One Operator, Two Coordinate Descriptions
A linear operator is a map from a vector space to itself. Its matrix, however, depends on the basis used to record input and output coordinates. Changing the basis changes that matrix even though the underlying map does not change. The main task is to keep track of the coordinate conversions on both sides of the operator: first convert the input coordinates into the basis in which the operator is known, and then convert the output coordinates into the desired basis.
Let \(V\) be a finite-dimensional vector space over \(\mathbb{R}\), and let \(\mathcal{A}=(\mathbf{a}_1,\dots,\mathbf{a}_n)\) and \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) be ordered bases of \(V\). For \(\mathbf{v}\in V\), the coordinate column \([\mathbf{v}]_{\mathcal{A}}\) records the coefficients of \(\mathbf{v}\) in basis \(\mathcal{A}\). A linear operator \(T:V\to V\) has a matrix \([T]_{\mathcal{A}}\) in basis \(\mathcal{A}\), defined by the rule \([T(\mathbf{v})]_{\mathcal{A}}=[T]_{\mathcal{A}}[\mathbf{v}]_{\mathcal{A}}\) for every \(\mathbf{v}\in V\). The same rule defines \([T]_{\mathcal{B}}\). These are two matrices representing the same map, not two different operators.
We use the direction convention that \(P_{\mathcal{B}\leftarrow\mathcal{A}}\) converts \(\mathcal{A}\)-coordinates into \(\mathcal{B}\)-coordinates: \([\mathbf{v}]_{\mathcal{B}}=P_{\mathcal{B}\leftarrow\mathcal{A}}[\mathbf{v}]_{\mathcal{A}}\). Reversing the arrow gives the inverse conversion, since converting from \(\mathcal{A}\) to \(\mathcal{B}\) and back leaves the coordinate column unchanged.
The Conjugation Formula and Its Direction
To find the operator matrix in basis \(\mathcal{B}\), begin with an input column in \(\mathcal{B}\)-coordinates. The operator matrix \([T]_{\mathcal{A}}\) requires an \(\mathcal{A}\)-coordinate input, so first use \(P_{\mathcal{A}\leftarrow\mathcal{B}}\). After applying the operator, its output is still expressed in \(\mathcal{A}\)-coordinates; use \(P_{\mathcal{B}\leftarrow\mathcal{A}}\) to convert it back. This gives the formula below.
Proof. Fix any \(\mathbf{v}\in V\). By the definition of the transition matrix, \([\mathbf{v}]_{\mathcal{A}}=P_{\mathcal{A}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Applying the definition of the operator matrix in basis \(\mathcal{A}\), then converting the output coordinates to basis \(\mathcal{B}\), gives $$ [T(\mathbf{v})]_{\mathcal{B}} =P_{\mathcal{B}\leftarrow\mathcal{A}}[T(\mathbf{v})]_{\mathcal{A}} =P_{\mathcal{B}\leftarrow\mathcal{A}}[T]_{\mathcal{A}}[\mathbf{v}]_{\mathcal{A}} =P_{\mathcal{B}\leftarrow\mathcal{A}}[T]_{\mathcal{A}}P_{\mathcal{A}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}. $$ By definition, \([T(\mathbf{v})]_{\mathcal{B}}=[T]_{\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Thus the two matrices \([T]_{\mathcal{B}}\) and \(P_{\mathcal{B}\leftarrow\mathcal{A}}[T]_{\mathcal{A}}P_{\mathcal{A}\leftarrow\mathcal{B}}\) give the same output for every coordinate column \([\mathbf{v}]_{\mathcal{B}}\). In particular, they agree on each standard coordinate column, so their corresponding columns agree and the matrices are equal. The inverse form follows from \(P_{\mathcal{A}\leftarrow\mathcal{B}}=P_{\mathcal{B}\leftarrow\mathcal{A}}^{-1}\). \(\square\)
The arrow directions are part of the formula, not decoration. The matrix immediately to the right of \([T]_{\mathcal{A}}\) converts the input from \(\mathcal{B}\)-coordinates to \(\mathcal{A}\)-coordinates. The matrix to the left converts the output from \(\mathcal{A}\)-coordinates to \(\mathcal{B}\)-coordinates. In particular, if \(P=P_{\mathcal{B}\leftarrow\mathcal{A}}\), then the formula is \([T]_{\mathcal{B}}=P[T]_{\mathcal{A}}P^{-1}\), not \(P^{-1}[T]_{\mathcal{A}}P\).
