Lab 5 Interactive: Coordinates, Matrices, and Change of Basis

Explore coordinate vectors, matrix representations of linear transformations, change of basis, similarity, and diagonalization. The goal is to see that coordinates are a language, not the object itself.

1. Coordinates in a basis of \(\mathbb R^2\)

Enter basis vectors \(b_1,b_2\) and a vector \(v\). The tool solves \(P_{\mathcal B}c=v\), where \(P_{\mathcal B}=[b_1\ b_2]\).

Basis matrix \(P_{\mathcal B}\)

Vector \(v\)

2. Geometry of coordinates

Blue and orange vectors are the basis vectors. The black vector is \(v\). The coordinate vector tells how many copies of each basis vector are needed to build \(v\).

3. Reconstruct a vector from coordinates

Enter \([v]_{\mathcal B}\). The tool computes \(v=P_{\mathcal B}[v]_{\mathcal B}\).

Coordinate column \([v]_{\mathcal B}\)

4. Change coordinates from \(\mathcal B\) to \(\mathcal C\)

The change-of-coordinate matrix is \(P_{\mathcal C}^{-1}P_{\mathcal B}\). It converts \([v]_{\mathcal B}\) into \([v]_{\mathcal C}\).

New basis matrix \(P_{\mathcal C}\)

5. Matrix of a linear transformation in chosen bases

Let \(T:\mathbb R^2\to\mathbb R^2\) be given in standard coordinates by \(T(x)=Ax\). The matrix with input basis \(\mathcal B\) and output basis \(\mathcal C\) is

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

Standard matrix \(A\)

6. Similarity: same map, same basis for input and output

When the same basis is used for domain and codomain, the new matrix is \(A_{\mathcal B}=P_{\mathcal B}^{-1}AP_{\mathcal B}\). Similar matrices have the same determinant, trace, and eigenvalues.

7. Diagonalization check

If the basis vectors are eigenvectors of \(A\), then \(P_{\mathcal B}^{-1}AP_{\mathcal B}\) is diagonal.

8. Independent-study practice tasks with answers

Task A. Let \(b_1=(1,1)\), \(b_2=(-1,2)\), and \(v=(2,8)\). Find \([v]_{\mathcal B}\).
Answer: \([v]_{\mathcal B}=(4,2)^T\).
Task B. If \([v]_{\mathcal B}=(3,-1)^T\) for the same basis, find \(v\).
Answer: \(v=3b_1-b_2=(4,1)^T\).
Task C. For \(A=\begin{bmatrix}2&1\\0&3\end{bmatrix}\), compute \([T]_{\mathcal C\leftarrow\mathcal B}\) when \(\mathcal C\) is the standard basis.
Answer: \([T]_{\mathcal C\leftarrow\mathcal B}=AP_{\mathcal B}=\begin{bmatrix}3&0\\3&6\end{bmatrix}\).
Similar practice. Use \(\mathcal C=((1,1),(1,-1))\).
Answer: \([T]_{\mathcal C\leftarrow\mathcal B}=P_{\mathcal C}^{-1}AP_{\mathcal B}=\begin{bmatrix}3&3\\0&-3\end{bmatrix}\).
AI companion. Ask an AI assistant to explain why \([v]_{\mathcal B}\) and \([T]_{\mathcal C\leftarrow\mathcal B}\) are different kinds of coordinate objects. Check whether it clearly names the basis of the domain and codomain.

9. Workspace