Computing Orthonormal Coordinates in Rn

Coordinates Depend on the Basis

A vector in \(\mathbb{R}^n\) has standard coordinates: its entries give its components along the standard basis vectors \(\mathbf{e}_1,\dots,\mathbf{e}_n\). A different basis describes the same vector using different coefficients. When the basis is orthonormal, those coefficients can be computed directly by inner products, and the vector can be reconstructed by adding the corresponding multiples of the basis vectors. This tutorial focuses on carrying out both calculations accurately.

Throughout, \(\mathbb{R}^n\) has the standard dot product \(\mathbf{x}\cdot\mathbf{y}=\mathbf{x}^{\top}\mathbf{y}\). An ordered list \(\mathcal{B}=(\mathbf{q}_1,\dots,\mathbf{q}_n)\) is an orthonormal basis of \(\mathbb{R}^n\) when it is a basis and satisfies \(\mathbf{q}_i\cdot\mathbf{q}_j=1\) for \(i=j\) and \(0\) for \(i\ne j\). The coordinate vector \([\mathbf{x}]_{\mathcal{B}}\) is the column of coefficients \(c_1,\dots,c_n\) in the unique expression \(\mathbf{x}=c_1\mathbf{q}_1+\cdots+c_n\mathbf{q}_n\). In general, these coefficients are not the entries of \(\mathbf{x}\) in standard coordinates.

Proof. Because \(\mathcal{B}\) is a basis, there are unique coefficients \(c_1,\dots,c_n\) such that \(\mathbf{x}=\sum_{j=1}^{n}c_j\mathbf{q}_j\). Fix an index \(i\) and take the dot product with \(\mathbf{q}_i\). Bilinearity gives \(\mathbf{q}_i\cdot\mathbf{x}=\sum_{j=1}^{n}c_j(\mathbf{q}_i\cdot\mathbf{q}_j)\). Orthonormality makes every term zero except the term \(j=i\), which is \(c_i(\mathbf{q}_i\cdot\mathbf{q}_i)=c_i\). Therefore \(c_i=\mathbf{q}_i\cdot\mathbf{x}\). Substituting these coefficients into the basis expansion gives the displayed coordinate vector and reconstruction formula. \(\square\)

The theorem provides a direct procedure: take one dot product per basis vector, place the results in the same order as the basis list, and use those coefficients to reconstruct the vector. The order matters. Swapping two basis vectors swaps the corresponding entries of the coordinate vector as well.

Worked Coordinates in a Nonstandard Basis

The Basis-Matrix Calculation

For an orthonormal basis, place its vectors as the columns of a matrix: \(\mathbf{Q}=\begin{bmatrix}\mathbf{q}_1&\cdots&\mathbf{q}_n\end{bmatrix}\). Then \(\mathbf{Q}\mathbf{c}\) forms the linear combination of the columns with coefficient column \(\mathbf{c}\). The transpose \(\mathbf{Q}^{\top}\) places the inner products with those columns into a column. This gives a compact version of the coordinate theorem.

Proof. The entry in row \(i\), column \(j\) of \(\mathbf{Q}^{\top}\mathbf{Q}\) is \(\mathbf{q}_i\cdot\mathbf{q}_j\). It equals \(1\) when \(i=j\) and \(0\) otherwise, so \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\). The \(i\)-th entry of \(\mathbf{Q}^{\top}\mathbf{x}\) is \(\mathbf{q}_i\cdot\mathbf{x}\), which the coordinate theorem identifies as the \(i\)-th coordinate. Hence \(\mathbf{Q}^{\top}\mathbf{x}=[\mathbf{x}]_{\mathcal{B}}\), and the reconstruction formula gives \(\mathbf{x}=\mathbf{Q}[\mathbf{x}]_{\mathcal{B}}\). To establish the other matrix identity, apply \(\mathbf{Q}\mathbf{Q}^{\top}\) to any \(\mathbf{x}\). The formulas just proved give \(\mathbf{Q}\mathbf{Q}^{\top}\mathbf{x}=\mathbf{Q}[\mathbf{x}]_{\mathcal{B}}=\mathbf{x}\). Since this holds for every column \(\mathbf{x}\in\mathbb{R}^n\), the matrix acts as the identity on every standard basis vector and therefore equals \(\mathbf{I}_n\). Thus \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). \(\square\)

Both matrix identities are useful checks on an orthonormal basis matrix. They also explain why the transpose appears in the coordinate calculation: \(\mathbf{Q}^{\top}\mathbf{x}\) extracts coefficients, whereas \(\mathbf{Q}\) uses those coefficients to rebuild the vector.

Reconstructing from a Coordinate List

The direction of the calculation can start with coordinates instead of a vector. Given \(\mathbf{c}=(c_1,\dots,c_n)^{\top}\), the vector it represents in \(\mathcal{B}\) is \(\mathbf{x}=\mathbf{Q}\mathbf{c}=\sum_{i=1}^{n}c_i\mathbf{q}_i\). After reconstructing, one can verify that the coordinates really were the entries of \(\mathbf{c}\): for each \(j\), \(\mathbf{q}_j\cdot\mathbf{x}=\sum_{i=1}^{n}c_i(\mathbf{q}_j\cdot\mathbf{q}_i)=c_j\). The orthonormality removes all terms except the \(i=j\) term.

A Useful Consistency Check

For any two vectors \(\mathbf{x},\mathbf{y}\in\mathbb{R}^n\), orthonormal coordinates also preserve dot products: \[ \mathbf{x}\cdot\mathbf{y} =[\mathbf{x}]_{\mathcal{B}}^{\top}[\mathbf{y}]_{\mathcal{B}}. \] Indeed, writing each vector as a sum of its basis components gives \(\mathbf{x}\cdot\mathbf{y}=\sum_i\sum_j c_i d_j(\mathbf{q}_i\cdot\mathbf{q}_j)=\sum_i c_i d_i\), where \(c_i\) and \(d_i\) are the respective coordinates. This is the coordinate form of Parseval's identity. In particular, the squared length is the sum of the squares of the coordinates. It is a useful check, but it does not replace checking the signs and order of the individual coefficients.

Common mistake: multiplying in the wrong order. In the first example, the basis matrix is \(\mathbf{Q}=\tfrac{1}{\sqrt{2}}\begin{bmatrix}1&-1\\1&1\end{bmatrix}\). A tempting but wrong calculation is to find coordinates as \(\mathbf{Q}\mathbf{x}\), which gives \(\tfrac{1}{\sqrt{2}}(2,4)^{\top}=(\sqrt{2},2\sqrt{2})^{\top}\). Those are not the coordinates: the columns of \(\mathbf{Q}\) are the basis vectors, so coordinates are found with the transpose: \[ \mathbf{Q}^{\top}\mathbf{x} =\frac{1}{\sqrt{2}} \begin{bmatrix}1&1\\-1&1\end{bmatrix} \begin{bmatrix}3\\1\end{bmatrix} = \begin{bmatrix}2\sqrt{2}\\-\sqrt{2}\end{bmatrix}. \] Multiplying this correct coordinate column by \(\mathbf{Q}\) reconstructs \((3,1)^{\top}\). Keep the roles distinct: transpose to extract coordinates, basis matrix to reconstruct.