Computing the Formula in Coordinates
Worked Example: From the Standard Basis to a New Basis
Let \(V=\mathbb{R}^2\), let \(\mathcal{A}\) be the standard basis, and let \(\mathcal{B}=((1,1),(1,-1))\). Suppose \([T]_{\mathcal{A}}=\begin{bmatrix}2&1\\0&3\end{bmatrix}\). Find \([T]_{\mathcal{B}}\) and check the result.
The matrix whose columns are the vectors of \(\mathcal{B}\) converts \(\mathcal{B}\)-coordinates to standard coordinates. Its inverse converts in the other direction:
Indeed, multiplying these two matrices in either order gives \(\mathbf{I}_2\). Apply the formula, first multiplying the operator matrix by the input conversion matrix:
Now convert the output coordinates:
To check the columns directly, write \(\mathbf{b}_1=(1,1)\) and \(\mathbf{b}_2=(1,-1)\). In standard coordinates, \(T(\mathbf{b}_1)=(3,3)=3\mathbf{b}_1\), so the first column is \((3,0)^\top\). Also, \(T(\mathbf{b}_2)=(1,-3)=-\mathbf{b}_1+2\mathbf{b}_2\), so the second column is \((-1,2)^\top\). These are exactly the columns of the computed matrix.
A matrix identity gives another check: the formula implies \([T]_{\mathcal{A}}P_{\mathcal{A}\leftarrow\mathcal{B}}=P_{\mathcal{A}\leftarrow\mathcal{B}}[T]_{\mathcal{B}}\). Both sides equal $$ \begin{bmatrix}3&1\\3&-3\end{bmatrix}. $$
Worked Example: Converting in the Reverse Direction
Let \(\mathcal{A}\) be the standard basis of \(\mathbb{R}^2\), and let \(\mathcal{B}=((1,0),(1,1))\). Suppose \([T]_{\mathcal{A}}=\begin{bmatrix}2&0\\1&3\end{bmatrix}\). Find \([T]_{\mathcal{B}}\).
The basis vectors of \(\mathcal{B}\), placed in columns, give the conversion from \(\mathcal{B}\)-coordinates to \(\mathcal{A}\)-coordinates. Inverting that matrix gives the reverse conversion:
Their products in either order are \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Thus
First multiply the operator matrix by \(P_{\mathcal{A}\leftarrow\mathcal{B}}\), then multiply on the left:
Check by applying \(T\) to the basis vectors. Since \(\mathbf{b}_1=(1,0)\), its image is \((2,1)=\mathbf{b}_1+\mathbf{b}_2\), giving the first column \((1,1)^\top\). Since \(\mathbf{b}_2=(1,1)\), its image is \((2,4)=-2\mathbf{b}_1+4\mathbf{b}_2\), giving the second column \((-2,4)^\top\). Both columns agree with the formula.
Using a Transition Matrix That Is Given Directly
Sometimes a transition matrix is supplied as a coordinate conversion, rather than computed from basis vectors. The arrow still determines the order of the factors. This example uses \(P_{\mathcal{C}\leftarrow\mathcal{B}}\) directly, with the operator initially represented in \(\mathcal{B}\).
Worked Example: Converting Between Two Nonstandard Bases
Suppose \(P_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \([T]_{\mathcal{B}}=\begin{bmatrix}1&0\\1&2\end{bmatrix}\). Find \([T]_{\mathcal{C}}\), and verify that the given transition direction is consistent with bases in \(\mathbb{R}^2\).
