One Map, Different Coordinate Matrices
A linear map does not change when we choose different bases to describe its inputs and outputs. Its matrix does change, because a matrix records the map in coordinates. To convert that matrix correctly, we must follow what happens to an input: first express it in the original domain coordinates, then apply the original matrix, and finally express the output in the new codomain coordinates. The order of these three operations determines the order of the matrix factors.
Work over \(\mathbb{R}\), and let \(V\) and \(W\) be finite-dimensional vector spaces. Suppose \(\mathcal{B}\) and \(\mathcal{B}'\) are ordered bases of \(V\), and \(\mathcal{C}\) and \(\mathcal{C}'\) are ordered bases of \(W\). For a vector \(\mathbf{x}\in V\), the coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) is the column of coefficients that expresses \(\mathbf{x}\) as a linear combination of the ordered basis \(\mathcal{B}\). A transition matrix converts one such coordinate column into another.
The arrow records the direction of conversion: the coordinates on the right of the arrow are the input coordinates, and those on the left are the output coordinates. If \(T:V\to W\) is linear, its coordinate matrix from \(\mathcal{B}\) to \(\mathcal{C}\), written \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\), is defined by \([T(\mathbf{x})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\) for every \(\mathbf{x}\in V\).
Transition Matrices Compose in Coordinate Order
When coordinates are converted more than once, the first conversion acts first on the coordinate column, so its matrix is the rightmost factor. This observation gives a useful rule for chaining transition matrices.
Proof. Let \(\mathbf{x}\) be any vector in the space. Applying the definition of each transition matrix in sequence gives \([\mathbf{x}]_{\mathcal{E}}=\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{D}}[\mathbf{x}]_{\mathcal{D}}=\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{D}}\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). The direct conversion also gives \([\mathbf{x}]_{\mathcal{E}}=\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). Every column in the coordinate space can occur as \([\mathbf{x}]_{\mathcal{B}}\), since \(\mathcal{B}\) is a basis. Therefore the two matrices act identically on every column, and their columns, obtained by applying them to the standard coordinate columns, agree. The matrices are equal. \(\square\)
Deriving the Matrix Conversion Formula
Let \(T:V\to W\) be linear. Write \(\mathbf{A}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}\) for its original coordinate matrix. Given a new input coordinate column \([\mathbf{x}]_{\mathcal{B}'}\), the map’s original matrix cannot be applied to it immediately: \(\mathbf{A}\) expects coordinates in \(\mathcal{B}\). The transition matrix \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\) supplies those. After applying \(\mathbf{A}\), the output is in \(\mathcal{C}\)-coordinates, so \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}\) converts it into the required new coordinates.
Proof. Assume the stated bases and map, and let \(\mathbf{x}\in V\) be arbitrary. Set \(\mathbf{u}=[\mathbf{x}]_{\mathcal{B}'}\). The input conversion gives \([\mathbf{x}]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\mathbf{u}\). By the definition of the original coordinate matrix, \([T(\mathbf{x})]_{\mathcal{C}}=[T]_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). Converting the output gives $$ [T(\mathbf{x})]_{\mathcal{C}'} = \mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}} [T]_{\mathcal{C}\leftarrow\mathcal{B}} \mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'} \mathbf{u}. $$ By the definition of the new coordinate matrix, the same left-hand side is \([T]_{\mathcal{C}'\leftarrow\mathcal{B}'}\mathbf{u}\). Every \(\mathcal{B}'\)-coordinate column \(\mathbf{u}\) represents a vector in \(V\), so these equalities hold for every column. Two matrices that give the same output for every coordinate column have the same columns, by evaluating them on the standard coordinate columns. Hence the matrices are equal, as claimed. \(\square\)
The rightmost factor changes the input coordinates before the map acts. The leftmost factor changes the output coordinates after the map acts. The formula makes no assumption that the domain and codomain have the same basis, or even the same dimension; the factors have the dimensions needed for the successive coordinate conversions.
Worked Conversions
Worked Example: Changing Both the Domain and Codomain Bases
Let \(V=W=\mathbb{R}^2\), let the old domain and codomain bases be standard, and suppose $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix}2&1\\1&1\end{bmatrix}. $$ Take \(\mathcal{B}'=((1,1),(1,-1))\) and \(\mathcal{C}'=((1,0),(1,1))\). Find \([T]_{\mathcal{C}'\leftarrow\mathcal{B}'}\).
The columns of the matrix whose columns are the vectors in \(\mathcal{B}'\) give their standard coordinates, so $$ \mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'} = \begin{bmatrix}1&1\\1&-1\end{bmatrix}. $$ For the codomain, the basis vectors in \(\mathcal{C}'\) have standard-coordinate columns \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\). To convert standard coordinates into \(\mathcal{C}'\)-coordinates, use its inverse: $$ \mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}} = \begin{bmatrix}1&-1\\0&1\end{bmatrix}, $$ since multiplying this matrix by \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\) gives \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\).
Apply the factors from right to left: $$ \begin{bmatrix}2&1\\1&1\end{bmatrix} \begin{bmatrix}1&1\\1&-1\end{bmatrix} = \begin{bmatrix}3&1\\2&0\end{bmatrix}, \qquad \begin{bmatrix}1&-1\\0&1\end{bmatrix} \begin{bmatrix}3&1\\2&0\end{bmatrix} = \begin{bmatrix}1&1\\2&0\end{bmatrix}. $$ Therefore the new matrix is \(\begin{bmatrix}1&1\\2&0\end{bmatrix}\).
