer to expressing vectors and matrices relative to a new set of basis vectors. A basis is a set of linearly independent vectors that span the vector space. Changing the basis requires a transformation that converts coordinates from the old basis to the new basis.
To perform a change of basis, you need to formulate a matrix `P`, where each column of `P` represents the coordinates of the new basis vectors in terms of the old basis.
In our problem, we are given: \( \textbf{e}_1=\begin{pmatrix} 1 \ 1 \ 0 \end{pmatrix}, \textbf{e}_2=\begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix}, \textbf{e}_3=\begin{pmatrix} 0 \ 0 \ 1 \end{pmatrix} \.\)
Constructing matrix `P` with these vectors as columns results in: \[ P = \begin{pmatrix} 1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
This matrix `P` enables us to convert vectors and matrices between the two bases.

matrix representation

Matrices represent linear transformations acting on vectors in vector spaces. If you have a vector space and a linear transformation, your task is to describe how that transformation acts on a basis of the space. The matrix representation of this transformation varies depending on the choice of basis.
Given the matrix \[ A = \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 5 \end{pmatrix} \,\] it represents a linear transformation with respect to the original basis.
To represent the same linear transformation in a new basis, we use the relationship \( A' = P^{-1} A P \) where `A'` is the matrix in the new basis, `P` is the change of basis matrix, and `P^{-1}` is its inverse.
This operation is known as a similarity transformation, which we will describe in more detail later.

vector transformation

Vector transformations involve expressing a given vector with respect to a different basis. Given a vector in the original basis, say \( \textbf{v} = \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} \), we can transform it to the new basis by multiplying it with the inverse of the change of basis matrix `P^{-1}`.
From our earlier section, we determined: \[ P^{-1} = \begin{pmatrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
Thus, the vector in the new basis, \( \textbf{v}' \), is given by: \[ \textbf{v}' = P^{-1} \textbf{v} = \begin{pmatrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 \ 2 \ 3 \end{pmatrix} = \begin{pmatrix} 1.5 \ -0.5 \ 3 \end{pmatrix} \]
This process re-expresses the original vector in terms of the new basis vectors.

inverse matrix

The inverse of a matrix `P`, denoted as `P^{-1}`, is a matrix that reverses the transformation effected by `P`. More formally, if \( P P^{-1} = P^{-1} P = I \) where `I` is the identity matrix, then `P^{-1}` is the inverse of `P`.
In our case, finding \( P^{-1} \) involves calculating the inverse of: \[ P = \begin{pmatrix} 1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} \] which results in: \[ P^{-1} = \begin{pmatrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{pmatrix} \]
This inverse matrix is crucial for performing both the vector transformation and the similarity transformation of the matrix.

similarity transformation

A similarity transformation helps us re-express a matrix in a different basis. For a given matrix `A`, when we change bases using matrix `P`, the corresponding transformation in the new basis is given by \( A' = P^{-1} A P \).
In our problem, we calculated the intermediate product \( P^{-1} A \) first: \[ P^{-1} A = \begin{pmatrix} 0.5 & 0.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \ 1 & 2 & 0 \ 0 & 0 & 5 \end{pmatrix} = \begin{pmatrix} 1.5 & 1.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 5 \end{pmatrix} \]
Then, performing the final multiplication \( A' = (P^{-1} A) P \), we get: \[ A' = \begin{pmatrix} 1.5 & 1.5 & 0 \ 0.5 & -0.5 & 0 \ 0 & 0 & 5 \end{pmatrix} \begin{pmatrix} 1 & 1 & 0 \ 1 & -1 & 0 \ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 5 \end{pmatrix} \]
Thus, matrix `A'` is the transformed matrix representing the linear transformation in the new basis.