Simultaneous Diagonalization of Commuting Symmetric Matrices

One Orthonormal Basis for Two Matrices

A real symmetric matrix can be diagonalized in an orthonormal basis. For two matrices, the stronger question is whether one orthonormal basis works for both at once. The key condition is that the matrices commute: their products in the two possible orders are equal. The reason this condition matters is that it forces one matrix to preserve the eigenspaces of the other. Within each such eigenspace, the second matrix can be diagonalized without leaving the space.

Throughout, let \(\mathbf{A},\mathbf{B}\in\mathbb{R}^{n\times n}\), with \(n\geq 1\). A matrix \(\mathbf{A}\) is symmetric if \(\mathbf{A}^{\top}=\mathbf{A}\), and the matrices commute if \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\). An orthonormal eigenbasis for a matrix is an orthonormal basis of \(\mathbb{R}^n\) whose vectors are eigenvectors of that matrix. A common orthonormal eigenbasis consists of vectors that are eigenvectors of both matrices.

For an eigenvalue \(\lambda\) of \(\mathbf{A}\), its eigenspace is \(E_{\lambda}=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\}\). An eigenspace is a subspace, and its vectors are exactly the vectors on which \(\mathbf{A}\) acts as multiplication by \(\lambda\). If an eigenvalue is repeated, its eigenspace can have dimension greater than one. That larger space is where the second matrix has room to act while preserving the first matrix’s eigenvalue equation.

Why Commutation Preserves Eigenspaces

Proof. Assume \(\mathbf{x}\in E_{\lambda}\), so \(\mathbf{A}\mathbf{x}=\lambda\mathbf{x}\). Using commutation gives \(\mathbf{A}(\mathbf{B}\mathbf{x})=\mathbf{B}(\mathbf{A}\mathbf{x})=\mathbf{B}(\lambda\mathbf{x})=\lambda(\mathbf{B}\mathbf{x})\). This is the defining equation for membership in \(E_{\lambda}\). It also covers the case \(\mathbf{B}\mathbf{x}=\mathbf{0}\), since \(\mathbf{A}\mathbf{0}=\lambda\mathbf{0}\), so \(\mathbf{B}\mathbf{x}\) still belongs to the eigenspace. \(\square\)

The lemma gives invariance, not necessarily that \(\mathbf{B}\mathbf{x}=\mathbf{x}\) or that \(\mathbf{B}\mathbf{x}\) is a scalar multiple of \(\mathbf{x}\). If \(E_{\lambda}\) has dimension greater than one, \(\mathbf{B}\) may send one vector in that space to a different vector in the same space. Symmetry of \(\mathbf{B}\) is what lets us choose an orthonormal eigenbasis for its action within each preserved eigenspace.

Proof. First suppose that \(\mathbf{A}\) and \(\mathbf{B}\) commute. By the real Spectral Theorem, \(\mathbb{R}^n\) is the orthogonal direct sum of the eigenspaces of \(\mathbf{A}\). For each eigenvalue \(\lambda\) of \(\mathbf{A}\), the lemma shows that \(E_{\lambda}\) is invariant under \(\mathbf{B}\), so the restriction of \(\mathbf{B}\) to \(E_{\lambda}\) maps that space into itself. This restriction is symmetric: for \(\mathbf{x},\mathbf{y}\in E_{\lambda}\), the standard inner product satisfies \(\langle\mathbf{B}\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{B}\mathbf{y}\rangle\), because \(\mathbf{B}^{\top}=\mathbf{B}\). Applying the real Spectral Theorem to this symmetric restriction gives an orthonormal basis of \(E_{\lambda}\) consisting of eigenvectors of \(\mathbf{B}\). Every vector in \(E_{\lambda}\) is already an eigenvector of \(\mathbf{A}\), so this is a common orthonormal eigenbasis for that eigenspace. Taking these bases over all eigenspaces of \(\mathbf{A}\) gives an orthonormal basis of \(\mathbb{R}^n\): the eigenspaces are mutually orthogonal, and together they span \(\mathbb{R}^n\). Thus the combined basis is common to \(\mathbf{A}\) and \(\mathbf{B}\). Conversely, suppose there is a common orthonormal eigenbasis, and let \(\mathbf{Q}\) be the matrix whose columns are its vectors. Then \(\mathbf{Q}^{\top}\mathbf{A}\mathbf{Q}=\mathbf{D}_A\) and \(\mathbf{Q}^{\top}\mathbf{B}\mathbf{Q}=\mathbf{D}_B\) are diagonal matrices. Diagonal matrices commute, so \(\mathbf{D}_A\mathbf{D}_B=\mathbf{D}_B\mathbf{D}_A\). Since \(\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\), this implies \(\mathbf{A}\mathbf{B}=\mathbf{Q}\mathbf{D}_A\mathbf{D}_B\mathbf{Q}^{\top}=\mathbf{Q}\mathbf{D}_B\mathbf{D}_A\mathbf{Q}^{\top}=\mathbf{B}\mathbf{A}\). Therefore the matrices commute. \(\square\)

Constructing the Shared Basis

The proof gives a practical construction. First find the eigenspaces of one matrix, say \(\mathbf{A}\). Commutation guarantees that each is preserved by \(\mathbf{B}\). Then, within each eigenspace, find an orthonormal eigenbasis for the restriction of \(\mathbf{B}\). The resulting vectors are eigenvectors of both matrices. This method does not require the eigenvalues of \(\mathbf{A}\) or \(\mathbf{B}\) to be distinct. In particular, repeated eigenvalues of \(\mathbf{A}\) create larger invariant spaces on which to diagonalize \(\mathbf{B}\).

This example shows why it is useful to work with eigenspaces rather than individual eigenvectors of \(\mathbf{A}\). The standard basis vectors in \(E_1\) are not eigenvectors of \(\mathbf{B}\), but their whole plane is invariant, and within that plane we can choose the needed eigenvectors.

When Symmetry Is Not Enough

The theorem requires both matrices to be symmetric and to commute. Symmetry alone does not guarantee a common eigenbasis. If the matrices do not commute, the invariant-eigenspace argument can fail: an eigenvector of one matrix may be sent outside its eigenspace by the other.

Common Pitfalls

Do not infer commutation from symmetry. A common incorrect calculation is to use \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{A}^{\top}\mathbf{B}^{\top}\). The correct transpose rule is \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}^{\top}\mathbf{A}^{\top}\). For symmetric matrices this says \((\mathbf{A}\mathbf{B})^{\top}=\mathbf{B}\mathbf{A}\); it does not say \(\mathbf{A}\mathbf{B}=\mathbf{B}\mathbf{A}\). In the example above, the two products are explicitly different.

Do not require distinct eigenvalues. The theorem makes no such assumption. When \(\mathbf{A}\) has a repeated eigenvalue, its eigenspace may have dimension greater than one; commutation preserves that entire space, and symmetry of the restriction of \(\mathbf{B}\) supplies an orthonormal eigenbasis within it. Conversely, having matching eigenvalues is not the criterion: the condition in the theorem is commutation.

The shared basis makes the action of each matrix easy to interpret: every basis vector is scaled by one eigenvalue for \(\mathbf{A}\) and another for \(\mathbf{B}\). In matrix form, the same orthogonal matrix \(\mathbf{Q}\) diagonalizes both. The result is useful whenever two symmetric transformations act on the same space and their order of application does not matter.