The Inverse of a Change-of-Basis Matrix

Why Reversing a Basis Change Gives an Inverse

A change-of-basis matrix changes the coordinates used to describe a vector; it does not change the vector itself. If one matrix takes coordinates in a basis \(\mathcal{B}\) and returns coordinates in a basis \(\mathcal{C}\), then reversing the direction must recover the original coordinates. The precise statement is that the two conversion matrices are inverses. The direction of each matrix matters: its input coordinates are named on the right of the arrow, and its output coordinates on the left.

Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\), 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 \(V\). A basis is an ordered list such that every vector in \(V\) has a unique expression as a linear combination of its listed vectors. The coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of coefficients in the expression of \(\mathbf{v}\) using \(\mathcal{B}\). The same definition applies to \([\mathbf{v}]_{\mathcal{C}}\).

In particular, \(P_{\mathcal{B}\leftarrow\mathcal{C}}\) has the opposite direction: it takes \(\mathcal{C}\)-coordinates as input and returns \(\mathcal{B}\)-coordinates. Both matrices are \(n\times n\), since both bases have \(n\) vectors. The arrow notation records which coordinate system is the input and which is the output; it is useful to read it aloud before multiplying a matrix by a coordinate vector.

The Inverse Theorem

Proof. Write \(P=P_{\mathcal{C}\leftarrow\mathcal{B}}\) and \(Q=P_{\mathcal{B}\leftarrow\mathcal{C}}\). Take any column \(\mathbf{x}\in\mathbb{F}^n\). Since \(\mathcal{B}\) is a basis, there is a vector \(\mathbf{v}\in V\) with \([\mathbf{v}]_{\mathcal{B}}=\mathbf{x}\): explicitly, \(\mathbf{v}=x_1\mathbf{b}_1+\cdots+x_n\mathbf{b}_n\). By the definition of \(P\), \(P\mathbf{x}=[\mathbf{v}]_{\mathcal{C}}\). Applying the reverse conversion then gives \(QP\mathbf{x}=Q[\mathbf{v}]_{\mathcal{C}}=[\mathbf{v}]_{\mathcal{B}}=\mathbf{x}\). This holds for every \(\mathbf{x}\), so \(QP=I_n\): the two matrices agree on each standard coordinate vector, and these are the columns of the identity matrix.

For the other product, take any \(\mathbf{y}\in\mathbb{F}^n\). Because \(\mathcal{C}\) is a basis, it is the coordinate vector \([\mathbf{w}]_{\mathcal{C}}\) of \(\mathbf{w}=y_1\mathbf{c}_1+\cdots+y_n\mathbf{c}_n\). The reverse conversion gives \(Q\mathbf{y}=[\mathbf{w}]_{\mathcal{B}}\), and converting back gives \(PQ\mathbf{y}=P[\mathbf{w}]_{\mathcal{B}}=[\mathbf{w}]_{\mathcal{C}}=\mathbf{y}\). Thus \(PQ=I_n\) as well. These two identities show that each conversion undoes the other, so \(Q=P^{-1}\). \(\square\)

The two products are both worth checking: the first tests whether a \(\mathcal{B}\)-coordinate column returns to itself after conversion to \(\mathcal{C}\) and back, while the second tests the reverse starting point. The proof uses only the definition of the coordinate conversions and the fact that coordinates in a basis are unique.

Proof. Suppose \(Q_1\) and \(Q_2\) both perform this conversion. For any \(\mathbf{y}\in\mathbb{F}^n\), choose \(\mathbf{v}\in V\) with \([\mathbf{v}]_{\mathcal{C}}=\mathbf{y}\), which is possible because \(\mathcal{C}\) is a basis. Both matrices must return \([\mathbf{v}]_{\mathcal{B}}\), so \(Q_1\mathbf{y}=Q_2\mathbf{y}\). This holds for every \(\mathbf{y}\), and therefore the matrices have the same columns and are equal. \(\square\)

Computing Both Directions

Checking a Conversion on a Vector

A Second Pair of Bases

Common Pitfall: Transpose Is Not the Reverse Conversion

A common error is to assume that reversing a basis change means transposing its matrix. In the first example, transposing the forward matrix gives the incorrect candidate \(\begin{bmatrix}0&1\\2&-1\end{bmatrix}\), whereas the actual reverse matrix is \(\begin{bmatrix}\tfrac12&1\\\tfrac12&0\end{bmatrix}\). The inverse theorem requires the reverse conversion to multiply the forward conversion to give the identity; taking a transpose does not guarantee that property.

Test the incorrect candidate using the correct input direction. The vector \((4,2)^{\top}\) has \(\mathcal{C}\)-coordinates \((2,2)^{\top}\), so those—not its \(\mathcal{B}\)-coordinates—must be supplied to a proposed reverse matrix. The mistaken matrix produces

but the correct \(\mathcal{B}\)-coordinates are \((3,1)^{\top}\). Reconstructing the mistaken output in \(\mathcal{B}\) gives \(2(1,1)^{\top}+2(1,-1)^{\top}=(4,0)^{\top}\), not \((4,2)^{\top}\). This check exposes both the incorrect matrix and the importance of using the coordinate system named by the input side of the arrow.

For any pair of bases, a reliable procedure is to find each column by expressing the input basis vector in the output basis, then check both matrix products and, when converting a particular vector, reconstruct it from the output coordinates. The matrix inverse is not a change in the vector: it is the operation that reverses the coordinate description.