Lab 5. Coordinates, Matrices of Linear Transformations, and Change of Basis: Independent Study

This lab accompanies Chapter 5: Coordinates, Matrices, and Change of Basis.

The goal is to make the central story of Chapter 5 computational:

A basis is a coordinate system.
A vector has different coordinate columns in different bases.
A linear transformation has different matrix representations in different input/output bases.
A change of basis is a translation between coordinate languages.
Similar matrices represent the same linear map written in different bases.

This is an independent-study lab. Each main question includes a worked solution and a similar practice question so that students can check their work as they go.

Python practice notebook

You may also use the Jupyter notebook version for longer Python practice:

Interactive lab

Study guide and worked questions

Question 1. Coordinates in a nonstandard basis

Let

\[ b_1=\begin{bmatrix}1\\1\end{bmatrix},\qquad b_2=\begin{bmatrix}-1\\2\end{bmatrix},\qquad \mathcal B=(b_1,b_2), \]

and let

\[ x=\begin{bmatrix}2\\8\end{bmatrix}. \]

Find $[x]_{\mathcal B}$.

Solution

Place the basis vectors into the columns of the basis matrix

\[ P_{\mathcal B}=\begin{bmatrix}1&-1\\1&2\end{bmatrix}. \]

The coordinate vector $[x]_{\mathcal B}$ is the solution of

\[ P_{\mathcal B}[x]_{\mathcal B}=x. \]

Thus

\[ c_1\begin{bmatrix}1\\1\end{bmatrix} +c_2\begin{bmatrix}-1\\2\end{bmatrix} =\begin{bmatrix}2\\8\end{bmatrix}. \]

This gives

\[ c_1-c_2=2,\qquad c_1+2c_2=8. \]

Solving gives $c_1=4$ and $c_2=2$. Hence

\[ [x]_{\mathcal B}=\begin{bmatrix}4\\2\end{bmatrix}. \]

Similar practice

For the same basis, find $[y]_{\mathcal B}$ for

\[ y=\begin{bmatrix}5\\7\end{bmatrix}. \]

Answer: Solve $c_1-c_2=5$ and $c_1+2c_2=7$. Then $c_2=\frac{2}{3}$ and $c_1=\frac{17}{3}$, so

\[ [y]_{\mathcal B}=\begin{bmatrix}17/3\\2/3\end{bmatrix}. \]

Question 2. A coordinate vector is not the vector itself

Using the basis from Question 1, compute the standard vector whose $\mathcal B$-coordinates are

\[ [v]_{\mathcal B}=\begin{bmatrix}3\\-1\end{bmatrix}. \]

Solution

To reconstruct the actual vector $v$, multiply by the basis matrix:

\[ v=P_{\mathcal B}[v]_{\mathcal B} =\begin{bmatrix}1&-1\\1&2\end{bmatrix} \begin{bmatrix}3\\-1\end{bmatrix} =\begin{bmatrix}4\\1\end{bmatrix}. \]

Thus the coordinate column $\begin{bmatrix}3\\-1\end{bmatrix}$ is not the same thing as the vector $\begin{bmatrix}4\\1\end{bmatrix}$. It is a description of that vector in the basis $\mathcal B$.

Similar practice

Using the same basis, reconstruct the vector with coordinates

\[ [w]_{\mathcal B}=\begin{bmatrix}2\\5\end{bmatrix}. \]

Answer:

\[ w=2b_1+5b_2 =2\begin{bmatrix}1\\1\end{bmatrix} +5\begin{bmatrix}-1\\2\end{bmatrix} =\begin{bmatrix}-3\\12\end{bmatrix}. \]

Question 3. Matrix of a linear transformation in chosen bases

Let $T:\mathbb R^2\to\mathbb R^2$ be defined in standard coordinates by

\[ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}. \]

Let

\[ \mathcal B=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}-1\\2\end{bmatrix} \right), \qquad \mathcal C=\left( \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix} \right). \]

Find $[T]_{\mathcal C\leftarrow\mathcal B}$.

