Why These Errors Are Easy to Make
For real matrices, the transpose is the adjoint used in tests for normality. For complex matrices, that shortcut fails: the adjoint also conjugates every entry. Losing that conjugation can change the outcome of a unitarity check or an eigenvector calculation. A separate but related error is to treat normality as if it forced eigenvalues to have modulus one, or as if any diagonalization were enough to establish normality.
Throughout, matrices and vectors are over \(\mathbb{C}\), and vectors are columns. For a complex number \(z=a+bi\), its conjugate is \(\overline{z}=a-bi\). The conjugate transpose of a matrix is obtained by transposing it and conjugating each entry. A square matrix \(\mathbf{U}\) is unitary if \(\mathbf{U}^{*}\mathbf{U}=\mathbf{U}\mathbf{U}^{*}=\mathbf{I}\). A square matrix \(\mathbf{A}\) is normal if \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\).
The definition of normality compares two products, so a reliable test is to compute both of them, using \(\mathbf{A}^{*}\) rather than \(\mathbf{A}^{\top}\). Unitarity is a stronger-looking but different equation: it asks that the adjoint be the inverse. Normality does not require either product to equal the identity.
Conjugation Rules That Prevent Sign Errors
Proof. The \((i,j)\) entry of \((c\mathbf{A})^{*}\) is \(\overline{ca_{ji}}=\overline{c}\,\overline{a_{ji}}\), which is the \((i,j)\) entry of \(\overline{c}\mathbf{A}^{*}\). Conjugation distributes over addition, so the \((i,j)\) entries of \((\mathbf{A}+\mathbf{B})^{*}\) and \(\mathbf{A}^{*}+\mathbf{B}^{*}\) also agree. For the product rule, suppose \(\mathbf{A}\in\mathbb{C}^{m\times n}\) and \(\mathbf{B}\in\mathbb{C}^{n\times p}\). The \((i,j)\) entry of \((\mathbf{A}\mathbf{B})^{*}\) is \(\overline{\sum_{k=1}^{n}a_{jk}b_{ki}}=\sum_{k=1}^{n}\overline{b_{ki}}\,\overline{a_{jk}}\). This is the \((i,j)\) entry of \(\mathbf{B}^{*}\mathbf{A}^{*}\). Thus all three identities hold. \(\square\)
Two details deserve particular attention. First, a scalar moves through the adjoint with a conjugate: \(i\) becomes \(-i\). Second, the order of factors reverses. The product rule is also why \((\mathbf{U}^{*}\mathbf{A}\mathbf{U})^{*}=\mathbf{U}^{*}\mathbf{A}^{*}\mathbf{U}\) when \(\mathbf{U}\) is unitary.
Worked Example: A Transpose Is Not an Adjoint
Consider the complex matrix \(\mathbf{U}=\tfrac{1}{\sqrt{2}}\begin{bmatrix}1&i\\i&1\end{bmatrix}\). Check whether it is unitary, and compare the correct calculation with the transpose-only calculation.
Conjugate and transpose to obtain \(\mathbf{U}^{*}=\tfrac{1}{\sqrt{2}}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\). Then
In the other order, \(\mathbf{U}\mathbf{U}^{*}\) has diagonal entries \((1+i(-i))/2=1\) and off-diagonal entries \((-i+i)/2=0\), so it too equals \(\mathbf{I}_2\). Thus \(\mathbf{U}\) is unitary. But \(\mathbf{U}^{\top}=\mathbf{U}\), and the incorrect transpose-only test gives
The false conclusion that \(\mathbf{U}\) is not unitary came from using \(\mathbf{U}^{\top}\) in place of \(\mathbf{U}^{*}\). The correct products above verify unitarity directly.
Worked Example: Conjugating a Scalar Factor
Let \(\mathbf{A}=\begin{bmatrix}1&i\\0&2\end{bmatrix}\). Compute \((i\mathbf{A})^{*}\), and compare it with the result of incorrectly leaving \(i\) unchanged.
First, \(\mathbf{A}^{*}=\begin{bmatrix}1&0\\-i&2\end{bmatrix}\) and \(i\mathbf{A}=\begin{bmatrix}i&-1\\0&2i\end{bmatrix}\). Taking the adjoint of this last matrix gives
If one incorrectly writes \((i\mathbf{A})^{*}=i\mathbf{A}^{*}\), the result is instead \(\begin{bmatrix}i&0\\1&2i\end{bmatrix}\), which differs entry by entry. The correction follows from \(\overline{i}=-i\). As a check, the \((2,1)\) entry of the adjoint must be the conjugate of the \((1,2)\) entry \(-1\) of \(i\mathbf{A}\), and is therefore \(-1\), as computed.
What Normality Says About Eigenvectors
An eigenvalue \(\lambda\) of \(\mathbf{A}\) is a scalar for which there is a nonzero vector \(\mathbf{v}\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). A useful diagnostic for a normal matrix is that the same vector is an eigenvector of its adjoint, with conjugate eigenvalue. This statement does not say that \(\lambda=\overline{\lambda}\); that extra conclusion would require additional hypotheses.
