ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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The same operator has two different matrices in two ordered bases

Example

Let $T:F^2\to F^2$ be $T(x,y)=(x,2y)$ . In the standard ordered basis $\mathcal E=(e_0,e_1)$ ,

$[T]_{\mathcal E}^{\mathcal E}=\begin{pmatrix}1&0\\0&2\end{pmatrix}.$

For $\mathcal B=(e_0+e_1,e_1)$ , the same operator has

$[T]_{\mathcal B}^{\mathcal B}=\begin{pmatrix}1&0\\1&2\end{pmatrix}.$

Facts & Assumptions

Given: The displayed operator and ordered bases.

[L1]

A matrix of an operator records the coordinate columns of its basis-vector images, and basis change acts by the two-sided formula (Coordinate columns $[v]_{\mathcal B}$ and matrices $[T]_{\mathcal B}^{\mathcal C}$ of linear maps relative to ordered bases, $[T]_{\mathcal B'}^{\mathcal C'}=P_{\mathcal C'\leftarrow\mathcal C}[T]_{\mathcal B}^{\mathcal C}P_{\mathcal B\leftarrow\mathcal B'}$ ).

Verification

technique · direct

1.1

The standard images are $T(e_0)=e_0$ and $T(e_1)=2e_1$ . The transition matrices are $P_{\mathcal E\leftarrow\mathcal B}=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ and $P_{\mathcal B\leftarrow\mathcal E}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$ , whose products in either order are $I_2$ .

givenL1

2.1

Directly, $T(e_0+e_1)=e_0+2e_1=(e_0+e_1)+e_1$ and $T(e_1)=2e_1$ , giving the columns $(1,1)$ and $(0,2)$ . Equivalently, $P_{\mathcal B\leftarrow\mathcal E}[T]_{\mathcal E}^{\mathcal E}P_{\mathcal E\leftarrow\mathcal B}=\begin{pmatrix}1&0\\1&2\end{pmatrix}$ .

step 1.1L1∎

Depends on

Used by

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Dependency tree · next 3 levels

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Sources

S. Axler, Linear Algebra Done Right, 4th ed., Example 3.83 (standard reference, not scraped)