Solution

The matrix $[T]_{\mathcal C\leftarrow\mathcal B}$ takes $\mathcal B$-coordinates as input and returns $\mathcal C$-coordinates as output. The formula is

\[ [T]_{\mathcal C\leftarrow\mathcal B}=P_{\mathcal C}^{-1}AP_{\mathcal B}. \]

Since $\mathcal C$ is the standard basis, $P_{\mathcal C}=I$. Therefore

\[ [T]_{\mathcal C\leftarrow\mathcal B}=AP_{\mathcal B} =\begin{bmatrix}2&1\\0&3\end{bmatrix} \begin{bmatrix}1&-1\\1&2\end{bmatrix} =\begin{bmatrix}3&0\\3&6\end{bmatrix}. \]

The columns have a direct meaning:

\[ T(b_1)=\begin{bmatrix}3\\3\end{bmatrix},\qquad T(b_2)=\begin{bmatrix}0\\6\end{bmatrix}. \]

Similar practice

For the same $A$ and $\mathcal B$, let

\[ \mathcal C=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right). \]

Compute $[T]_{\mathcal C\leftarrow\mathcal B}$.

Answer:

\[ P_{\mathcal C}=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\qquad [T]_{\mathcal C\leftarrow\mathcal B}=P_{\mathcal C}^{-1}AP_{\mathcal B} =\begin{bmatrix}3&3\\0&-3\end{bmatrix}. \]

Question 4. Change of coordinates from one basis to another

Let

\[ \mathcal B=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}-1\\2\end{bmatrix} \right), \qquad \mathcal C=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right). \]

Suppose

\[ [v]_{\mathcal B}=\begin{bmatrix}4\\2\end{bmatrix}. \]

Find $[v]_{\mathcal C}$.

Solution

First reconstruct the standard vector:

\[ v=P_{\mathcal B}[v]_{\mathcal B} =\begin{bmatrix}1&-1\\1&2\end{bmatrix} \begin{bmatrix}4\\2\end{bmatrix} =\begin{bmatrix}2\\8\end{bmatrix}. \]

Then solve

\[ P_{\mathcal C}[v]_{\mathcal C}=v. \]

Equivalently,

\[ [v]_{\mathcal C}=P_{\mathcal C}^{-1}P_{\mathcal B}[v]_{\mathcal B}. \]

A direct computation gives

\[ [v]_{\mathcal C}=\begin{bmatrix}5\\-3\end{bmatrix}. \]

Similar practice

With the same bases, convert

\[ [w]_{\mathcal B}=\begin{bmatrix}1\\3\end{bmatrix} \]

to $\mathcal C$-coordinates.

Answer: First $w=P_{\mathcal B}[w]_{\mathcal B}=(-2,7)^T$. Solving $P_{\mathcal C}[w]_{\mathcal C}=w$ gives

\[ [w]_{\mathcal C}=\begin{bmatrix}5/2\\-9/2\end{bmatrix}. \]

Question 5. Similarity and same transformation in a new basis

Let

\[ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}, \qquad P=\begin{bmatrix}1&1\\0&1\end{bmatrix}. \]

Compute the matrix of the same transformation in the basis whose basis matrix is $P$.

Solution

For a linear map $T:\mathbb R^2\to\mathbb R^2$ with the same basis used in the domain and codomain, the new matrix is

\[ A_{\mathcal B}=P^{-1}AP. \]

Here

\[ P^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}. \]

Thus

\[ A_{\mathcal B} =P^{-1}AP =\begin{bmatrix}2&0\\0&3\end{bmatrix}. \]

This tells us that the chosen basis is an eigenbasis. In that basis, the transformation only scales coordinates.

Similar practice

Let

\[ A=\begin{bmatrix}4&2\\1&3\end{bmatrix}, \qquad P=\begin{bmatrix}2&1\\1&-1\end{bmatrix}. \]

Compute $P^{-1}AP$.

Answer:

\[ P^{-1}AP=\begin{bmatrix}5&0\\0&2\end{bmatrix}. \]

Question 6. AI companion activity

Ask an AI assistant:

Explain the difference between a vector, its coordinate vector, and the matrix of a linear transformation. Give an example where the same vector has two different coordinate columns.

Then check its answer using these criteria:

Does it distinguish the vector from its coordinates?
Does it specify which basis is being used?
Does it use the formula $v=P_{\mathcal B}[v]_{\mathcal B}$ correctly?
Does it distinguish $[T]_{\mathcal C\leftarrow\mathcal B}$ from $[v]_{\mathcal B}$?

A good answer should explicitly say: coordinates depend on the basis, but the underlying vector or linear transformation does not.

