Worked Examples of Commuting Symmetric Matrices

What a Common Eigenbasis Accomplishes

Each real symmetric matrix can be diagonalized in an orthonormal basis. For a pair of matrices, the important question is whether one orthonormal basis works for both. If it does, the same orthogonal matrix \(\mathbf{Q}\) puts both matrices into diagonal form: \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) and \(\mathbf{Q}^{\top}\mathbf{B}\mathbf{Q}\) are diagonal. In these coordinates, each basis vector is an eigenvector of both matrices, and the two matrices act by scaling the same directions.

Throughout, \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\) are real symmetric matrices. An orthonormal list has unit vectors that are pairwise orthogonal; an orthonormal basis is such a list that spans \(\mathbb{R}^n\). A matrix \(\mathbf{Q}\) is orthogonal when \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\); its columns then form an orthonormal basis. The matrices commute when \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\).

Why Commutation Preserves Eigenspaces

An eigenspace gives all vectors associated with one eigenvalue, including the zero vector. The key calculation is that if \(\mathbf{A}\) and \(\mathbf{B}\) commute, applying \(\mathbf{A}\) to an eigenvector of \(\mathbf{B}\) cannot take it out of that eigenspace.

Proof. Assume \(\mathbf{x}\in E_\lambda\), so \(\mathbf{B}\mathbf{x}=\lambda\mathbf{x}\). Using commutation and the fact that scalar multiplication can be moved outside a matrix product, we obtain \(\mathbf{B}(\mathbf{A}\mathbf{x})=\mathbf{A}(\mathbf{B}\mathbf{x})=\mathbf{A}(\lambda\mathbf{x})=\lambda(\mathbf{A}\mathbf{x})\). By the definition of \(E_\lambda\), this says \(\mathbf{A}\mathbf{x}\in E_\lambda\). If \(\mathbf{A}\mathbf{x}=\mathbf{0}\), the conclusion still holds, since \(\mathbf{B}\mathbf{0}=\lambda\mathbf{0}\). \(\square\)

This gives a practical construction. First split the space into the mutually orthogonal eigenspaces of \(\mathbf{B}\), using the Spectral Theorem for real symmetric matrices. Commutation says each such space is preserved by \(\mathbf{A}\). The restriction of \(\mathbf{A}\) to each eigenspace remains symmetric with respect to the inherited dot product, so the Spectral Theorem provides an orthonormal eigenbasis for that restriction. Combining those bases gives vectors that are eigenvectors of \(\mathbf{A}\) and of \(\mathbf{B}\).

Proof. First suppose that the matrices commute. By the Spectral Theorem, \(\mathbb{R}^n\) is the orthogonal direct sum of the eigenspaces of \(\mathbf{B}\). For any eigenvalue \(\lambda\), the lemma shows that \(E_\lambda\) is invariant under \(\mathbf{A}\). The restriction \(\mathbf{A}|_{E_\lambda}\) is a linear map from \(E_\lambda\) to itself. It is symmetric: for \(\mathbf{x},\mathbf{y}\in E_\lambda\), symmetry of \(\mathbf{A}\) gives \(\langle \mathbf{A}\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{A}\mathbf{y}\rangle\). The Spectral Theorem applied on this finite-dimensional real inner-product space gives an orthonormal basis of \(E_\lambda\) consisting of eigenvectors of \(\mathbf{A}|_{E_\lambda}\). Every vector in \(E_\lambda\) is also an eigenvector of \(\mathbf{B}\) with eigenvalue \(\lambda\). Combining these orthonormal bases over all eigenspaces of \(\mathbf{B}\) produces a common orthonormal eigenbasis of \(\mathbb{R}^n\).

Conversely, suppose \((\mathbf{q}_1,\dots,\mathbf{q}_n)\) is a common orthonormal eigenbasis, and let \(\mathbf{Q}\) have these vectors as its columns. Then \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), and for diagonal matrices \(\mathbf{D}_A,\mathbf{D}_B\) we have \(\mathbf{A}=\mathbf{Q}\mathbf{D}_A\mathbf{Q}^{\top}\) and \(\mathbf{B}=\mathbf{Q}\mathbf{D}_B\mathbf{Q}^{\top}\). Diagonal matrices commute, so \(\mathbf{A}\mathbf{B}=\mathbf{Q}\mathbf{D}_A\mathbf{D}_B\mathbf{Q}^{\top}=\mathbf{Q}\mathbf{D}_B\mathbf{D}_A\mathbf{Q}^{\top}=\mathbf{B}\mathbf{A}\). Thus the matrices commute. \(\square\)

Worked Examples: Constructing and Checking a Shared Basis

When Commutation Fails

Common Mistakes and Verification

Mistake: assuming any eigenbasis of one matrix works for the other. In the repeated-eigenspace example, \(\mathbf{e}_1\) is an eigenvector of \(\mathbf{B}\), but it is not an eigenvector of \(\mathbf{A}\): \(\mathbf{A}\mathbf{e}_1=(2,1,0)^{\top}\), which is not a scalar multiple of \(\mathbf{e}_1\). The correction is to use the invariant eigenspace of \(\mathbf{B}\), then diagonalize the restriction of \(\mathbf{A}\) within it.

When checking a proposed answer, keep the eigenvalues in the same order as the columns of \(\mathbf{Q}\). If \(\mathbf{A}\mathbf{q}_j=\alpha_j\mathbf{q}_j\), the \(j\)-th diagonal entry of \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}\) must be \(\alpha_j\), not an eigenvalue assigned to a different column. A reliable check is to multiply \(\mathbf{A}\mathbf{q}_j\) and \(\mathbf{B}\mathbf{q}_j\) explicitly before writing either diagonal matrix. Also verify \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\) and, separately, \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\).

These checks answer distinct questions. Commutation is the criterion for whether a common orthonormal eigenbasis exists for two real symmetric matrices. Orthonormality verifies the coordinate change, and the individual eigenvector equations verify that the diagonal entries in both representations are correct.