One Map, Two Coordinate Descriptions
A change of basis can affect both the coordinates of an input vector and the matrix used to represent a map. These changes must be coordinated: converting the input but applying the matrix for the wrong bases can produce coordinates that do not represent the map’s output. The central task here is to track the input conversion, the map, and the output conversion in the correct order, then verify that the resulting coordinates describe the same vector as a direct calculation in the original bases.
Let \(V\) and \(W\) be finite-dimensional vector spaces over \(\mathbb{R}\), with ordered bases \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) and \(\mathcal{B}'\) for \(V\), and ordered bases \(\mathcal{C}\) and \(\mathcal{C}'\) for \(W\). For \(\mathbf{v}\in V\), the coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients expressing \(\mathbf{v}\) as a linear combination of the vectors in \(\mathcal{B}\). A basis gives each vector exactly one such coordinate column.
Direction matters. The matrix \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\) converts \(\mathcal{B}'\)-coordinates to \(\mathcal{B}\)-coordinates, so it is the matrix that must act on a new-basis input before using a map matrix written in the old bases. The output then needs its own conversion, from \(\mathcal{C}\)-coordinates to \(\mathcal{C}'\)-coordinates.
The Compatibility Formula
Proof. Take any \(\mathbf{v}\in V\). First convert its \(\mathcal{B}'\)-coordinates to \(\mathcal{B}\)-coordinates. The definition of the old matrix of \(T\) then gives \([T(\mathbf{v})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}[\mathbf{v}]_{\mathcal{B}'}\). Converting the output coordinates to \(\mathcal{C}'\) gives \([T(\mathbf{v})]_{\mathcal{C}'}=\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}[T]_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}[\mathbf{v}]_{\mathcal{B}'}\). By the definition of the new matrix, the left side is also \([T]_{\mathcal{C}'\leftarrow\mathcal{B}'}[\mathbf{v}]_{\mathcal{B}'}\). These equalities hold for every coordinate column \([\mathbf{v}]_{\mathcal{B}'}\). In particular, they hold for each standard coordinate column, so the corresponding matrix columns agree. This proves the formula. \(\square\)
The formula also explains how to check a combined calculation. Starting from the input coordinates in \(\mathcal{B}'\), either apply the new matrix directly, or convert to \(\mathcal{B}\), apply the old matrix, and convert the output to \(\mathcal{C}'\). The theorem says the resulting \(\mathcal{C}'\)-coordinate columns must agree.
Worked Example: Convert the Input and Change Both Bases
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have standard-coordinate matrix \(\mathbf{A}=\begin{bmatrix}1&2\\0&1\end{bmatrix}\). Use the standard basis \(\mathcal{B}=\mathcal{C}\), the new domain basis \(\mathcal{B}'=((1,1),(1,-1))\), and the new codomain basis \(\mathcal{C}'=((1,0),(1,1))\). Find the output when the input has \(\mathcal{B}'\)-coordinates \(\begin{bmatrix}2\\-1\end{bmatrix}\), using both coordinate routes.
The input conversion matrix has the new domain basis vectors as its columns in standard coordinates: \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\). The codomain conversion matrix from \(\mathcal{C}'\) to the standard basis is \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{C}'}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), so \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). Apply the formula:
Using the new matrix directly gives \(\begin{bmatrix}2&0\\1&-1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}4\\3\end{bmatrix}\). Now check by the old-basis route. The input in standard coordinates is \(\begin{bmatrix}1&1\\1&-1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}1\\3\end{bmatrix}\). The old map matrix gives \(\begin{bmatrix}1&2\\0&1\end{bmatrix}\begin{bmatrix}1\\3\end{bmatrix}=\begin{bmatrix}7\\3\end{bmatrix}\). Converting this output to \(\mathcal{C}'\)-coordinates gives \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}\begin{bmatrix}7\\3\end{bmatrix}=\begin{bmatrix}4\\3\end{bmatrix}\), as before.
Check in the original vector space: The new-coordinate answer represents \(4(1,0)+3(1,1)=(7,3)\). Directly, \(T(1,3)=(1+2\cdot3,3)=(7,3)\). Both routes therefore give the same output vector.
When the Map Is an Operator
An operator is a map from a vector space to itself. If the same basis change is used for its input and output, the two change-of-basis matrices are inverses: one converts into the new basis and the other converts back. Thus, if \(\mathcal{B}\) and \(\mathcal{B}'\) are bases of \(V\), then \([T]_{\mathcal{B}'\leftarrow\mathcal{B}'}=\mathbf{P}_{\mathcal{B}'\leftarrow\mathcal{B}}[T]_{\mathcal{B}\leftarrow\mathcal{B}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\). This is a special case of the theorem, not a separate rule with a different order.
Worked Example: One Basis Change for an Operator
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) have standard matrix \(\mathbf{D}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\). Change to the basis \(\mathcal{B}'=((1,1),(1,-1))\), and apply the resulting matrix to the new-coordinate input \(\begin{bmatrix}2\\0\end{bmatrix}\).
