Unitary Diagonalization Criteria for Complex Matrices

Normality and Unitary Diagonalization

Diagonalization expresses a matrix using its eigenvalues and eigenvectors. Unitary diagonalization is a more restrictive form: it requires the eigenvectors to be chosen as an orthonormal basis. That extra requirement matters in complex inner product spaces, where orthonormal coordinates preserve lengths and inner products. The exact test is normality, a condition stated directly in terms of a matrix and its adjoint.

Throughout, let \(\mathbf{A}\in\mathbb{C}^{n\times n}\), with \(n\geq 1\). The adjoint of a matrix is its conjugate transpose, denoted \(\mathbf{A}^{*}\). A square matrix \(\mathbf{U}\) is unitary if \(\mathbf{U}^{*}\mathbf{U}=\mathbf{U}\mathbf{U}^{*}=\mathbf{I}_n\). Its columns then form an orthonormal basis of \(\mathbb{C}^n\).

The columns of \(\mathbf{U}\) are the eigenvectors used in this representation. Indeed, from \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\) and \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_n\), we get \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{D}\). If the \(j\)th diagonal entry of \(\mathbf{D}\) is \(\lambda_j\), the \(j\)th column \(\mathbf{u}_j\) of \(\mathbf{U}\) satisfies \(\mathbf{A}\mathbf{u}_j=\lambda_j\mathbf{u}_j\). The additional condition is that these eigenvectors are orthonormal, not merely linearly independent.

The Unitary Diagonalization Criterion

Proof. First suppose that \(\mathbf{A}\) is unitarily diagonalizable, so \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\) for a unitary \(\mathbf{U}\) and a diagonal \(\mathbf{D}\). By the adjoint laws, \(\mathbf{A}^{*}=\mathbf{U}\mathbf{D}^{*}\mathbf{U}^{*}\). Therefore

Every diagonal matrix commutes with its adjoint: if its diagonal entries are \(d_1,\dots,d_n\), then both \(\mathbf{D}^{*}\mathbf{D}\) and \(\mathbf{D}\mathbf{D}^{*}\) have diagonal entries \(|d_1|^2,\dots,|d_n|^2\). Hence \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\), so \(\mathbf{A}\) is normal.

Conversely, suppose that \(\mathbf{A}\) is normal. By the Complex Spectral Theorem, \(\mathbb{C}^n\) has an orthonormal basis \(\mathbf{u}_1,\dots,\mathbf{u}_n\) consisting of eigenvectors of \(\mathbf{A}\). Let \(\mathbf{U}\) be the matrix with these vectors as columns, and let \(\mathbf{D}\) be diagonal with their corresponding eigenvalues on the diagonal. Orthonormality gives \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_n\), and because the \(n\) vectors form a basis, \(\mathbf{U}\) is unitary. The eigenvector equations give \(\mathbf{A}\mathbf{U}=\mathbf{U}\mathbf{D}\). Multiplying on the right by \(\mathbf{U}^{*}\) yields \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\). Thus \(\mathbf{A}\) is unitarily diagonalizable. Both directions hold, proving the equivalence. \(\square\)

The first direction is a direct computation. The second uses the Complex Spectral Theorem, which is why the field matters: the theorem guarantees an orthonormal eigenbasis for a normal matrix over \(\mathbb{C}\). The diagonal entries may repeat; the argument requires an orthonormal eigenbasis, not distinct eigenvalues.

Worked Applications

Common Pitfall: Using the Transpose in Place of the Adjoint

For a complex matrix, the adjoint is the conjugate transpose, not merely the transpose. A tempting but wrong calculation is to take a complex symmetric matrix, observe \(\mathbf{A}^{\top}=\mathbf{A}\), and conclude that \(\mathbf{A}^{\top}\mathbf{A}=\mathbf{A}\mathbf{A}^{\top}\) makes it normal. Normality is defined using \(\mathbf{A}^{*}\), so that calculation does not test the required condition.

For a concrete correction, let \(\mathbf{B}=\begin{bmatrix}1&i\\i&0\end{bmatrix}\). It is symmetric, so the transpose-based comparison gives equal products automatically. But its adjoint is \(\mathbf{B}^{*}=\begin{bmatrix}1&-i\\-i&0\end{bmatrix}\), and the correct products are

They are unequal, so \(\mathbf{B}\) is not normal and cannot be unitarily diagonalized. The conjugation of each entry is essential to the criterion.

How to Use the Criterion

For a complex square matrix, the criterion separates two questions. Ordinary diagonalizability asks whether there is some eigenbasis, equivalently a representation \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) with \(\mathbf{P}\) invertible. Unitary diagonalizability asks whether that eigenbasis can be chosen orthonormal. Normality answers the second question exactly. It does not say that every diagonalizable matrix is normal, as the second example demonstrates.

In practice, when the goal is to decide unitary diagonalizability, compute \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\) and compare them. Equality proves normality and, by the theorem, guarantees a unitary diagonalization. Inequality rules one out. The test concerns the matrix as a whole; checking a few eigenvectors or finding a diagonalization with an arbitrary invertible matrix does not replace it.