Why Basis Changes Compose
A vector in a finite-dimensional vector space does not change when we describe it using a different basis, but its coordinate column generally does. If a conversion takes coordinates from basis \(\mathcal{B}\) to basis \(\mathcal{C}\), and a second conversion takes them from \(\mathcal{C}\) to basis \(\mathcal{D}\), the combined conversion takes them from \(\mathcal{B}\) directly to \(\mathcal{D}\). The order of the matrices records the order of these actions: the conversion applied first is the rightmost factor.
Throughout, let \(V\) be a finite-dimensional vector space over \(\mathbb{R}\), of dimension \(n\). A basis is an ordered list \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) that represents each vector in \(V\) uniquely as a linear combination of its listed vectors. The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in that unique representation. Every transition matrix below is \(n\times n\), since each basis has \(n\) vectors.
In particular, the columns of \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) are the \(\mathcal{C}\)-coordinates of the vectors in the ordered list \(\mathcal{B}\). To see why, apply the defining conversion to each basis vector \(\mathbf{b}_j\). Its \(\mathcal{B}\)-coordinate column is the \(j\)-th standard coordinate column, so the product with the transition matrix selects its \(j\)-th column. This is useful for constructing a transition matrix, but composition can be established directly from the definition.
The Composition Rule
Proof. Let \(\mathbf{v}\in V\). By the definition of the transition matrix from \(\mathcal{B}\) to \(\mathcal{C}\), and then the definition of the transition matrix from \(\mathcal{C}\) to \(\mathcal{D}\), $$ [\mathbf{v}]_{\mathcal{D}} = \mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}} = \mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}} \mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}. $$ The direct conversion also gives \([\mathbf{v}]_{\mathcal{D}}=\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Thus the two matrices in the claimed equality give the same output for every column \([\mathbf{v}]_{\mathcal{B}}\). Every column in \(\mathbb{R}^n\) is the coordinate column of some vector in \(V\): given a column \(\mathbf{x}\), take the vector whose \(\mathcal{B}\)-coordinates are \(\mathbf{x}\). Therefore the two matrices act equally on every column of \(\mathbb{R}^n\). In particular, they agree on each standard coordinate column, which means their corresponding columns are equal. Hence the matrices are equal. \(\square\)
The factor on the right acts first. This is the same convention used when multiplying matrices by a coordinate column: the rightmost matrix converts \(\mathcal{B}\)-coordinates into \(\mathcal{C}\)-coordinates, after which the left matrix converts the result into \(\mathcal{D}\)-coordinates. Both factors are \(n\times n\), but matching dimensions alone does not determine the correct order.
Worked Example: Converting Through Three Bases in R²
Take \(V=\mathbb{R}^2\), let \(\mathcal{B}\) be the standard basis, and use the ordered bases \(\mathcal{C}=((1,1),(1,-1))\) and \(\mathcal{D}=((2,1),(1,1))\). Convert the column \([\mathbf{v}]_{\mathcal{B}}=\begin{bmatrix}3\\-1\end{bmatrix}\) to \(\mathcal{D}\)-coordinates by way of \(\mathcal{C}\).
The matrix whose columns are the vectors of \(\mathcal{C}\) is \(\begin{bmatrix}1&1\\1&-1\end{bmatrix}\). Since \(\mathcal{B}\) is standard, \([\mathbf{v}]_{\mathcal{B}}\) is the usual column of entries of \(\mathbf{v}\), and the conversion from \(\mathcal{B}\) to \(\mathcal{C}\) is \(\begin{bmatrix}1&1\\1&-1\end{bmatrix}^{-1}\). The inverse is \(\begin{bmatrix}\tfrac12&\tfrac12\\\tfrac12&-\tfrac12\end{bmatrix}\), as multiplying it by the basis matrix gives \(\mathbf{I}_2\). Thus
The basis matrix for \(\mathcal{D}\) is \(\begin{bmatrix}2&1\\1&1\end{bmatrix}\), whose inverse is \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\), since their product is \(\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Multiplying this inverse by the basis matrix for \(\mathcal{C}\) gives the conversion from \(\mathcal{C}\) to \(\mathcal{D}\):
Now apply this conversion to the intermediate coordinates. The product of the two transition matrices is
Check: the vector with \(\mathcal{B}\)-coordinates \(\begin{bmatrix}3\\-1\end{bmatrix}\) is \((3,-1)\). Reconstructing from the claimed \(\mathcal{D}\)-coordinates gives \(4(2,1)-5(1,1)=(8-5,4-5)=(3,-1)\), the original vector. The direct transition matrix in this example is \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\), and it also gives \(\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\begin{bmatrix}3\\-1\end{bmatrix}=\begin{bmatrix}4\\-5\end{bmatrix}\).
Composing a Longer Chain
The same reasoning applies when a conversion passes through more than one intermediate basis. For ordered bases \(\mathcal{B}_0,\mathcal{B}_1,\dots,\mathcal{B}_k\) of \(V\), each adjacent transition changes coordinates from one basis to the next. The whole chain is obtained by multiplying the adjacent transition matrices from left to right in the order of the destinations, with the first conversion at the far right.
