Transition Matrices and Coordinate Direction

Coordinates Depend on the Basis

A vector does not change when we describe it using a different basis, but its coordinate column generally does. A transition matrix records how to translate between these descriptions. The key to using one correctly is to keep track of its direction: the columns of the matrix from a new basis to an old basis are the new basis vectors written in old coordinates.

Let \(V\) be a finite-dimensional vector space over \(\mathbb{R}\), and let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_n)\) be ordered bases of \(V\). The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in the unique expression \(\mathbf{v}=a_1\mathbf{b}_1+\cdots+a_n\mathbf{b}_n\). The ordering matters: changing the order of a basis changes the coordinate columns and the transition matrix.

For example, \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\) takes a coordinate column with subscript \(\mathcal{C}\) as input and produces one with subscript \(\mathcal{B}\). The reverse conversion is \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\). The subscripts are a useful safeguard against treating a coordinate vector as though it were the vector itself.

Building and Using a Transition Matrix

Proof. Write \([\mathbf{v}]_{\mathcal{C}}=(d_1,\dots,d_n)^{\top}\). By the definition of coordinates, \(\mathbf{v}=\sum_{j=1}^{n}d_j\mathbf{c}_j\). Write each basis vector \(\mathbf{c}_j\) in the basis \(\mathcal{B}\); its coefficient column is \([\mathbf{c}_j]_{\mathcal{B}}\), the \(j\)th column of \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\). Therefore the \(\mathcal{B}\)-coordinate column of \(\sum_{j=1}^{n}d_j\mathbf{c}_j\) is \(\sum_{j=1}^{n}d_j[\mathbf{c}_j]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\), proving the conversion formula. Applying the same formula in the reverse direction gives \([\mathbf{v}]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\). Substitution of the first formula into the second yields \([\mathbf{v}]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\). This holds for every coordinate column, including each standard coordinate column, so the product is \(\mathbf{I}_n\). Reversing the roles of the bases gives the product in the opposite order as \(\mathbf{I}_n\) as well. Thus the two transition matrices are inverses. \(\square\)

The inverse relationship is not an extra computational trick: it follows because both coordinate systems describe the same vector, and each basis gives a unique coordinate column. In a calculation, however, it is often faster to solve a small system than to compute an inverse explicitly.

Changing Coordinates Through an Intermediate Basis

Suppose there are three ordered bases \(\mathcal{B},\mathcal{C},\mathcal{D}\). To convert from \(\mathcal{D}\)-coordinates to \(\mathcal{B}\)-coordinates, first convert from \(\mathcal{D}\) to \(\mathcal{C}\), then from \(\mathcal{C}\) to \(\mathcal{B}\). Matrix multiplication acts right to left, so the matrix for the first conversion appears on the right.

Proof. Let \(\mathbf{v}\) be any vector in the space. The transition formula gives \([\mathbf{v}]_{\mathcal{C}}=\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{D}}[\mathbf{v}]_{\mathcal{D}}\), and then \([\mathbf{v}]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}[\mathbf{v}]_{\mathcal{C}}\). Substituting the first equality into the second gives \([\mathbf{v}]_{\mathcal{B}}=\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{D}}[\mathbf{v}]_{\mathcal{D}}\). This holds for every \(\mathbf{v}\), and in particular for vectors whose \(\mathcal{D}\)-coordinate columns are the standard coordinate columns. The columns of the product are therefore exactly the \(\mathcal{B}\)-coordinates of the \(\mathcal{D}\) basis vectors, which define \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{D}}\). \(\square\)

A Basis Change in a Polynomial Space

Common Direction and Multiplication Errors

Using the forward matrix for the reverse conversion is not valid. In the first example, \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\) converts \(\mathcal{C}\)-coordinates to \(\mathcal{B}\)-coordinates. If \([\mathbf{v}]_{\mathcal{B}}=(1,2,3)^{\top}\), applying that same matrix again gives \[ \begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix} \begin{bmatrix}1\\2\\3\end{bmatrix} = \begin{bmatrix}3\\5\\3\end{bmatrix}, \] which is not the desired \(\mathcal{C}\)-coordinate column. The correction is to use the inverse transition matrix: \[ \begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix} \begin{bmatrix}1\\2\\3\end{bmatrix} = \begin{bmatrix}2\\-1\\3\end{bmatrix}. \] Multiplying this result by \(\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\) recovers \((1,2,3)^{\top}\).

Do not reverse the order in a composition. In the second example, the input has \(\mathcal{D}\)-coordinates. The rightmost matrix must therefore be \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{D}}\), because that is the first conversion performed. Writing \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{D}}\mathbf{P}_{\mathcal{B}\leftarrow\mathcal{C}}\) would try to apply a matrix expecting \(\mathcal{C}\)-coordinates directly to a \(\mathcal{D}\)-coordinate column. Checking the subscripts at each multiplication catches this mismatch before any arithmetic is done.

When constructing a transition matrix, write the destination and source bases first, then place the source basis vectors’ coordinates in the destination basis as columns. When converting back, use the reverse transition matrix—or the inverse—and confirm the answer by converting it back or reconstructing the vector.