Three Conditions, Three Different Questions
For a complex matrix, transposing its entries is not enough: conjugation is part of the adjoint. This matters when comparing matrices that preserve inner products, have real eigenvalues, or can be diagonalized in an orthonormal basis. Normal, self-adjoint, and unitary matrices are closely related, but their defining equations are different, and their eigenvalues answer different questions.
Throughout, let \(\mathbf{A}\in\mathbb{C}^{n\times n}\), where \(n\geq 1\). The conjugate transpose \(\mathbf{A}^{*}\) is obtained by transposing the matrix and then taking the complex conjugate of every entry. A matrix \(\mathbf{Q}\) is unitary when \(\mathbf{Q}^{*}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{*}=\mathbf{I}_n\). The three classes are defined as follows.
Normality compares two products; it does not require either product to be the identity. Self-adjointness compares the matrix directly with its adjoint. Unitarity requires the products to be the identity. Those differences already show why the names are not interchangeable.
Which Classes Are Contained in Which?
Both self-adjoint and unitary matrices are normal. For a self-adjoint matrix, \(\mathbf{A}^{*}=\mathbf{A}\), so both products in the normality condition equal \(\mathbf{A}^{2}\). For a unitary matrix, both products equal \(\mathbf{I}_n\). Neither class is contained in the other: a self-adjoint matrix need not be unitary, and a unitary matrix need not be self-adjoint.
Proof. If \(\mathbf{A}^{*}=\mathbf{A}\), then \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}=\mathbf{A}^{2}\) and \(\mathbf{A}\mathbf{A}^{*}=\mathbf{A}\mathbf{A}=\mathbf{A}^{2}\). Thus \(\mathbf{A}\) is normal. If \(\mathbf{A}\) is unitary, then \(\mathbf{A}^{*}\mathbf{A}=\mathbf{I}_n=\mathbf{A}\mathbf{A}^{*}\), so it is normal as well. To see that neither class is contained in the other, consider \(\begin{bmatrix}2&0\\0&-1\end{bmatrix}\), which is self-adjoint but not unitary, and \(\begin{bmatrix}i&0\\0&1\end{bmatrix}\), which is unitary but not self-adjoint; their properties are checked explicitly below. \(\square\)
Worked Example: Self-Adjoint but Not Unitary
Let \(\mathbf{A}=\begin{bmatrix}2&0\\0&-1\end{bmatrix}\). Since the entries are real and the matrix is diagonal, its conjugate transpose is \(\mathbf{A}^{*}=\begin{bmatrix}2&0\\0&-1\end{bmatrix}=\mathbf{A}\). Thus \(\mathbf{A}\) is self-adjoint.
Multiplying gives
The products agree, so \(\mathbf{A}\) is normal. They are not \(\mathbf{I}_2=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), so \(\mathbf{A}\) is not unitary. This example belongs to the self-adjoint and normal classes, but not the unitary class.
Worked Example: Unitary but Not Self-Adjoint
Let \(\mathbf{U}=\begin{bmatrix}i&0\\0&1\end{bmatrix}\). Conjugation changes \(i\) to \(-i\), so \(\mathbf{U}^{*}=\begin{bmatrix}-i&0\\0&1\end{bmatrix}\). Direct multiplication gives
Here \((-i)i=1\) and \(i(-i)=1\), so \(\mathbf{U}\) is unitary and therefore normal. But \(\mathbf{U}^{*}\ne\mathbf{U}\), because \(-i\ne i\). It is not self-adjoint.
Check: the two products are both \(\mathbf{I}_2\), while the first diagonal entries of \(\mathbf{U}\) and \(\mathbf{U}^{*}\) differ. These computations verify both claims directly.
Eigenvalues Separate the Classes
An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). For self-adjoint matrices, every eigenvalue is real. For unitary matrices, every eigenvalue has absolute value \(1\). Normality alone imposes neither restriction: a normal matrix can have any complex number as an eigenvalue.
For the following argument, use the standard inner product on \(\mathbb{C}^n\), with the convention that it is linear in the first slot. In particular, \(\langle \lambda\mathbf{v},\mathbf{v}\rangle=\lambda\|\mathbf{v}\|^2\), while \(\langle\mathbf{v},\lambda\mathbf{v}\rangle=\overline{\lambda}\|\mathbf{v}\|^2\).