Proof. For \(k=1\), the formula states that \(\mathbf{P}_{\mathcal{B}_1\leftarrow\mathcal{B}_0}\) is itself the conversion from \(\mathcal{B}_0\) to \(\mathcal{B}_1\), which is its definition. Suppose the formula holds for a chain ending at \(\mathcal{B}_{k-1}\). By the two-basis composition theorem, $$ \mathbf{P}_{\mathcal{B}_k\leftarrow\mathcal{B}_0} = \mathbf{P}_{\mathcal{B}_k\leftarrow\mathcal{B}_{k-1}} \mathbf{P}_{\mathcal{B}_{k-1}\leftarrow\mathcal{B}_0}. $$ Substitute the formula for the second factor from the induction hypothesis. The result is exactly the stated product, with the final conversion on the left. This proves the formula for every \(k\geq1\) by induction. \(\square\)
Grouping adjacent factors does not change the result because matrix multiplication is associative. For compatible matrices, the \((i,j)\)-entry of either \((\mathbf{A}\mathbf{B})\mathbf{C}\) or \(\mathbf{A}(\mathbf{B}\mathbf{C})\) is \(\sum_r\sum_s a_{ir}b_{rs}c_{sj}\); the finite sums can be grouped in either order. Associativity allows a long product to be evaluated in convenient stages, but it does not permit rearranging the factors.
Worked Example: Four Bases in a Conversion Chain
Use the bases \(\mathcal{B},\mathcal{C},\mathcal{D}\) from the preceding example and add \(\mathcal{E}=((1,0),(1,1))\). Find \(\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{B}}\) by converting through both \(\mathcal{C}\) and \(\mathcal{D}\).
The basis matrix for \(\mathcal{E}\) is \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\), with inverse \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). Multiplying this inverse by the basis matrix for \(\mathcal{D}\) gives \(\mathbf{P}_{\mathcal{E}\leftarrow\mathcal{D}}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\). The previous example gave \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}=\begin{bmatrix}0&2\\1&-3\end{bmatrix}\). Their product is
Multiplying by the first transition matrix completes the chain:
For the same vector with \(\mathcal{B}\)-coordinates \(\begin{bmatrix}3\\-1\end{bmatrix}\), the intermediate columns are \(\begin{bmatrix}1\\2\end{bmatrix}\) in \(\mathcal{C}\), then \(\begin{bmatrix}4\\-5\end{bmatrix}\) in \(\mathcal{D}\), and finally \(\begin{bmatrix}1&0\\1&1\end{bmatrix}\begin{bmatrix}4\\-5\end{bmatrix}=\begin{bmatrix}4\\-1\end{bmatrix}\) in \(\mathcal{E}\).
Check: the \(\mathcal{E}\)-coordinates reconstruct \(4(1,0)-1(1,1)=(3,-1)\), as required. Also, the direct matrix \(\begin{bmatrix}1&-1\\0&1\end{bmatrix}\) applied to \(\begin{bmatrix}3\\-1\end{bmatrix}\) gives \(\begin{bmatrix}4\\-1\end{bmatrix}\).
Common Mistake: Reversing the Factors
A frequent error is to write \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}\) for a conversion from \(\mathcal{B}\) through \(\mathcal{C}\) to \(\mathcal{D}\). This reverses the action order. In the first example, that incorrect product would be
Applying this wrong product to \(\begin{bmatrix}3\\-1\end{bmatrix}\) gives \(\begin{bmatrix}2\\-4\end{bmatrix}\), not the correct \(\mathcal{D}\)-coordinates \(\begin{bmatrix}4\\-5\end{bmatrix}\). The first matrix expects \(\mathcal{C}\)-coordinates as input, but the reversed product applies it to the original \(\mathcal{B}\)-coordinates. A reliable check is to read each factor from right to left and verify that its input basis matches the coordinates currently held.
Composition is useful whenever a vector must be expressed in a basis reached by several intermediate coordinate systems. The intermediate steps may be convenient to compute or already available; the theorem guarantees that their product performs the same conversion as a direct transition. To keep the notation unambiguous, write every factor with its source and destination, then check that each destination matches the next source. Finally, test the result on a vector by reconstructing it from the claimed target coordinates.
Check Your Understanding
For each question, write the conversion direction explicitly before multiplying any matrices.
- If coordinates change from \(\mathcal{B}\) to \(\mathcal{C}\) and then to \(\mathcal{D}\), which transition matrix is the rightmost factor?
- Suppose \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) and \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}=\begin{bmatrix}2&0\\1&1\end{bmatrix}\). Compute the matrix that converts \(\mathcal{B}\)-coordinates directly to \(\mathcal{D}\)-coordinates.
- For the bases in the first worked example, convert \(\begin{bmatrix}0\\2\end{bmatrix}\) from \(\mathcal{B}\)-coordinates to \(\mathcal{D}\)-coordinates through \(\mathcal{C}\), and verify by reconstructing the vector.
- Explain why the theorem requires the bases in the chain to be bases of the same vector space.
- For a chain of four bases, write the product that converts coordinates from the first basis to the fourth.
