Orthogonal Diagonalization of Quadratic Forms

From a Quadratic Form to Orthogonal Coordinates

A quadratic form in \(\mathbb{R}^n\) can look complicated when written in the original coordinates, particularly when its expression contains mixed terms such as \(x_1x_2\). If its matrix is real and symmetric, the Spectral Theorem supplies an orthonormal basis of eigenvectors. Using those vectors as coordinate axes removes all mixed terms: the form becomes a sum of squared new coordinates, each multiplied by an eigenvalue.

For \(\mathbf{x}\in\mathbb{R}^n\) and a real symmetric matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\), write the quadratic form as \(q(\mathbf{x})=\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\). An orthogonal matrix \(\mathbf{Q}\) is a square matrix whose columns are orthonormal; equivalently, \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). Its columns can be used as a new orthonormal coordinate system.

The distinction between the two coordinate vectors matters: \(\mathbf{x}=\mathbf{Q}\mathbf{y}\) expresses the original vector using the new coordinates, whereas \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\) extracts those coordinates from the original vector. Substitution into the form gives the matrix \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\). This is a congruence transformation, and in this setting it will be diagonal.

The Spectral Theorem and the Form of the Answer

Proof. The Spectral Theorem for finite-dimensional real symmetric matrices states that \(\mathbf{A}\) has an orthonormal basis of real eigenvectors. This is the earlier result being used here; it requires that \(\mathbf{A}\) be real and symmetric. Put those unit eigenvectors into the columns of \(\mathbf{Q}\), and put their corresponding eigenvalues \(\lambda_1,\dots,\lambda_n\) along the diagonal of \(\mathbf{D}\). Orthonormality gives \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and the eigenvector equations give \(\mathbf{A}\mathbf{Q}=\mathbf{Q}\mathbf{D}\). Multiplying on the left by \(\mathbf{Q}^{\top}\) proves \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{D}\). Now substitute \(\mathbf{x}=\mathbf{Q}\mathbf{y}\) into the quadratic form: \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}=(\mathbf{Q}\mathbf{y})^{\top}\mathbf{A}(\mathbf{Q}\mathbf{y})=\mathbf{y}^{\top}(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q})\mathbf{y}=\mathbf{y}^{\top}\mathbf{D}\mathbf{y}=\sum_{i=1}^{n}\lambda_i y_i^2\). For the converse, if \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\), then \(\mathbf{A}^{\top}=(\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top})^{\top}=\mathbf{Q}\mathbf{D}^{\top}\mathbf{Q}^{\top}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}=\mathbf{A}\), since \(\mathbf{D}\) is diagonal and hence symmetric. \(\square\)

The numbers \(\lambda_i\) are the eigenvalues, and \(y_i\) is the coordinate of \(\mathbf{x}\) in the direction of the corresponding unit eigenvector. Thus each term in the new expression is the eigenvalue weight times the square of a coordinate. A zero eigenvalue contributes a zero term. If an eigenvalue is repeated, any orthonormal basis of its eigenspace can be used for the corresponding columns of \(\mathbf{Q}\); the diagonal entries for those columns are the same repeated eigenvalue.

Worked Examples

A Reliable Calculation and a Common Pitfall

For a concrete symmetric matrix, the calculation follows a short sequence. First find its eigenvalues and an eigenvector for each eigenvalue. Next normalize the eigenvectors and place them as columns of \(\mathbf{Q}\). For a symmetric matrix, eigenvectors belonging to distinct eigenvalues are orthogonal; when an eigenvalue is repeated, choose an orthonormal basis within its eigenspace. Then form \(\mathbf{D}\) from the eigenvalues in the same order as the columns of \(\mathbf{Q}\), and use \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\) to obtain the new coordinates.

Common mistake: reversing the two copies of \(\mathbf{Q}\). The coordinate substitution \(\mathbf{x}=\mathbf{Q}\mathbf{y}\) gives \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\), not \(\mathbf{Q}\mathbf{A}\mathbf{Q}^{\top}\). For the first example, the correct product is \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\begin{bmatrix}5&0\\0&0\end{bmatrix}\). Using the reversed product instead gives \(\mathbf{Q}\mathbf{A}\mathbf{Q}^{\top}=\frac15\begin{bmatrix}9&12\\12&16\end{bmatrix}\), which is not diagonal. That calculation does not express the form in the coordinates \(\mathbf{y}=\mathbf{Q}^{\top}\mathbf{x}\); the correction is to substitute \(\mathbf{x}=\mathbf{Q}\mathbf{y}\) into \(\mathbf{x}^{\top}\mathbf{A}\mathbf{x}\) and preserve the resulting order.

What the Diagonal Expression Tells You

Orthogonal diagonalization identifies the directions along which the form has its separate squared-coordinate contributions. In the new coordinates, there are no products \(y_i y_j\) with \(i\ne j\); all contributions appear as \(\lambda_i y_i^2\). Returning to the original coordinates simply rotates or reflects those axes, since the columns of \(\mathbf{Q}\) are orthonormal. The eigenvalues are therefore not just entries produced by a matrix computation: they are the weights attached to the form along its orthogonal eigenvector directions.

This representation also provides a direct check on a result. The diagonal entries must match the eigenvalues corresponding to the columns of \(\mathbf{Q}\), and expanding \(\sum_i\lambda_i y_i^2\) must recover the original expression. These checks catch mismatched eigenvalue order, unnormalized eigenvectors, and an incorrectly oriented coordinate substitution. The signs of the weights can be used for further questions about the form, but the task here is the orthogonal coordinate representation itself.