Proof. Suppose first that \(\mathbf{A}\) is self-adjoint and \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for \(\mathbf{v}\ne\mathbf{0}\). The adjoint identity and \(\mathbf{A}^{*}=\mathbf{A}\) give \(\langle\mathbf{A}\mathbf{v},\mathbf{v}\rangle=\langle\mathbf{v},\mathbf{A}\mathbf{v}\rangle\). Substituting the eigenvalue equation into both sides yields \(\lambda\|\mathbf{v}\|^2=\overline{\lambda}\|\mathbf{v}\|^2\). Since \(\|\mathbf{v}\|^2>0\), \(\lambda=\overline{\lambda}\), so \(\lambda\) is real. Now suppose \(\mathbf{A}\) is unitary. Then \(\|\mathbf{A}\mathbf{v}\|^2=\langle\mathbf{A}\mathbf{v},\mathbf{A}\mathbf{v}\rangle=\langle\mathbf{v},\mathbf{A}^{*}\mathbf{A}\mathbf{v}\rangle=\|\mathbf{v}\|^2\). The eigenvalue equation also gives \(\|\mathbf{A}\mathbf{v}\|^2=\|\lambda\mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2\). Again \(\|\mathbf{v}\|^2>0\), so \(|\lambda|^2=1\), hence \(|\lambda|=1\). \(\square\)
These are necessary restrictions on eigenvalues; for normal matrices they fit into a stronger description. The Complex Spectral Theorem states that every normal matrix in \(\mathbb{C}^{n\times n}\) can be written \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{*}\), where \(\mathbf{Q}\) is unitary and \(\mathbf{D}\) is diagonal. Conversely, every matrix of this form is normal: multiplying \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\), using \(\mathbf{Q}^{*}\mathbf{Q}=\mathbf{I}_n\), gives \(\mathbf{Q}\mathbf{D}^{*}\mathbf{D}\mathbf{Q}^{*}\) and \(\mathbf{Q}\mathbf{D}\mathbf{D}^{*}\mathbf{Q}^{*}\); diagonal matrices commute, so these products agree. The theorem is the earlier result on the complex spectral theorem.
Proof. The Complex Spectral Theorem gives unitary diagonalization for every normal matrix. Conversely, if \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{*}\) with \(\mathbf{D}\) diagonal and \(\mathbf{Q}\) unitary, the products \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\) are \(\mathbf{Q}\mathbf{D}^{*}\mathbf{D}\mathbf{Q}^{*}\) and \(\mathbf{Q}\mathbf{D}\mathbf{D}^{*}\mathbf{Q}^{*}\). Since diagonal matrices commute, \(\mathbf{A}\) is normal. If \(\mathbf{A}\) is self-adjoint, it is normal and hence has such a diagonalization. Its diagonal entries are eigenvalues, so the preceding proposition makes them real. Conversely, if \(\mathbf{D}\) is real diagonal, then \(\mathbf{D}^{*}=\mathbf{D}\), and \(\mathbf{A}^{*}=(\mathbf{Q}\mathbf{D}\mathbf{Q}^{*})^{*}=\mathbf{Q}\mathbf{D}^{*}\mathbf{Q}^{*}=\mathbf{A}\). If \(\mathbf{A}\) is unitary, it is normal and its diagonal entries in a unitary diagonalization are eigenvalues of absolute value \(1\). Conversely, if those entries have absolute value \(1\), then \(\mathbf{D}^{*}\mathbf{D}=\mathbf{D}\mathbf{D}^{*}=\mathbf{I}_n\), and multiplication gives \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}=\mathbf{I}_n\). Thus \(\mathbf{A}\) is unitary. \(\square\)
Worked Example: Normal but Neither Self-Adjoint nor Unitary
Consider \(\mathbf{N}=\begin{bmatrix}1+i&0\\0&2\end{bmatrix}\). Its conjugate transpose is \(\mathbf{N}^{*}=\begin{bmatrix}1-i&0\\0&2\end{bmatrix}\). The two products are
The equality holds because \((1-i)(1+i)=1-i^2=2\), so \(\mathbf{N}\) is normal. It is not self-adjoint because \(1+i\ne1-i\). It is not unitary because \(\mathbf{N}^{*}\mathbf{N}=\begin{bmatrix}2&0\\0&4\end{bmatrix}\ne\mathbf{I}_2\). Its eigenvalues are \(1+i\) and \(2\), illustrating that normal matrices need not have real eigenvalues or eigenvalues of absolute value \(1\).
Check: the diagonal entries of \(\mathbf{N}^{*}\mathbf{N}\) are \(|1+i|^2=2\) and \(|2|^2=4\), matching the displayed product.
