The Same Operator, Two Coordinate Systems
A linear operator does not change when we describe its input and output in a different basis; only its matrix representation changes. The formula for that change can look like a memorized algebraic rule, but its order is determined by the journey a coordinate vector takes: first convert from the chosen basis to standard coordinates, apply the operator there, then convert the result back. Following those steps explains both the formula and why reversing the order gives the wrong matrix.
Work over \(\mathbb{R}\), and let \(T:\mathbb{R}^n\to\mathbb{R}^n\) be linear. Write \(\mathcal{E}=(\mathbf{e}_1,\dots,\mathbf{e}_n)\) for the standard basis and \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) for another ordered basis. The matrix of \(T\) in standard coordinates is \(\mathbf{A}=[T]_{\mathcal{E}\leftarrow\mathcal{E}}\). The matrix of the same operator in \(\mathcal{B}\)-coordinates is \([T]_{\mathcal{B}\leftarrow\mathcal{B}}\), abbreviated here as \([T]_{\mathcal{B}}\).
The direction indicated by the subscripts matters. A column \(\mathbf{x}\) supplied to \(\mathbf{P}\) is a list of \(\mathcal{B}\)-coordinates; the output \(\mathbf{P}\mathbf{x}\) is the same vector expressed in standard coordinates. In contrast, \(\mathbf{P}^{-1}\) starts with standard coordinates and returns \(\mathcal{B}\)-coordinates. Neither matrix changes the vector itself.
Following the Coordinate Conversions
Let \(\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}\). To find the \(\mathcal{B}\)-coordinates of \(T(\mathbf{v})\), apply the stages in order. The first conversion gives the standard-coordinate input \(\mathbf{P}\mathbf{x}\). The standard matrix \(\mathbf{A}\) then gives the standard-coordinate output \(\mathbf{A}\mathbf{P}\mathbf{x}\). Finally, \(\mathbf{P}^{-1}\) converts that output back to \(\mathcal{B}\)-coordinates.
Proof. Take any \(\mathbf{x}\in\mathbb{R}^n\) as a \(\mathcal{B}\)-coordinate vector, and let \(\mathbf{v}\) be the vector it represents. By the definition of \(\mathbf{P}\), the standard-coordinate vector of \(\mathbf{v}\) is \(\mathbf{P}\mathbf{x}\). By the definition of \(\mathbf{A}\), the standard-coordinate vector of \(T(\mathbf{v})\) is \(\mathbf{A}\mathbf{P}\mathbf{x}\). Converting this output to \(\mathcal{B}\)-coordinates gives \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\mathbf{x}\). By the definition of the matrix of \(T\) in \(\mathcal{B}\), those same output coordinates are \([T]_{\mathcal{B}\leftarrow\mathcal{B}}\mathbf{x}\). These two expressions agree for every \(\mathbf{x}\). Two matrices that give the same output for every input have equal columns, since their \(j\)th columns are their outputs on \(\mathbf{e}_j\). Therefore \([T]_{\mathcal{B}\leftarrow\mathcal{B}}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). \(\square\)
In the product, the rightmost matrix acts first. Thus \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) is not an arbitrary rearrangement: it records, from right to left, conversion into standard coordinates, application of \(T\), and conversion back. The outside matrices are inverse coordinate conversions because the input and output of the represented operator are both described in \(\mathcal{B}\).
Proof. The theorem gives \(\mathbf{C}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). Multiplying on the left by \(\mathbf{P}\) yields \(\mathbf{P}\mathbf{C}=\mathbf{P}\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{A}\mathbf{P}\). For the converse, suppose \(\mathbf{P}\mathbf{C}=\mathbf{A}\mathbf{P}\). Since \(\mathbf{P}\) is invertible, multiplying on the left by \(\mathbf{P}^{-1}\) gives \(\mathbf{C}=\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\). This equation determines a single \(\mathbf{C}\), proving uniqueness as well. \(\square\)
The consistency equation is a useful check: applying \(T\) and then converting the output to standard coordinates must agree with first converting the input to standard coordinates and then applying the standard matrix. Both routes give the same standard-coordinate output.
Computing the Formula in Examples
Worked Example: A Basis in Which the Operator Acts by Separate Scales
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have standard matrix \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), and take \(\mathcal{B}=((1,1),(1,-1))\). Find \([T]_{\mathcal{B}}\).
The basis vectors are the columns of \(\mathbf{P}\). Its inverse is shown by direct multiplication:
First multiply \(\mathbf{A}\mathbf{P}\). The columns of this product are the standard-coordinate outputs on the basis vectors:
Now convert those output columns back to \(\mathcal{B}\)-coordinates:
Check: use \(\mathbf{x}=(2,-1)^{\top}\) as \(\mathcal{B}\)-coordinates. Then \(\mathbf{P}\mathbf{x}=(1,3)^{\top}\), and \(\mathbf{A}\mathbf{P}\mathbf{x}=(5,7)^{\top}\). Converting back gives \(\mathbf{P}^{-1}(5,7)^{\top}=(6,-1)^{\top}\). The proposed matrix also gives \([T]_{\mathcal{B}}\mathbf{x}=(6,-1)^{\top}\), as required.
