One Operator, Different Coordinate Matrices
A linear operator on \(\mathbb{R}^2\) is a map from \(\mathbb{R}^2\) to itself. Its action on vectors does not depend on how we describe those vectors, but its matrix does depend on the basis used for coordinates. This distinction matters in calculations: an operator may have one matrix in the standard basis and a different matrix in another basis, while still producing exactly the same output vector.
We will use the operator \(T:\mathbb{R}^2\to\mathbb{R}^2\) defined by \(T(x,y)=(2x+y,x+2y)\). In standard coordinates, its matrix is \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\). Indeed, applying this matrix to the column \(\begin{bmatrix}x\\y\end{bmatrix}\) gives \(\begin{bmatrix}2x+y\\x+2y\end{bmatrix}\). The goal is to represent this same \(T\) using two different ordered bases.
The order in the basis is important: the first column records the image of the first basis vector, and the second column records the image of the second. To find a column, apply \(T\) to that basis vector and then express the result as a linear combination of the basis vectors. The coefficients, in their basis order, are the column entries.
The Basis-Change Formula and Why It Works
For an ordered basis \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), its basis matrix is \(\mathbf{P}_{\mathcal{B}}=\begin{bmatrix}\mathbf{b}_1&\mathbf{b}_2\end{bmatrix}\), whose columns are the basis vectors written in standard coordinates. It reconstructs a vector from its basis coordinates: \(\mathbf{x}=\mathbf{P}_{\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). Since \(\mathcal{B}\) is a basis, every vector has a unique coordinate vector, so \(\mathbf{P}_{\mathcal{B}}\) is invertible and \([\mathbf{x}]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{x}\).
Proof. Let \(\mathbf{c}\in\mathbb{R}^2\) be any coordinate column and let \(\mathbf{x}=\mathbf{P}_{\mathcal{B}}\mathbf{c}\). Then \(\mathbf{x}\) has \(\mathcal{B}\)-coordinates \(\mathbf{c}\). Applying \(T\) in standard coordinates gives \(T(\mathbf{x})=\mathbf{A}\mathbf{x}=\mathbf{A}\mathbf{P}_{\mathcal{B}}\mathbf{c}\). Converting this output back to \(\mathcal{B}\)-coordinates gives \([T(\mathbf{x})]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}\mathbf{c}\). This holds for every \(\mathbf{c}\), so the matrix taking input \(\mathcal{B}\)-coordinates to output \(\mathcal{B}\)-coordinates is \(\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}\). For the column statement, let \(\mathbf{e}_j\) be the \(j\)th standard coordinate vector. The \(j\)th column of \([T]_{\mathcal{B}}\) is \([T]_{\mathcal{B}}\mathbf{e}_j\). Since \(\mathbf{P}_{\mathcal{B}}\mathbf{e}_j=\mathbf{b}_j\), the formula just established gives \([T]_{\mathcal{B}}\mathbf{e}_j=[T(\mathbf{b}_j)]_{\mathcal{B}}\). Thus each column records the coordinates of the corresponding basis vector’s image. \(\square\)
The order of the three factors reflects the steps on a vector: first convert its basis coordinates to standard coordinates using \(\mathbf{P}_{\mathcal{B}}\), then apply \(\mathbf{A}\), then convert the output back using \(\mathbf{P}_{\mathcal{B}}^{-1}\). The formula is not an instruction to change the operator itself; it is a way to express the same action using a different coordinate system.
Computing the Matrix in an Adapted Basis
Worked Example: The Operator in an Eigenvector Basis
Take the ordered basis \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\), where \(\mathbf{b}_1=(1,1)\) and \(\mathbf{b}_2=(1,-1)\). Find \([T]_{\mathcal{B}}\), then check its action on \(\mathbf{x}=(2,0)\).
Apply \(T\) to the first basis vector:
So the first column is \(\begin{bmatrix}3\\0\end{bmatrix}\). For the second basis vector,
So the second column is \(\begin{bmatrix}0\\1\end{bmatrix}\), and \([T]_{\mathcal{B}}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\). For \(\mathbf{x}=(2,0)\), write \(\mathbf{x}=c_1\mathbf{b}_1+c_2\mathbf{b}_2\). The coordinate equations are \(c_1+c_2=2\) and \(c_1-c_2=0\), giving \(c_1=1\) and \(c_2=1\). Thus \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}1\\1\end{bmatrix}\), and the matrix gives \([T(\mathbf{x})]_{\mathcal{B}}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\begin{bmatrix}1\\1\end{bmatrix}=\begin{bmatrix}3\\1\end{bmatrix}\).
