Two-by-Two Complex Normal Matrices

Normality in Two Dimensions

For a complex matrix, normality is defined by a comparison involving the conjugate transpose, not merely the transpose. In general, expanding that comparison can be cumbersome. For a two-by-two matrix, however, it reduces to two scalar conditions. Those conditions make normality quick to test and also reveal how the entries constrain eigenvalues and eigenvectors.

Throughout, let \(\mathbf{A}\in\mathbb{C}^{2\times2}\). The conjugate transpose \(\mathbf{A}^{*}\) is obtained by transposing the matrix and conjugating each entry. A matrix is normal when \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\). A matrix \(\mathbf{U}\in\mathbb{C}^{2\times2}\) is unitary when \(\mathbf{U}^{*}\mathbf{U}=\mathbf{U}\mathbf{U}^{*}=\mathbf{I}_2\). A unitary diagonalization is an equation \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\), where \(\mathbf{D}\) is diagonal.

Write the entries of \(\mathbf{A}\) as \(a,b,c,d\), and set \(x=(a-d)/2\). The quantity \(x\) measures the difference between the diagonal entries. The off-diagonal entries and this difference determine the normality test.

Proof. The conjugate transpose is \(\mathbf{A}^{*}=\begin{bmatrix}\overline{a}&\overline{c}\\\overline{b}&\overline{d}\end{bmatrix}\). Multiplying both ways gives

Equality of the upper-left entries is equivalent to \(|c|^2=|b|^2\), which is equivalent to \(|b|=|c|\) because both moduli are nonnegative. When this holds, the lower-right entries also agree. The lower-left entry of each product is the complex conjugate of its upper-right entry, so equality of the upper-right entries guarantees equality of the lower-left entries as well. Thus the remaining condition is \(\overline{a}b+\overline{c}d=a\overline{c}+b\overline{d}\). Rearranging, this is \(b(\overline{a}-\overline{d})=\overline{c}(a-d)\). Since \(a-d=2x\), it is equivalent to \(b\overline{x}=\overline{c}x\). These conditions are therefore necessary and sufficient for the two products to agree. \(\square\)

The centered form is also useful for interpreting the test. Write \(\mu=(a+d)/2\). Then \(\mathbf{A}=\mu\mathbf{I}_2+\begin{bmatrix}x&b\\c&-x\end{bmatrix}\). The scalar part \(\mu\mathbf{I}_2\) does not appear in either condition. If the diagonal entries are equal, then \(x=0\), so normality reduces simply to \(|b|=|c|\). If the matrix is upper triangular, \(c=0\), and normality forces \(|b|=0\), hence \(b=0\): an upper triangular normal matrix of this size must be diagonal.

Eigenvalues from the Entries

The eigenvalues are the roots of the characteristic equation \(\det(\mathbf{A}-\lambda\mathbf{I}_2)=0\). For a two-by-two matrix, that equation can be written directly in terms of the same centered entries used in the normality test.

To see the formula, expand the determinant. Since \(a=\mu+x\) and \(d=\mu-x\), we have \((a-\lambda)(d-\lambda)-bc=(\mu-\lambda+x)(\mu-\lambda-x)-bc=(\mu-\lambda)^2-x^2-bc\). Solving the resulting quadratic equation gives the stated roots. Over \(\mathbb{C}\), every complex number has a square root; choosing the other square root only interchanges the two values.

For a normal matrix, the Complex Spectral Theorem established in the earlier material on normal operators says that a finite-dimensional complex normal operator has an orthonormal basis of eigenvectors. In matrix terms, \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\) for a unitary \(\mathbf{U}\) and diagonal \(\mathbf{D}\). The examples below compute such diagonalizations explicitly, rather than relying on the theorem to supply the vectors.

Repeated Eigenvalues and a Triangular Case

Proof. By the Complex Spectral Theorem, there is a unitary \(\mathbf{U}\) and a diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{U}\mathbf{D}\mathbf{U}^{*}\), with the eigenvalues of \(\mathbf{A}\) on the diagonal of \(\mathbf{D}\). If the eigenvalue is repeated, then \(\mathbf{D}=\begin{bmatrix}\lambda&0\\0&\lambda\end{bmatrix}=\lambda\mathbf{I}_2\). Therefore \(\mathbf{A}=\lambda\mathbf{U}\mathbf{I}_2\mathbf{U}^{*}=\lambda\mathbf{I}_2\), because a unitary matrix satisfies \(\mathbf{U}\mathbf{U}^{*}=\mathbf{I}_2\). \(\square\)

The entrywise test gives a quick way to recognize this in the triangular case, without computing eigenvectors.

A Common Mistake: Checking Only Part of the Products

A tempting but incorrect shortcut is to compare only the off-diagonal entries of \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\). In the last example those entries match exactly, but the matrices are still unequal: their upper-left entries are \(4\) and \(6\), and their lower-right entries are \(6\) and \(4\). Equality of two matrices requires equality in every position.

The correction is to use both parts of the criterion. First compare the diagonal entries, which forces \(|b|=|c|\). Then compare one off-diagonal pair, which forces \(b\overline{x}=\overline{c}x\). The conjugate symmetry of the remaining pair then follows automatically. This order is efficient, but it does not permit skipping the diagonal condition.

Another useful check is to compute the conjugate transpose before multiplying. Replacing \(\mathbf{A}^{*}\) by \(\mathbf{A}^{\top}\) drops complex conjugation and can change the products. For instance, the entry \(1+i\) in \(\mathbf{B}\) becomes \(1-i\) in \(\mathbf{B}^{*}\). Keeping the conjugates visible helps prevent a false normality test.

What the Two-by-Two Test Provides

For a general two-by-two complex matrix, the entrywise criterion supplies a direct answer about normality, while the centered characteristic polynomial supplies the eigenvalues. If the matrix passes the test, the Complex Spectral Theorem guarantees a unitary diagonalization; the worked examples show how to verify one by finding eigenvectors, checking orthogonality, normalizing, and multiplying back. When the eigenvalue is repeated, normality forces the matrix to be a scalar multiple of the identity. These conclusions depend on checking the full normality condition, not just one set of entries.