Worked Example: A Nondiagonal Matrix in the New Basis
Let \(\mathbf{A}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\), and choose \(\mathcal{B}=((1,1),(0,1))\). Compute the matrix of the operator in \(\mathcal{B}\)-coordinates.
The basis-column matrix and its inverse are
Multiply the standard matrix by the basis-column matrix, then convert the output columns back:
Check: for \(\mathbf{x}=(2,-1)^{\top}\), the standard-coordinate input is \(\mathbf{P}\mathbf{x}=(2,1)^{\top}\). Applying \(\mathbf{A}\) gives \((3,2)^{\top}\), whose \(\mathcal{B}\)-coordinates are \(\mathbf{P}^{-1}(3,2)^{\top}=(3,-1)^{\top}\). The computed new-basis matrix gives \(\begin{bmatrix}2&1\\0&1\end{bmatrix}(2,-1)^{\top}=(3,-1)^{\top}\), confirming the result.
Worked Example: Following Three Coordinate Conversions
Let \(\mathbf{A}=\begin{bmatrix}1&1&0\\0&2&0\\0&0&3\end{bmatrix}\), and let \(\mathcal{B}=((1,1,0),(0,1,0),(0,0,1))\). Find \([T]_{\mathcal{B}}\), then check it on an input.
The change-of-coordinates matrix has the three basis vectors as columns. Solving for the inverse, or multiplying to check, gives
First compute each column of \(\mathbf{A}\mathbf{P}\):
Convert its columns back to \(\mathcal{B}\)-coordinates:
Check: take \(\mathbf{x}=(1,-1,2)^{\top}\). The standard-coordinate input is \(\mathbf{P}\mathbf{x}=(1,0,2)^{\top}\). Applying \(\mathbf{A}\) gives \((1,0,6)^{\top}\), and converting back gives \(\mathbf{P}^{-1}(1,0,6)^{\top}=(1,-1,6)^{\top}\). The computed matrix gives \([T]_{\mathcal{B}}\mathbf{x}=(1,-1,6)^{\top}\), the same result.
Common Mistake: Reversing the Conversions
A tempting but incorrect expression is \(\mathbf{P}\mathbf{A}\mathbf{P}^{-1}\), which reverses the roles of the two coordinate conversions. In the second example, this gives
This cannot be the matrix of \(T\) in \(\mathcal{B}\). Its first column claims that \(T(\mathbf{b}_1)\) has \(\mathcal{B}\)-coordinates \((0,-2)^{\top}\). But \(\mathbf{b}_1=(1,1)^{\top}\), so direct application gives \(T(\mathbf{b}_1)=\mathbf{A}\mathbf{b}_1=(2,2)^{\top}=2\mathbf{b}_1\), whose \(\mathcal{B}\)-coordinates are \((2,0)^{\top}\). The correction is to start with the basis coordinates, so \(\mathbf{P}\) must act first on the right, and to finish by converting back, so \(\mathbf{P}^{-1}\) must act last on the left.
One practical check is to compare columns: the \(j\)th column of \([T]_{\mathcal{B}}\) must be the \(\mathcal{B}\)-coordinate vector of \(T(\mathbf{b}_j)\). This follows directly from the definition of a matrix: its \(j\)th column is its output on the \(j\)th standard coordinate vector, which represents \(\mathbf{b}_j\) when the input is expressed in \(\mathcal{B}\)-coordinates.
Why the Formula Matters
The formula explains how to translate an operator’s matrix without changing the operator. The matrix \(\mathbf{A}\) answers questions using standard-coordinate inputs and outputs; \([T]_{\mathcal{B}}\) answers the same questions using \(\mathcal{B}\)-coordinate inputs and outputs. A reliable derivation therefore begins by identifying what coordinates each matrix accepts and produces, rather than trying to memorize the placement of the inverse.
It also gives two complementary ways to verify a calculation. For an individual vector, compare \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\mathbf{x}\) with the result of converting, applying \(T\), and converting back. For a whole matrix, check \(\mathbf{P}[T]_{\mathcal{B}}=\mathbf{A}\mathbf{P}\). Both checks express the same requirement: the coordinate description must agree with the original operator at every stage.
Check Your Understanding
In each question, keep track of whether a vector is written in standard coordinates or in basis coordinates.
- State what coordinate conversion \(\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{B}}\) performs, and what conversion \(\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{B}}^{-1}\) performs.
- For \(\mathbf{A}=\begin{bmatrix}2&0\\0&1\end{bmatrix}\) and \(\mathbf{P}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), compute \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\).
- Explain why the rightmost matrix in \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}\) acts first.
- Verify the coordinate consistency equation \(\mathbf{P}[T]_{\mathcal{B}}=\mathbf{A}\mathbf{P}\) for one of the worked examples.
- For the basis in the second worked example, use \(T(\mathbf{b}_2)\) to determine the second column of \([T]_{\mathcal{B}}\) directly.