Check: Reconstructing this output gives \(3\mathbf{b}_1+\mathbf{b}_2=3(1,1)+(1,-1)=(4,2)\). Directly, \(T(2,0)=(2\cdot2+0,\;2+2\cdot0)=(4,2)\), so the coordinate calculation agrees with the operator.
Computing the Matrix in a Second Basis
Worked Example: A Different Basis for the Same Operator
Now use \(\mathcal{C}=(\mathbf{c}_1,\mathbf{c}_2)\), where \(\mathbf{c}_1=(1,0)\) and \(\mathbf{c}_2=(1,1)\). Find \([T]_{\mathcal{C}}\), then apply it to the same vector \(\mathbf{x}=(4,2)\) used below for a direct comparison.
For the first column, \(T(\mathbf{c}_1)=T(1,0)=(2,1)\). To express this in the basis \(\mathcal{C}\), solve \(a(1,0)+b(1,1)=(2,1)\). The second coordinate gives \(b=1\); the first gives \(a+b=2\), so \(a=1\). Hence \([T(\mathbf{c}_1)]_{\mathcal{C}}=\begin{bmatrix}1\\1\end{bmatrix}\). For the second column, \(T(\mathbf{c}_2)=T(1,1)=(3,3)=0\mathbf{c}_1+3\mathbf{c}_2\), giving \([T(\mathbf{c}_2)]_{\mathcal{C}}=\begin{bmatrix}0\\3\end{bmatrix}\). Therefore \([T]_{\mathcal{C}}=\begin{bmatrix}1&0\\1&3\end{bmatrix}\).
The vector \(\mathbf{x}=(4,2)\) has \(\mathcal{C}\)-coordinates \(\begin{bmatrix}2\\2\end{bmatrix}\), because \(2\mathbf{c}_1+2\mathbf{c}_2=(4,2)\). Applying the matrix gives
Check: Reconstruct from these output coordinates: \(2\mathbf{c}_1+8\mathbf{c}_2=2(1,0)+8(1,1)=(10,8)\). Direct application gives \(T(4,2)=(2\cdot4+2,\;4+2\cdot2)=(10,8)\), as required.
The two matrices look different, but both describe \(T\). In \(\mathcal{B}\), the input coordinates \(\begin{bmatrix}3\\1\end{bmatrix}\) for \((4,2)\) become \(\begin{bmatrix}9\\1\end{bmatrix}\), which reconstructs to \(9(1,1)+(1,-1)=(10,8)\). In \(\mathcal{C}\), the input coordinates \(\begin{bmatrix}2\\2\end{bmatrix}\) become \(\begin{bmatrix}2\\8\end{bmatrix}\), which reconstructs to that same output. Coordinates differ; the vector in \(\mathbb{R}^2\) does not.
Changing Directly Between the Two Coordinate Representations
The basis-change formula can also be used to convert the matrix in \(\mathcal{B}\) into the matrix in \(\mathcal{C}\). Define \(\mathbf{Q}=\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{P}_{\mathcal{B}}\). This matrix takes \(\mathcal{B}\)-coordinates to \(\mathcal{C}\)-coordinates: reconstruct in standard coordinates with \(\mathbf{P}_{\mathcal{B}}\), then convert to \(\mathcal{C}\)-coordinates with \(\mathbf{P}_{\mathcal{C}}^{-1}\).
