Why Transition Matrix Columns Are Basis Coordinates

What a Transition-Matrix Column Records

A transition matrix packages a change in coordinates into a single matrix multiplication. Its columns are not arbitrary entries assembled from the two bases: each one records how a particular vector in the source basis is written using the target basis. The key is to apply the coordinate conversion to the source basis vectors one at a time. For the \(j\)th source vector, its source-coordinate vector is especially simple: it is the \(j\)th standard unit vector.

Work over \(\mathbb{R}\), and let \(V\) be a finite-dimensional vector space of dimension \(n\). Let \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) and \(\mathcal{C}=(\mathbf{c}_1,\dots,\mathbf{c}_n)\) be ordered bases of \(V\). For \(\mathbf{v}\in V\), the coordinate vector \([\mathbf{v}]_{\mathcal{B}}\in\mathbb{R}^n\) is the column of coefficients in the unique expression of \(\mathbf{v}\) as a linear combination of the vectors in \(\mathcal{B}\). The definition of \([\mathbf{v}]_{\mathcal{C}}\) is the same, using \(\mathcal{C}\).

For \(j\in\{1,\dots,n\}\), let \(\mathbf{e}_j\in\mathbb{R}^n\) denote the column with a \(1\) in position \(j\) and zeroes elsewhere. Since \(\mathbf{b}_j\) has coefficient \(1\) on itself and coefficient \(0\) on every other vector in \(\mathcal{B}\), its coordinate vector in that basis is \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\). Multiplying any matrix by \(\mathbf{e}_j\) selects its \(j\)th column. These two facts give the column formula.

Proof. Fix an index \(j\) with \(1\le j\le n\). By the definition of the transition matrix, applying it to the \(\mathcal{B}\)-coordinate vector of \(\mathbf{b}_j\) gives the \(\mathcal{C}\)-coordinate vector of that same vector: \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{b}_j]_{\mathcal{B}}=[\mathbf{b}_j]_{\mathcal{C}}\). The source coordinates are \([\mathbf{b}_j]_{\mathcal{B}}=\mathbf{e}_j\). For any matrix \(\mathbf{P}\), the product \(\mathbf{P}\mathbf{e}_j\) is its \(j\)th column, because the \(j\)th column is multiplied by \(1\) and every other column by \(0\). Therefore the \(j\)th column of \(\mathbf{P}_{\mathcal{C}\leftarrow\mathcal{B}}\) is \([\mathbf{b}_j]_{\mathcal{C}}\). This holds for every \(j\), proving the displayed matrix formula. \(\square\)

This argument is also a practical construction rule: find the target-basis coordinates of each source-basis vector, then put those coordinate columns in the same order as the source basis. The order matters. Column \(j\) belongs to \(\mathbf{b}_j\), not to \(\mathbf{c}_j\) unless the bases happen to align in a special way.

Why the Columns Give the Conversion for Every Vector

The column formula explains more than what happens to the basis vectors. Every \(\mathbf{v}\in V\) has a unique source-basis expression \(\mathbf{v}=x_1\mathbf{b}_1+\cdots+x_n\mathbf{b}_n\), where \(x_j\in\mathbb{R}\). Its source-coordinate vector is \((x_1,\dots,x_n)^{\top}\). Because the target-coordinate map is linear, the target coordinates of this sum are the same linear combination of the target-coordinate vectors of the \(\mathbf{b}_j\). The matrix product by the column matrix performs exactly that combination.

Proof. Write \([\mathbf{v}]_{\mathcal{B}}=(x_1,\dots,x_n)^{\top}\). By the definition of coordinates, \(\mathbf{v}=\sum_{j=1}^{n}x_j\mathbf{b}_j\). By the definition of the matrix whose columns are \([\mathbf{b}_j]_{\mathcal{C}}\), its product with this coordinate vector is \(\sum_{j=1}^{n}x_j[\mathbf{b}_j]_{\mathcal{C}}\). Each vector \(\mathbf{b}_j\) is the linear combination of the \(\mathcal{C}\)-basis vectors specified by \([\mathbf{b}_j]_{\mathcal{C}}\). Substituting those expressions into \(\sum_{j=1}^{n}x_j\mathbf{b}_j\) shows that its \(\mathcal{C}\)-coordinates are precisely \(\sum_{j=1}^{n}x_j[\mathbf{b}_j]_{\mathcal{C}}\). Thus the matrix product equals \([\mathbf{v}]_{\mathcal{C}}\), as required. \(\square\)

In this proposition, the columns do the work of carrying each source-basis vector into target coordinates; the entries of \([\mathbf{v}]_{\mathcal{B}}\) specify how much of each column to use. We can see both steps explicitly in examples.

Computing the Columns in \(\mathbb{R}^2\)

Computing the Columns in \(\mathbb{R}^3\)

The Same Column Rule for Polynomial Bases

The argument does not depend on vectors being lists of real numbers. A polynomial space has vectors too: its elements are polynomials, and coordinates still mean coefficients in an ordered basis. The unit-vector reasoning is unchanged, so the same column formula applies.

A Common Mistake: Putting the Coordinates in Rows

The column rule is easy to reverse when writing a matrix by inspection. A common incorrect calculation is to write each \([\mathbf{b}_j]_{\mathcal{C}}\) as a row. In the polynomial example, that would give the transpose of the correct matrix: \(\begin{bmatrix}1&0&0\\-1&1&0\\0&-1&1\end{bmatrix}\). But applying this incorrect matrix to \([1]_{\mathcal{B}}=(1,0,0)^{\top}\) gives \((1,-1,0)^{\top}\), which is not \([1]_{\mathcal{C}}=(1,0,0)^{\top}\). The correction is to place the coordinate vectors as columns. Then multiplication by \(\mathbf{e}_j\) selects exactly the target coordinates of \(\mathbf{b}_j\).

A reliable check is to multiply the proposed transition matrix by each standard unit vector. The result in position \(j\) must be \([\mathbf{b}_j]_{\mathcal{C}}\). A second check is to use each resulting coordinate column to reconstruct \(\mathbf{b}_j\) from \(\mathcal{C}\). These checks test both the placement of entries and the meaning of the coordinates.