The conversion matrix from the new basis to the standard basis is \(\mathbf{P}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\). Its inverse is \(\mathbf{P}^{-1}=\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), as multiplying these matrices in either order gives \(\mathbf{I}_2\). The new matrix is \(\mathbf{P}^{-1}\mathbf{D}\mathbf{P}\):
Applying it to the specified input gives \(\begin{bmatrix}\frac52&-\frac12\\-\frac12&\frac52\end{bmatrix}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}5\\-1\end{bmatrix}\). For the direct check, the input vector in standard coordinates is \(\mathbf{P}\begin{bmatrix}2\\0\end{bmatrix}=\begin{bmatrix}2\\2\end{bmatrix}\). The original operator gives \(\mathbf{D}\begin{bmatrix}2\\2\end{bmatrix}=\begin{bmatrix}4\\6\end{bmatrix}\). Converting that output to \(\mathcal{B}'\)-coordinates gives \(\mathbf{P}^{-1}\begin{bmatrix}4\\6\end{bmatrix}=\begin{bmatrix}5\\-1\end{bmatrix}\).
Check: Converting the new-coordinate output back gives \(\mathbf{P}\begin{bmatrix}5\\-1\end{bmatrix}=\begin{bmatrix}4\\6\end{bmatrix}\), which is exactly the standard-coordinate output.
Different Dimensions and Independent Bases
The domain and codomain need not have the same dimension, and their basis changes need not have any relation to each other. The dimensions determine the matrix sizes: the old map matrix has one column for each domain basis vector and one row for each codomain basis vector. In the conversion formula, the rightmost matrix changes the domain coordinates and the leftmost matrix changes the codomain coordinates. The products are defined because each adjacent pair uses matching coordinate spaces.
Worked Example: A Map from R² to R³
Let \(T:\mathbb{R}^2\to\mathbb{R}^3\) have standard-coordinate matrix \(\mathbf{A}=\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\). Use domain basis \(\mathcal{B}'=((1,1),(1,0))\) and codomain basis \(\mathcal{C}'=((1,1,0),(0,1,1),(0,0,1))\), with standard bases as the old bases. Find the new matrix and apply it to \(\begin{bmatrix}2\\-1\end{bmatrix}\).
The conversion matrices are \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}=\begin{bmatrix}1&1\\1&0\end{bmatrix}\) and \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{C}'}=\begin{bmatrix}1&0&0\\1&1&0\\0&1&1\end{bmatrix}\). The inverse codomain conversion matrix is \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}=\begin{bmatrix}1&0&0\\-1&1&0\\1&-1&1\end{bmatrix}\); multiplying it by \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{C}'}\) gives \(\mathbf{I}_3\). First compute the map matrix after the domain conversion:
Now convert its output coordinates to \(\mathcal{C}'\):
The new matrix applied to the input gives \(\begin{bmatrix}1&1\\0&-1\\2&2\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}\). For the direct route, the input in standard coordinates is \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}1\\2\end{bmatrix}\). The old matrix gives \(\begin{bmatrix}1&0\\0&1\\1&1\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}\). Converting to \(\mathcal{C}'\)-coordinates yields \(\begin{bmatrix}1&0&0\\-1&1&0\\1&-1&1\end{bmatrix}\begin{bmatrix}1\\2\\3\end{bmatrix}=\begin{bmatrix}1\\1\\2\end{bmatrix}\).
Check in \(\mathbb{R}^3\): The new output coordinates represent \(1(1,1,0)+1(0,1,1)+2(0,0,1)=(1,2,3)\), agreeing with the direct output.
Common Pitfall: Reversing a Conversion
A common error is to use the conversion matrix in the direction opposite to the one needed. In the first example, \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\) converts new input coordinates to old coordinates, while \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}\) converts old output coordinates to new coordinates. The correct order is \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}\mathbf{A}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\). Using the reversed-direction matrices in their places would instead give \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{C}'}\mathbf{A}\mathbf{P}_{\mathcal{B}'\leftarrow\mathcal{B}}\), which is generally a different matrix.
For the first example, that incorrect product is \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix}\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix} =\begin{bmatrix}2&-1\\\frac12&-\frac12\end{bmatrix}\). Applied to \(\begin{bmatrix}2\\-1\end{bmatrix}\), it gives \(\begin{bmatrix}5\\\frac32\end{bmatrix}\), not the correct \(\begin{bmatrix}4\\3\end{bmatrix}\). The correction is to write down the coordinate route before multiplying: new input to old input, apply the old map, then old output to new output. That route determines the matrix order without guesswork.
For any combined basis-change calculation, the final check should return to the underlying vector space. Convert the input to the basis used by the old map, compute its output, and express that output in the requested codomain basis. Matching coordinate columns alone is not enough if they are coordinates in different bases; matching the actual output vector confirms that the bases and conversion directions were handled consistently.
Check Your Understanding
For each question, keep track of which basis the input and output coordinates use.
- State the order of the three matrices in the formula for changing both the domain and codomain bases of a map.
- For \(\mathbf{P}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), what does \(\mathbf{P}\) convert, and what does \(\mathbf{P}^{-1}\) convert?
- In the first worked example, verify by expressing \(\begin{bmatrix}4\\3\end{bmatrix}\) in \(\mathcal{C}'\) that it represents \((7,3)\).
- Explain why a map from \(\mathbb{R}^2\) to \(\mathbb{R}^3\) has a \(3\times2\) coordinate matrix, regardless of the chosen bases.
- Describe the direct calculation that checks a new-coordinate output against the old map matrix.