Proof. By the basis-change formula, \([T]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{C}}\) and \([T]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}\). Since \(\mathbf{Q}=\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{P}_{\mathcal{B}}\), its inverse is \(\mathbf{Q}^{-1}=\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{P}_{\mathcal{C}}\). Substituting both expressions gives \[ \mathbf{Q}[T]_{\mathcal{B}}\mathbf{Q}^{-1} =(\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{P}_{\mathcal{B}}) (\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{B}}) (\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{P}_{\mathcal{C}}) =\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{C}} =[T]_{\mathcal{C}}. \] The adjacent factors \(\mathbf{P}_{\mathcal{B}}\mathbf{P}_{\mathcal{B}}^{-1}\) and \(\mathbf{P}_{\mathcal{B}}^{-1}\mathbf{P}_{\mathcal{B}}\) are identity matrices, which justifies the cancellations in the product. \(\square\)
Worked Example: Converting the First Matrix into the Second
Use the bases above to verify the direct conversion formula. Their basis matrices are \(\mathbf{P}_{\mathcal{B}}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\) and \(\mathbf{P}_{\mathcal{C}}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\). The inverse of \(\mathbf{P}_{\mathcal{C}}\) is \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}\), since multiplying it by \(\mathbf{P}_{\mathcal{C}}\) gives \(\mathbf{I}_2\). Thus
Its inverse is \(\mathbf{Q}^{-1}=\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\); multiplication gives \(\mathbf{Q}\mathbf{Q}^{-1}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Using \([T]_{\mathcal{B}}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\), calculate in two stages:
To check the coordinate conversion on the same vector, \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}3\\1\end{bmatrix}\) for \(\mathbf{x}=(4,2)\). Then \(\mathbf{Q}[\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}0&2\\1&-1\end{bmatrix}\begin{bmatrix}3\\1\end{bmatrix}=\begin{bmatrix}2\\2\end{bmatrix}=[\mathbf{x}]_{\mathcal{C}}\). The \(\mathcal{B}\)-matrix gives \([T(\mathbf{x})]_{\mathcal{B}}=\begin{bmatrix}9\\1\end{bmatrix}\); converting this output gives \(\mathbf{Q}\begin{bmatrix}9\\1\end{bmatrix}=\begin{bmatrix}2\\8\end{bmatrix}\). The \(\mathcal{C}\)-matrix gives the same result directly: \(\begin{bmatrix}1&0\\1&3\end{bmatrix}\begin{bmatrix}2\\2\end{bmatrix}=\begin{bmatrix}2\\8\end{bmatrix}\). Both coordinate calculations reconstruct \(T(\mathbf{x})=(10,8)\).
Common Mistake: Reversing the Conversion Order
A frequent error is to write \(\mathbf{P}_{\mathcal{C}}\mathbf{A}\mathbf{P}_{\mathcal{C}}^{-1}\) instead of \(\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{C}}\). For the basis \(\mathcal{C}\) here, the incorrect product is
This cannot represent the stated action in \(\mathcal{C}\)-coordinates. Applied to \([\mathbf{x}]_{\mathcal{C}}=\begin{bmatrix}2\\2\end{bmatrix}\), it gives \(\begin{bmatrix}6\\4\end{bmatrix}\), which reconstructs as \(6\mathbf{c}_1+4\mathbf{c}_2=(10,4)\), not \(T(4,2)=(10,8)\). The correction is to follow the vector: convert input coordinates to standard coordinates first, apply the standard matrix, and convert the output back. That order gives \(\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{A}\mathbf{P}_{\mathcal{C}}\).
A reliable check is to choose one concrete vector, calculate its input coordinates in the proposed basis, apply the proposed coordinate matrix, and reconstruct the output. This tests not just the arithmetic but also whether the basis-change factors act in the intended direction.
Check Your Understanding
For each question, keep track of whether a vector is in standard coordinates or basis coordinates.
- For \(T(x,y)=(2x+y,x+2y)\) and \(\mathcal{C}=((1,0),(1,1))\), compute \(T(\mathbf{c}_1)\) and \(T(\mathbf{c}_2)\), then form \([T]_{\mathcal{C}}\).
- For \(\mathcal{B}=((1,1),(1,-1))\), find the \(\mathcal{B}\)-coordinates of \((6,2)\), apply \([T]_{\mathcal{B}}\), and reconstruct the output.
- Explain why the \(j\)th column of \([T]_{\mathcal{B}}\) is \([T(\mathbf{b}_j)]_{\mathcal{B}}\).
- If \(\mathbf{Q}=\mathbf{P}_{\mathcal{C}}^{-1}\mathbf{P}_{\mathcal{B}}\), state what coordinate vector \(\mathbf{Q}\) converts and write the formula for \([T]_{\mathcal{C}}\).
- Identify the error in \(\mathbf{P}_{\mathcal{B}}\mathbf{A}\mathbf{P}_{\mathcal{B}}^{-1}\) as a formula for \([T]_{\mathcal{B}}\), and describe the correct order of operations.