The inverse transition matrix is
Therefore the operator matrix in \(\mathcal{C}\) is
The first two factors multiply to \(\begin{bmatrix}2&2\\1&2\end{bmatrix}\). Multiplying by the inverse transition matrix gives $$ [T]_{\mathcal{C}}=\begin{bmatrix}2&0\\1&1\end{bmatrix}. $$
For a concrete realization, take \(\mathcal{B}=((1,0),(0,1))\) and \(\mathcal{C}=((1,0),(-1,1))\). The matrix whose columns are the \(\mathcal{C}\)-basis vectors is \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}\), whose inverse is exactly \(P_{\mathcal{C}\leftarrow\mathcal{B}}\). Check the conversion identity using the operator matrices: $$ P_{\mathcal{C}\leftarrow\mathcal{B}}[T]_{\mathcal{B}} =\begin{bmatrix}2&2\\1&2\end{bmatrix} =[T]_{\mathcal{C}}P_{\mathcal{C}\leftarrow\mathcal{B}}. $$
Common Mistake: Reversing the Transition
Suppose \(P=P_{\mathcal{B}\leftarrow\mathcal{A}}\). A common incorrect calculation is \(P^{-1}[T]_{\mathcal{A}}P\), which uses the conversion from \(\mathcal{A}\) to \(\mathcal{B}\) on the input side, where the input must instead be converted from \(\mathcal{B}\) to \(\mathcal{A}\). The second worked example makes the error visible. There, $$ P_{\mathcal{B}\leftarrow\mathcal{A}}[T]_{\mathcal{A}}P_{\mathcal{A}\leftarrow\mathcal{B}} =\begin{bmatrix}1&-2\\1&4\end{bmatrix}, \qquad P_{\mathcal{A}\leftarrow\mathcal{B}}[T]_{\mathcal{A}}P_{\mathcal{B}\leftarrow\mathcal{A}} =\begin{bmatrix}3&0\\1&2\end{bmatrix}. $$
The incorrect matrix has first column \((3,1)^\top\), which would claim \(T(\mathbf{b}_1)=3\mathbf{b}_1+\mathbf{b}_2=(4,1)\). But the given operator matrix sends \(\mathbf{b}_1=(1,0)\) to \((2,1)\), not \((4,1)\). The correction is to start with the input: a \(\mathcal{B}\)-coordinate column must first be converted to \(\mathcal{A}\)-coordinates, so \(P_{\mathcal{A}\leftarrow\mathcal{B}}\) belongs on the right. After applying \([T]_{\mathcal{A}}\), convert the output back, placing \(P_{\mathcal{B}\leftarrow\mathcal{A}}\) on the left.
Why the Formula Matters
The formula separates the operator itself from the coordinate choices used to describe it. When one changes basis for an operator, the input and output use the same basis change, but on opposite sides of the matrix: one conversion prepares the input, and the other converts the output. The two transition matrices are inverses, so the resulting relationship has the form \(P[T]_{\mathcal{A}}P^{-1}\). Remembering the coordinate path is safer than memorizing the placement of \(P\) and \(P^{-1}\).
For any proposed calculation, three checks are especially useful:
- Read the arrow on each transition matrix: it must match the coordinates being converted.
- Apply the computed matrix to each basis coordinate column and compare with the operator’s action on that basis vector.
- Check the equivalent identity \([T]_{\mathcal{A}}P_{\mathcal{A}\leftarrow\mathcal{B}}=P_{\mathcal{A}\leftarrow\mathcal{B}}[T]_{\mathcal{B}}\).
Check Your Understanding
Use the coordinate path, rather than memorization alone, to answer each question.
- If \(P_{\mathcal{C}\leftarrow\mathcal{B}}\) is known, which transition matrix converts a \(\mathcal{C}\)-coordinate input to \(\mathcal{B}\)-coordinates?
- Write the formula for \([T]_{\mathcal{C}}\) in terms of \([T]_{\mathcal{B}}\) and \(P_{\mathcal{C}\leftarrow\mathcal{B}}\).
- Explain why a transition matrix appears on each side of the operator matrix.
- For the second worked example, use \([T]_{\mathcal{B}}\) to express \(T(\mathbf{b}_1)\) in \(\mathcal{B}\)-coordinates.
- Describe one direct check that can expose a transition-direction error.