Check: The first new input basis vector has old coordinates \(\begin{bmatrix}1\\1\end{bmatrix}\). The old matrix sends it to \(\begin{bmatrix}3\\2\end{bmatrix}\), which equals \(1(1,0)+2(1,1)\), so its \(\mathcal{C}'\)-coordinates are \(\begin{bmatrix}1\\2\end{bmatrix}\), the first column of the answer. The second input has old coordinates \(\begin{bmatrix}1\\-1\end{bmatrix}\); its output is \(\begin{bmatrix}1\\0\end{bmatrix}=1(1,0)+0(1,1)\), giving the second column \(\begin{bmatrix}1\\0\end{bmatrix}\). Both columns agree.
Worked Example: Changing Only the Domain Basis
Suppose \(V=W=\mathbb{R}^2\), both old bases are standard, and $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix}1&2\\0&1\end{bmatrix}. $$ Change the domain basis to \(\mathcal{B}'=((1,1),(0,1))\), while keeping the codomain basis standard.
The new input basis vectors are already in standard coordinates, so $$ \mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'} = \begin{bmatrix}1&0\\1&1\end{bmatrix}, \qquad \mathbf{P}_{\mathcal{C}\leftarrow\mathcal{C}}=\mathbf{I}_2 = \begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ The formula gives $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}'} = \begin{bmatrix}1&0\\0&1\end{bmatrix} \begin{bmatrix}1&2\\0&1\end{bmatrix} \begin{bmatrix}1&0\\1&1\end{bmatrix} = \begin{bmatrix}3&2\\1&1\end{bmatrix}. $$
Check: The first column is the standard-coordinate output \(T(1,1)=(1+2,1)=(3,1)\). The second is \(T(0,1)=(2,1)\). These are exactly the columns of the computed matrix, as required when the codomain basis stays standard.
Worked Example: Changing Only the Codomain Basis
Let both old bases be standard and suppose $$ [T]_{\mathcal{C}\leftarrow\mathcal{B}} = \begin{bmatrix}1&2\\2&1\end{bmatrix}. $$ Keep the domain basis, but use \(\mathcal{C}'=((1,1),(1,0))\) in the codomain.
The matrix whose columns are the new codomain basis vectors in standard coordinates is \(\begin{bmatrix}1&1\\1&0\end{bmatrix}\). Its inverse is \(\begin{bmatrix}0&1\\1&-1\end{bmatrix}\), as direct multiplication verifies. Thus $$ \mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}} = \begin{bmatrix}0&1\\1&-1\end{bmatrix}, \qquad \mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}}=\mathbf{I}_2. $$ The conversion is $$ [T]_{\mathcal{C}'\leftarrow\mathcal{B}} = \begin{bmatrix}0&1\\1&-1\end{bmatrix} \begin{bmatrix}1&2\\2&1\end{bmatrix} = \begin{bmatrix}2&1\\-1&1\end{bmatrix}. $$
Check: The old-coordinate output for the first standard input is \((1,2)\). In the new basis, \(2(1,1)-1(1,0)=(1,2)\), so its new coordinates are \((2,-1)\). For the second input the output is \((2,1)=(1,1)+(1,0)\), so its new coordinates are \((1,1)\). These are the two columns of the answer.
Common Mistake: Reversing an Output Conversion
A frequent error is to use the matrix whose columns are the new codomain basis vectors as though it converted old coordinates into new coordinates. In the first worked example, that basis matrix is \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\). It converts \(\mathcal{C}'\)-coordinates to standard coordinates, not standard coordinates to \(\mathcal{C}'\)-coordinates. Using it on the left would give the wrong result: $$ \begin{bmatrix}1&1\\0&1\end{bmatrix} \begin{bmatrix}3&1\\2&0\end{bmatrix} = \begin{bmatrix}5&1\\2&0\end{bmatrix}. $$ The first column claims that \(T(1,1)\) has \(\mathcal{C}'\)-coordinates \((5,2)\). But those coordinates reconstruct the standard vector \(5(1,0)+2(1,1)=(7,2)\), whereas the old matrix gives \(T(1,1)=(3,2)\). The claim fails when checked.
The correction is to use the inverse basis matrix for the output conversion: standard output coordinates must be converted to \(\mathcal{C}'\)-coordinates, so the factor is \(\mathbf{P}_{\mathcal{C}'\leftarrow\mathcal{C}}\). A reliable check is to read every factor by its arrow and verify that the output coordinates of one step match the input coordinates expected by the next.
Why the Formula Is Useful
The conversion formula separates three tasks that are easy to confuse: representing the input in the old domain basis, applying the map in its known coordinates, and representing the result in the new codomain basis. It also makes independent choices possible. A basis change in the domain affects the right side of the matrix product; a basis change in the codomain affects the left side. When both change, both factors are needed, and their positions follow from the order in which coordinate columns are acted on.
Check Your Understanding
For each question, keep track of the coordinate system expected at each stage.
- Write the matrix conversion formula when only the domain basis changes from \(\mathcal{B}\) to \(\mathcal{B}'\) and the codomain basis remains \(\mathcal{C}\).
- What does \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{B}'}\) do to a column of \(\mathcal{B}'\)-coordinates?
- If a basis matrix has columns equal to new basis vectors in standard coordinates, which coordinate conversion does that matrix perform?
- For the first worked example, verify the second output column by expressing \(T(1,-1)\) in the basis \(\mathcal{C}'\).
- Explain why the domain transition matrix multiplies on the right while the codomain transition matrix multiplies on the left.
