Following Coordinates Through Several Bases
A vector does not change when we describe it in a different basis; only its coordinate column changes. With several bases involved, the main challenge is keeping track of which coordinate system each column belongs to. The arrow in a transition-matrix subscript records that direction: the basis on the right is the starting coordinate system, and the basis on the left is the destination. Reading that direction consistently lets us move through multiple bases and then apply a linear map in the coordinates where its matrix is given.
Throughout, the vector space is \(\mathbb{R}^2\). An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\mathbf{b}_2)\) is an ordered list of vectors such that each vector in the space has a unique expression as a linear combination of \(\mathbf{b}_1,\mathbf{b}_2\). The coordinate column \([\mathbf{x}]_{\mathcal{B}}\) contains the coefficients in that expression: if \(\mathbf{x}=a_1\mathbf{b}_1+a_2\mathbf{b}_2\), then \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}a_1\\a_2\end{bmatrix}\). The order matters because the first coordinate multiplies the first basis vector and the second coordinate multiplies the second.
To see why the columns have this meaning, write \(\mathbf{x}=a_1\mathbf{b}_1+a_2\mathbf{b}_2\). The coordinates of \(\mathbf{x}\) in \(\mathcal{C}\) are \(a_1[\mathbf{b}_1]_{\mathcal{C}}+a_2[\mathbf{b}_2]_{\mathcal{C}}\). Placing those two coordinate columns into a matrix gives exactly the transition rule in the definition. For a sequence \(\mathcal{B}\to\mathcal{C}\to\mathcal{D}\), the matrices therefore act from right to left: first multiply by \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\), then by \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}\).
Why Transition Matrices Compose
Proof. Take any vector \(\mathbf{x}\) in the space. By the definition of transition matrices, \([\mathbf{x}]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\) and \([\mathbf{x}]_{\mathcal{D}}=\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}[\mathbf{x}]_{\mathcal{C}}\). Substitute the first equality into the second: \([\mathbf{x}]_{\mathcal{D}}=\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). The direct conversion also gives \([\mathbf{x}]_{\mathcal{D}}=\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\). These two matrix products act identically on every \(\mathcal{B}\)-coordinate column. In particular, they act identically on each standard coordinate column, so their corresponding columns agree and the matrices are equal. \(\square\)
The order in the formula is not optional. A coordinate column is multiplied on the left, and the matrix nearest the column performs the first conversion. This composition rule works for any number of successive bases: include one transition matrix for each step, with the first step at the far right.
A Three-Basis Conversion Followed by a Map
Worked Example: Convert Through Three Bases and Evaluate a Map
In \(\mathbb{R}^2\), let \(\mathcal{B}=((1,1),(1,-1))\), \(\mathcal{C}=((1,0),(1,1))\), and \(\mathcal{D}=((0,1),(1,1))\). Suppose \([\mathbf{x}]_{\mathcal{B}}=\begin{bmatrix}2\\-1\end{bmatrix}\). Convert first to \(\mathcal{C}\), then to \(\mathcal{D}\). Finally, evaluate a linear map \(T\) whose matrix in basis \(\mathcal{D}\) is \(\begin{bmatrix}1&2\\-1&1\end{bmatrix}\).
The columns of \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) are the \(\mathcal{C}\)-coordinates of the vectors in \(\mathcal{B}\). For \((1,1)\), the coordinates are \(\begin{bmatrix}0\\1\end{bmatrix}\), since \((1,1)=0(1,0)+1(1,1)\). For \((1,-1)\), solve \(a(1,0)+b(1,1)=(1,-1)\). The second component gives \(b=-1\), and the first gives \(a=2\). Thus
Next, express the vectors in \(\mathcal{C}\) in basis \(\mathcal{D}\). For \(c_1=(1,0)\), the equation \(a(0,1)+b(1,1)=(1,0)\) gives \(b=1\) and \(a=-1\). For \(c_2=(1,1)\), the coordinates are \(\begin{bmatrix}0\\1\end{bmatrix}\). Therefore
Composing the two conversions gives the direct transition matrix and the same final coordinates:
Now use the matrix of \(T\) in \(\mathcal{D}\), because the input column is currently in \(\mathcal{D}\)-coordinates:
Check: the original vector is \(2(1,1)-(1,-1)=(1,3)\). The output coordinates mean \(T(\mathbf{x})=4(0,1)-(1,1)=(-1,3)\). Converting \((-1,3)\) back into basis \(\mathcal{D}\) gives second coordinate \(-1\) from its first component, and first coordinate \(4\) from \(4+(-1)=3\). This recovers \(\begin{bmatrix}4\\-1\end{bmatrix}\), as required.
Using a Matrix to Evaluate a Linear Map
A linear map \(T:V\to W\) preserves sums and scalar multiples: \(T(a\mathbf{u}+b\mathbf{v})=aT(\mathbf{u})+bT(\mathbf{v})\). If \(\mathcal{B}=(\mathbf{b}_1,\ldots,\mathbf{b}_n)\) is a basis of \(V\) and \(\mathcal{C}\) is a basis of \(W\), the matrix \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\) has columns \([T(\mathbf{b}_j)]_{\mathcal{C}}\). Its purpose is to take input coordinates in \(\mathcal{B}\) to output coordinates in \(\mathcal{C}\).
