Orthogonal Changes of Coordinates in Orthonormal Bases

One Vector, Different Coordinate Columns

A vector does not change when we describe it in a different basis, but its coordinate column usually does. For orthonormal bases, the conversion between these coordinate columns is especially structured: it is multiplication by an orthogonal matrix. The direction of that multiplication matters. We will use \(Q_{\mathcal{C}\leftarrow\mathcal{B}}\) for the matrix that takes coordinates in \(\mathcal{B}\) to coordinates in \(\mathcal{C}\).

Throughout, the vector space is \(\mathbb{R}^n\) with the dot product \(\mathbf{u}\cdot\mathbf{v}=\mathbf{u}^{\top}\mathbf{v}\). An ordered basis \(\mathcal{B}=(\mathbf{b}_1,\dots,\mathbf{b}_n)\) is orthonormal if \(\mathbf{b}_i\cdot\mathbf{b}_j=1\) when \(i=j\), and \(\mathbf{b}_i\cdot\mathbf{b}_j=0\) when \(i\ne j\). For any basis \(\mathcal{B}\), the coordinate vector \([\mathbf{v}]_{\mathcal{B}}\) is the column of scalars satisfying \(\mathbf{v}=c_1\mathbf{b}_1+\cdots+c_n\mathbf{b}_n\), where \(c_i\) is its \(i\)th entry. The basis property guarantees that this column exists and is unique.

To see why the columns have this form, let \(\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}\). Then \(\mathbf{v}=\sum_{j=1}^{n}x_j\mathbf{b}_j\). Expressing each \(\mathbf{b}_j\) in \(\mathcal{C}\) and collecting the coefficients gives the \(\mathcal{C}\)-coordinate column as the same linear combination of the columns \([\mathbf{b}_j]_{\mathcal{C}}\). This fixes the matrix direction: the columns are source basis vectors expressed in the target basis.

Why the Change Matrix Is Orthogonal

A square real matrix \(Q\) is orthogonal if \(Q^{\top}Q=I_n\). Equivalently, its columns form an orthonormal basis of \(\mathbb{R}^n\). The next theorem explains why orthonormal coordinate systems lead exactly to such matrices.

Proof. Write \(Q=Q_{\mathcal{C}\leftarrow\mathcal{B}}\). Its \(j\)th column is \([\mathbf{b}_j]_{\mathcal{C}}\). For any vectors \(\mathbf{u}\) and \(\mathbf{w}\), their coordinates \(\mathbf{a}=[\mathbf{u}]_{\mathcal{C}}\) and \(\mathbf{d}=[\mathbf{w}]_{\mathcal{C}}\) satisfy \(\mathbf{u}=\sum_i a_i\mathbf{c}_i\) and \(\mathbf{w}=\sum_k d_k\mathbf{c}_k\). Therefore \(\mathbf{u}\cdot\mathbf{w}=\sum_i\sum_k a_i d_k(\mathbf{c}_i\cdot\mathbf{c}_k)=\sum_i a_i d_i=\mathbf{a}^{\top}\mathbf{d}\), since the vectors in \(\mathcal{C}\) are orthonormal. Applying this to \(\mathbf{b}_i,\mathbf{b}_j\), the dot product of columns \(i\) and \(j\) of \(Q\) is \(\mathbf{b}_i\cdot\mathbf{b}_j\). That equals \(1\) if \(i=j\) and \(0\) otherwise, because \(\mathcal{B}\) is orthonormal. Thus \(Q^{\top}Q=I_n\), so \(Q\) is orthogonal. Finally, the definition of its columns gives the stated coordinate conversion: if \(\mathbf{v}=\sum_j x_j\mathbf{b}_j\), then its \(\mathcal{C}\)-coordinates are \(\sum_j x_j[\mathbf{b}_j]_{\mathcal{C}}=Q\mathbf{x}\), where \(\mathbf{x}=[\mathbf{v}]_{\mathcal{B}}\). \(\square\)

The transpose reverses the conversion. This follows from the orthogonality just proved and the standard fact that an orthogonal matrix has inverse equal to its transpose. In coordinate language, this means that once the \(\mathcal{C}\)-coordinates are known, multiplying by \(Q^{\top}\) recovers the \(\mathcal{B}\)-coordinates.

Proof. The theorem gives \([\mathbf{v}]_{\mathcal{C}}=Q_{\mathcal{C}\leftarrow\mathcal{B}}[\mathbf{v}]_{\mathcal{B}}\), and \(Q_{\mathcal{C}\leftarrow\mathcal{B}}^{\top}Q_{\mathcal{C}\leftarrow\mathcal{B}}=I_n\). Multiply the coordinate equation on the left by \(Q_{\mathcal{C}\leftarrow\mathcal{B}}^{\top}\). The result is \([\mathbf{v}]_{\mathcal{B}}=Q_{\mathcal{C}\leftarrow\mathcal{B}}^{\top}[\mathbf{v}]_{\mathcal{C}}\). Since this holds for every coordinate column in \(\mathcal{C}\), the matrix that converts from \(\mathcal{C}\) to \(\mathcal{B}\) is the transpose. \(\square\)

Calculating a Change of Coordinates

For an orthonormal basis \(\mathcal{C}\), there is a useful direct formula: the \(i\)th coordinate of \(\mathbf{v}\) is \(\mathbf{v}\cdot\mathbf{c}_i\). Indeed, if \(\mathbf{v}=\sum_j d_j\mathbf{c}_j\), then \(\mathbf{v}\cdot\mathbf{c}_i=\sum_j d_j(\mathbf{c}_j\cdot\mathbf{c}_i)=d_i\). This is why dot products can produce the columns of a change matrix without solving a system.

Using the Transpose to Reverse the Change

A Three-Dimensional Calculation

Common Mistakes and Why the Structure Matters

Reversing the matrix direction. In the second example, the matrix \(Q_{\mathcal{C}\leftarrow\mathcal{B}}\) sends \((1,2)^{\top}\) to \((19/13,22/13)^{\top}\). A common error is to use \(Q^{\top}\) for that same forward conversion. But \(Q^{\top}(1,2)^{\top}=(31/65,142/65)^{\top}\), not \((19/13,22/13)^{\top}\). The correction is to track the arrow: \(Q_{\mathcal{C}\leftarrow\mathcal{B}}\) acts on \(\mathcal{B}\)-coordinates; its transpose acts on \(\mathcal{C}\)-coordinates to return to \(\mathcal{B}\).

Do not confuse coordinates with the vector itself. A coordinate column depends on its basis. The entries \((1,2)^{\top}\) in the second example mean \(\mathbf{b}_1+2\mathbf{b}_2\), not the standard-coordinate vector \((1,2)^{\top}\). The reconstruction check is a reliable way to catch this distinction.

Orthogonal coordinate changes are useful because they preserve dot products and lengths of coordinate columns. If \(\mathbf{x}=[\mathbf{u}]_{\mathcal{B}}\), \(\mathbf{y}=[\mathbf{w}]_{\mathcal{B}}\), and \(Q=Q_{\mathcal{C}\leftarrow\mathcal{B}}\), then \((Q\mathbf{x})^{\top}(Q\mathbf{y})=\mathbf{x}^{\top}Q^{\top}Q\mathbf{y}=\mathbf{x}^{\top}\mathbf{y}\). Thus changing between orthonormal coordinate systems does not alter the geometry encoded by the coordinates. The matrix changes the description, not the vectors or their relationships.