The Intersection of the Self-Adjoint and Unitary Classes
A matrix can be both self-adjoint and unitary. In that case, the two defining equations combine into a simple condition. If \(\mathbf{A}^{*}=\mathbf{A}\), then unitarity is equivalent to \(\mathbf{A}^{2}=\mathbf{I}_n\). The spectral description gives the same conclusion: a real number with absolute value \(1\) must be \(1\) or \(-1\).
Proof. If \(\mathbf{A}\) is both, then \(\mathbf{A}^{*}=\mathbf{A}\) and \(\mathbf{A}^{*}\mathbf{A}=\mathbf{I}_n\). Substitution gives \(\mathbf{A}^{2}=\mathbf{I}_n\). Conversely, if \(\mathbf{A}^{*}=\mathbf{A}\) and \(\mathbf{A}^{2}=\mathbf{I}_n\), then \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}^{2}=\mathbf{I}_n\) and \(\mathbf{A}\mathbf{A}^{*}=\mathbf{A}^{2}=\mathbf{I}_n\), so it is unitary as well as self-adjoint. By the theorem above, its diagonal entries are real and have absolute value \(1\); the only such real numbers are \(1\) and \(-1\). \(\square\)
Worked Example: Both Self-Adjoint and Unitary
Let \(\mathbf{S}=\begin{bmatrix}1&0\\0&-1\end{bmatrix}\). Then \(\mathbf{S}^{*}=\mathbf{S}\), and
Thus \(\mathbf{S}\) is self-adjoint and unitary, and therefore normal. Its eigenvalues \(1\) and \(-1\) are both real and have absolute value \(1\), as both conditions require.
A Common Conjugate-Transpose Mistake
For complex matrices, replacing the conjugate transpose by the ordinary transpose can reverse a classification. For \(\mathbf{U}=\begin{bmatrix}i&0\\0&1\end{bmatrix}\), a wrong calculation uses \(\mathbf{U}^{\top}=\mathbf{U}\) and concludes \(\mathbf{U}^{\top}\mathbf{U}=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\ne\mathbf{I}_2\), so \(\mathbf{U}\) is not unitary. The correction is to conjugate as well as transpose: \(\mathbf{U}^{*}=\begin{bmatrix}-i&0\\0&1\end{bmatrix}\), and \(\mathbf{U}^{*}\mathbf{U}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). The matrix is unitary, but not self-adjoint.
Another common error is to infer that every normal matrix is self-adjoint or unitary. The matrix \(\mathbf{N}\) above disproves both claims: its two products agree, but the shared product is not the identity and \(\mathbf{N}^{*}\ne\mathbf{N}\). Use the defining equation first, then check any claimed eigenvalue restriction against the relevant class.
| Class | Defining condition | Eigenvalue information |
|---|---|---|
| Normal | \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\) | May be any complex number |
| Self-adjoint | \(\mathbf{A}^{*}=\mathbf{A}\) | Every eigenvalue is real |
| Unitary | \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}=\mathbf{I}_n\) | Every eigenvalue has absolute value \(1\) |
| Both self-adjoint and unitary | \(\mathbf{A}^{*}=\mathbf{A}\) and \(\mathbf{A}^{2}=\mathbf{I}_n\) | Every eigenvalue is \(1\) or \(-1\) |
These distinctions are useful whenever a matrix is described through its eigenvalues or its action on vectors. A real spectrum alone is consistent with self-adjointness but does not establish it; unitary matrices are identified by their unit-circle eigenvalues when they are known to be normal; and normality gives orthonormal diagonalization without imposing either of those special eigenvalue restrictions. The equations, not the labels, determine which conclusion is justified.
Check Your Understanding
For each question, use the defining equations and state which matrix classes apply.
- For \(\mathbf{A}=\begin{bmatrix}3&0\\0&-2\end{bmatrix}\), compute \(\mathbf{A}^{*}\mathbf{A}\) and classify the matrix.
- For \(\mathbf{B}=\begin{bmatrix}-1&0\\0&i\end{bmatrix}\), compute its conjugate transpose and classify it.
- Give an example of a normal matrix that is neither self-adjoint nor unitary, and verify both product equations.
- Prove that if \(\mathbf{A}\) is both self-adjoint and unitary, then each eigenvalue is either \(1\) or \(-1\).
- Explain why normality alone does not force every eigenvalue to be real or to have absolute value \(1\).