Proof. Write \(\mathbf{x}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\), so \([\mathbf{x}]_{\mathcal{B}}=(a_1,\ldots,a_n)^\top\). Linearity gives \(T(\mathbf{x})=a_1T(\mathbf{b}_1)+\cdots+a_nT(\mathbf{b}_n)\). Taking \(\mathcal{C}\)-coordinates preserves this linear combination, so \([T(\mathbf{x})]_{\mathcal{C}}=a_1[T(\mathbf{b}_1)]_{\mathcal{C}}+\cdots+a_n[T(\mathbf{b}_n)]_{\mathcal{C}}\). The coordinate columns on the right are precisely the columns of \([T]_{\mathcal{C}\leftarrow\mathcal{B}}\). Multiplying that matrix by \((a_1,\ldots,a_n)^\top\) forms the same linear combination, proving the rule. \(\square\)
This rule explains the order of operations in a multi-step problem. First convert the input to the basis used for the map's matrix. Then multiply by that matrix. If the question asks for the output in another basis, convert the resulting output coordinates afterward. A map's matrix acts on coordinates; it is not itself the map's output vector.
Worked Example: Convert Backward and Check the Vector
Use the bases from the preceding example. Start with \([\mathbf{y}]_{\mathcal{D}}=\begin{bmatrix}3\\-2\end{bmatrix}\), and find \([\mathbf{y}]_{\mathcal{B}}\) by converting backward through \(\mathcal{C}\).
The inverse of \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}\) is \(\begin{bmatrix}-1&0\\1&1\end{bmatrix}\) itself, as multiplying it by itself gives the identity. Also, \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}^{-1}=\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\); multiplication in either order with \(\begin{bmatrix}0&2\\1&-1\end{bmatrix}\) gives \(\mathbf{I}_2\). Apply the inverse transitions in reverse order:
Check: reconstructing from \(\mathcal{B}\) gives \(-\tfrac12(1,1)-\tfrac32(1,-1)=(-2,1)\). Reconstructing from the original \(\mathcal{D}\)-coordinates gives \(3(0,1)-2(1,1)=(-2,1)\) as well. The two coordinate columns describe the same vector.
A Second Map Check in Three Bases
Worked Example: Compare Coordinate Evaluation with the Vector Formula
Define \(S:\mathbb{R}^2\to\mathbb{R}^2\) by \(S(x,y)=(2x+y,x-y)\). For \([\mathbf{z}]_{\mathcal{B}}=\begin{bmatrix}1\\2\end{bmatrix}\), convert to \(\mathcal{D}\)-coordinates, evaluate \(S\) using its matrix in \(\mathcal{D}\), and check using the formula for \(S\).
The direct transition matrix found above gives \([\mathbf{z}]_{\mathcal{D}}=\begin{bmatrix}0&-2\\1&1\end{bmatrix}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}-4\\3\end{bmatrix}\). To find the matrix of \(S\) in \(\mathcal{D}\), its first column is \([S(0,1)]_{\mathcal{D}}\). Since \(S(0,1)=(1,-1)=-2(0,1)+(1,1)\), this column is \(\begin{bmatrix}-2\\1\end{bmatrix}\). Its second column is \([S(1,1)]_{\mathcal{D}}\). Since \(S(1,1)=(3,0)=-3(0,1)+3(1,1)\), this column is \(\begin{bmatrix}-3\\3\end{bmatrix}\). Thus
Check: the input vector is \((1,1)+2(1,-1)=(3,-1)\). Applying the formula directly gives \(S(3,-1)=(5,4)\). In basis \(\mathcal{D}\), \(-1(0,1)+5(1,1)=(5,4)\), so its coordinates are indeed \(\begin{bmatrix}-1\\5\end{bmatrix}\).
Common Mistake: Reversing the Conversion
A frequent error is to apply a transition matrix whose direction does not match the coordinate column. In the first example, \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{C}}\) takes \(\mathcal{C}\)-coordinates to \(\mathcal{D}\)-coordinates. Applying it directly to \([\mathbf{x}]_{\mathcal{B}}\) would be invalid: the entries \(2,-1\) are coefficients relative to \(\mathcal{B}\), not \(\mathcal{C}\). Even when dimensions happen to match, an incorrect multiplication can produce a plausible-looking column without representing the intended conversion.
For example, the incorrect calculation \(\begin{bmatrix}-1&0\\1&1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}-2\\1\end{bmatrix}\) treats \(\mathcal{B}\)-coordinates as if they were \(\mathcal{C}\)-coordinates. The correction is to perform the first conversion before the second: \(\begin{bmatrix}-1&0\\1&1\end{bmatrix}\begin{bmatrix}0&2\\1&-1\end{bmatrix}\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}2\\1\end{bmatrix}\). The corrected column reconstructs the original vector \((1,3)\) in basis \(\mathcal{D}\), since \(2(0,1)+1(1,1)=(1,3)\). A reconstruction check is often the quickest way to catch a direction error.
Before multiplying, label the coordinate system of every column and the input and output basis of every matrix. For a chain of conversions, adjacent labels must match, and the rightmost matrix acts first. For a linear map, make one more check: the input coordinates must be in the basis used for the domain of the map matrix. Keeping these labels visible turns a long calculation into a sequence of individually checkable steps.
Check Your Understanding
For each question, keep track of the basis attached to every coordinate column and verify by reconstructing the vector where possible.
- In the first worked example, compute \(\mathbf{P}_{\mathcal{D}\leftarrow\mathcal{B}}[\mathbf{x}]_{\mathcal{B}}\) directly and compare with the two-step conversion.
- If \([\mathbf{v}]_{\mathcal{C}}\) is known, which transition matrix converts it to \(\mathcal{D}\)-coordinates?
- Explain why \([T]_{\mathcal{D}\leftarrow\mathcal{D}}\) cannot be applied directly to a column of \(\mathcal{B}\)-coordinates.
- For the backward-conversion example, reconstruct the vector using the intermediate \(\mathcal{C}\)-coordinates as well as the final \(\mathcal{B}\)-coordinates.
- State the order of the two transition matrices in a conversion from \(\mathcal{B}\) through \(\mathcal{C}\) to \(\mathcal{D}\), and explain which matrix acts first.