# Lab 5. Coordinates, Matrices of Linear Transformations, and Change of Basis: Independent Study {.unnumbered} This lab accompanies **Chapter 5: Coordinates, Matrices, and Change of Basis**. The goal is to make the central story of Chapter 5 computational: 1. A **basis** is a coordinate system. 2. A vector has different coordinate columns in different bases. 3. A linear transformation has different matrix representations in different input/output bases. 4. A change of basis is a translation between coordinate languages. 5. Similar matrices represent the same linear map written in different bases. This is an **independent-study lab**. Each main question includes a worked solution and a similar practice question so that students can check their work as they go. ## Python practice notebook You may also use the Jupyter notebook version for longer Python practice: - [Download Lab 5 independent-study notebook](lab-05-coordinates-matrices-change-of-basis-independent-study.ipynb) - [Open Lab 5 in Google Colab](https://colab.research.google.com/github/wanghemath/MATH5110/blob/main/Labs/lab-05-coordinates-matrices-change-of-basis-independent-study.ipynb) ## Interactive lab ```{=html} <iframe src="lab-05-interactive.html" width="100%" height="3600" style="border:1px solid #ddd; border-radius:12px;"> </iframe> ``` ## Study guide and worked questions ### Question 1. Coordinates in a nonstandard basis Let $$ b_1=\begin{bmatrix}1\\1\end{bmatrix},\qquad b_2=\begin{bmatrix}-1\\2\end{bmatrix},\qquad \mathcal B=(b_1,b_2), $$ and let $$ x=\begin{bmatrix}2\\8\end{bmatrix}. $$ Find $[x]_{\mathcal B}$. #### Solution Place the basis vectors into the columns of the basis matrix $$ P_{\mathcal B}=\begin{bmatrix}1&-1\\1&2\end{bmatrix}. $$ The coordinate vector $[x]_{\mathcal B}$ is the solution of $$ P_{\mathcal B}[x]_{\mathcal B}=x. $$ Thus $$ c_1\begin{bmatrix}1\\1\end{bmatrix} +c_2\begin{bmatrix}-1\\2\end{bmatrix} =\begin{bmatrix}2\\8\end{bmatrix}. $$ This gives $$ c_1-c_2=2,\qquad c_1+2c_2=8. $$ Solving gives $c_1=4$ and $c_2=2$. Hence $$ [x]_{\mathcal B}=\begin{bmatrix}4\\2\end{bmatrix}. $$ #### Similar practice For the same basis, find $[y]_{\mathcal B}$ for $$ y=\begin{bmatrix}5\\7\end{bmatrix}. $$ **Answer:** Solve $c_1-c_2=5$ and $c_1+2c_2=7$. Then $c_2=\frac{2}{3}$ and $c_1=\frac{17}{3}$, so $$ [y]_{\mathcal B}=\begin{bmatrix}17/3\\2/3\end{bmatrix}. $$ ### Question 2. A coordinate vector is not the vector itself Using the basis from Question 1, compute the standard vector whose $\mathcal B$-coordinates are $$ [v]_{\mathcal B}=\begin{bmatrix}3\\-1\end{bmatrix}. $$ #### Solution To reconstruct the actual vector $v$, multiply by the basis matrix: $$ v=P_{\mathcal B}[v]_{\mathcal B} =\begin{bmatrix}1&-1\\1&2\end{bmatrix} \begin{bmatrix}3\\-1\end{bmatrix} =\begin{bmatrix}4\\1\end{bmatrix}. $$ Thus the coordinate column $\begin{bmatrix}3\\-1\end{bmatrix}$ is not the same thing as the vector $\begin{bmatrix}4\\1\end{bmatrix}$. It is a description of that vector in the basis $\mathcal B$. #### Similar practice Using the same basis, reconstruct the vector with coordinates $$ [w]_{\mathcal B}=\begin{bmatrix}2\\5\end{bmatrix}. $$ **Answer:** $$ w=2b_1+5b_2 =2\begin{bmatrix}1\\1\end{bmatrix} +5\begin{bmatrix}-1\\2\end{bmatrix} =\begin{bmatrix}-3\\12\end{bmatrix}. $$ ### Question 3. Matrix of a linear transformation in chosen bases Let $T:\mathbb R^2\to\mathbb R^2$ be defined in standard coordinates by $$ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}. $$ Let $$ \mathcal B=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}-1\\2\end{bmatrix} \right), \qquad \mathcal C=\left( \begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix} \right). $$ Find $[T]_{\mathcal C\leftarrow\mathcal B}$. #### Solution The matrix $[T]_{\mathcal C\leftarrow\mathcal B}$ takes $\mathcal B$-coordinates as input and returns $\mathcal C$-coordinates as output. The formula is $$ [T]_{\mathcal C\leftarrow\mathcal B}=P_{\mathcal C}^{-1}AP_{\mathcal B}. $$ Since $\mathcal C$ is the standard basis, $P_{\mathcal C}=I$. Therefore $$ [T]_{\mathcal C\leftarrow\mathcal B}=AP_{\mathcal B} =\begin{bmatrix}2&1\\0&3\end{bmatrix} \begin{bmatrix}1&-1\\1&2\end{bmatrix} =\begin{bmatrix}3&0\\3&6\end{bmatrix}. $$ The columns have a direct meaning: $$ T(b_1)=\begin{bmatrix}3\\3\end{bmatrix},\qquad T(b_2)=\begin{bmatrix}0\\6\end{bmatrix}. $$ #### Similar practice For the same $A$ and $\mathcal B$, let $$ \mathcal C=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right). $$ Compute $[T]_{\mathcal C\leftarrow\mathcal B}$. **Answer:** $$ P_{\mathcal C}=\begin{bmatrix}1&1\\1&-1\end{bmatrix},\qquad [T]_{\mathcal C\leftarrow\mathcal B}=P_{\mathcal C}^{-1}AP_{\mathcal B} =\begin{bmatrix}3&3\\0&-3\end{bmatrix}. $$ ### Question 4. Change of coordinates from one basis to another Let $$ \mathcal B=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}-1\\2\end{bmatrix} \right), \qquad \mathcal C=\left( \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\end{bmatrix} \right). $$ Suppose $$ [v]_{\mathcal B}=\begin{bmatrix}4\\2\end{bmatrix}. $$ Find $[v]_{\mathcal C}$. #### Solution First reconstruct the standard vector: $$ v=P_{\mathcal B}[v]_{\mathcal B} =\begin{bmatrix}1&-1\\1&2\end{bmatrix} \begin{bmatrix}4\\2\end{bmatrix} =\begin{bmatrix}2\\8\end{bmatrix}. $$ Then solve $$ P_{\mathcal C}[v]_{\mathcal C}=v. $$ Equivalently, $$ [v]_{\mathcal C}=P_{\mathcal C}^{-1}P_{\mathcal B}[v]_{\mathcal B}. $$ A direct computation gives $$ [v]_{\mathcal C}=\begin{bmatrix}5\\-3\end{bmatrix}. $$ #### Similar practice With the same bases, convert $$ [w]_{\mathcal B}=\begin{bmatrix}1\\3\end{bmatrix} $$ to $\mathcal C$-coordinates. **Answer:** First $w=P_{\mathcal B}[w]_{\mathcal B}=(-2,7)^T$. Solving $P_{\mathcal C}[w]_{\mathcal C}=w$ gives $$ [w]_{\mathcal C}=\begin{bmatrix}5/2\\-9/2\end{bmatrix}. $$ ### Question 5. Similarity and same transformation in a new basis Let $$ A=\begin{bmatrix}2&1\\0&3\end{bmatrix}, \qquad P=\begin{bmatrix}1&1\\0&1\end{bmatrix}. $$ Compute the matrix of the same transformation in the basis whose basis matrix is $P$. #### Solution For a linear map $T:\mathbb R^2\to\mathbb R^2$ with the same basis used in the domain and codomain, the new matrix is $$ A_{\mathcal B}=P^{-1}AP. $$ Here $$ P^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}. $$ Thus $$ A_{\mathcal B} =P^{-1}AP =\begin{bmatrix}2&0\\0&3\end{bmatrix}. $$ This tells us that the chosen basis is an eigenbasis. In that basis, the transformation only scales coordinates. #### Similar practice Let $$ A=\begin{bmatrix}4&2\\1&3\end{bmatrix}, \qquad P=\begin{bmatrix}2&1\\1&-1\end{bmatrix}. $$ Compute $P^{-1}AP$. **Answer:** $$ P^{-1}AP=\begin{bmatrix}5&0\\0&2\end{bmatrix}. $$ ### Question 6. AI companion activity Ask an AI assistant: > Explain the difference between a vector, its coordinate vector, and the matrix of a linear transformation. Give an example where the same vector has two different coordinate columns. Then check its answer using these criteria: 1. Does it distinguish the vector from its coordinates? 2. Does it specify which basis is being used? 3. Does it use the formula $v=P_{\mathcal B}[v]_{\mathcal B}$ correctly? 4. Does it distinguish $[T]_{\mathcal C\leftarrow\mathcal B}$ from $[v]_{\mathcal B}$? A good answer should explicitly say: coordinates depend on the basis, but the underlying vector or linear transformation does not.