Proof. Set \(\mathbf{B}=\mathbf{A}-\lambda\mathbf{I}\). Expanding both products shows that \(\mathbf{B}^{*}\mathbf{B}\) and \(\mathbf{B}\mathbf{B}^{*}\) are equal: each is \(\mathbf{A}^{*}\mathbf{A}-\lambda\mathbf{A}^{*}-\overline{\lambda}\mathbf{A}+|\lambda|^{2}\mathbf{I}\), using normality of \(\mathbf{A}\). Thus \(\mathbf{B}\) is normal. The norm characterization of normality says that \(\|\mathbf{B}\mathbf{x}\|=\|\mathbf{B}^{*}\mathbf{x}\|\) for every \(\mathbf{x}\in\mathbb{C}^{n}\). Since \(\mathbf{B}\mathbf{v}=\mathbf{0}\), this gives \(\|\mathbf{B}^{*}\mathbf{v}\|=0\), and hence \(\mathbf{B}^{*}\mathbf{v}=\mathbf{0}\). But \(\mathbf{B}^{*}=\mathbf{A}^{*}-\overline{\lambda}\mathbf{I}\), so \(\mathbf{A}^{*}\mathbf{v}=\overline{\lambda}\mathbf{v}\). \(\square\)
This result is especially useful when checking a proposed eigenvector calculation: the conjugate belongs on the eigenvalue for the adjoint. It does not impose a modulus restriction on the eigenvalue. A modulus restriction comes from a different condition, such as unitarity.
Worked Example: Which Eigenvalues Have Modulus One?
Show that an eigenvalue of a unitary matrix has modulus one, then compare with a normal matrix that has an eigenvalue of a different modulus.
Suppose \(\mathbf{U}\) is unitary and \(\mathbf{U}\mathbf{v}=\lambda\mathbf{v}\) for \(\mathbf{v}\ne\mathbf{0}\). The identity \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}\) gives \(\|\mathbf{U}\mathbf{v}\|^{2}=\langle\mathbf{U}\mathbf{v},\mathbf{U}\mathbf{v}\rangle=\langle\mathbf{v},\mathbf{U}^{*}\mathbf{U}\mathbf{v}\rangle=\|\mathbf{v}\|^{2}\). On the other hand, \(\|\mathbf{U}\mathbf{v}\|=\|\lambda\mathbf{v}\|=|\lambda|\|\mathbf{v}\|\). Because \(\|\mathbf{v}\|>0\), these equations imply \(|\lambda|=1\).
Normality alone does not give this conclusion. Take \(\mathbf{D}=\begin{bmatrix}2&0\\0&i\end{bmatrix}\). Then \(\mathbf{D}^{*}=\begin{bmatrix}2&0\\0&-i\end{bmatrix}\), and
Thus \(\mathbf{D}\) is normal, but \(\mathbf{e}_1\) is an eigenvector with eigenvalue \(2\), whose modulus is \(2\), not \(1\). The calculation checks both claims: normality follows from the equal products, and \(\mathbf{D}\mathbf{e}_1=2\mathbf{e}_1\).
Diagonalizable Does Not Mean Normal
A matrix is diagonalizable over \(\mathbb{C}\) if it can be written \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for an invertible matrix \(\mathbf{P}\) and a diagonal matrix \(\mathbf{D}\). Normality is not guaranteed by this form. The distinction is that normal matrices have a unitary diagonalization: the Complex Spectral Theorem states that a complex square matrix is normal if and only if it is diagonalizable by a unitary matrix. The diagonalizing matrix in an arbitrary diagonalization need not be unitary.
Worked Example: A Diagonalizable Matrix That Is Not Normal
Consider \(\mathbf{A}=\begin{bmatrix}1&1\\0&2\end{bmatrix}\). Verify that it is diagonalizable, then test normality using the conjugate transpose.
Its columns of eigenvectors can be chosen as \(\begin{bmatrix}1\\0\end{bmatrix}\) for eigenvalue \(1\), and \(\begin{bmatrix}1\\1\end{bmatrix}\) for eigenvalue \(2\). Set \(\mathbf{P}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\), \(\mathbf{D}=\begin{bmatrix}1&0\\0&2\end{bmatrix}\), and \(\mathbf{P}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\). Multiplication verifies the diagonalization:
Since \(\mathbf{A}\) is real, \(\mathbf{A}^{*}=\mathbf{A}^{\top}=\begin{bmatrix}1&0\\1&2\end{bmatrix}\). The two products are
They are unequal, so \(\mathbf{A}\) is not normal. Direct multiplication confirms both results; for example, the top-left entries are \(1\) and \(2\). This example rules out the inference that diagonalizable implies normal.
A Reliable Diagnostic Checklist
When a complex-matrix calculation seems to conflict with normality or unitarity, check the operation before interpreting the result. For an adjoint, conjugate and transpose; for a scalar multiple, conjugate the scalar; for a product, reverse the order. Then distinguish the hypotheses behind any eigenvalue claim. Unitarity forces eigenvalues to have modulus one, while normality by itself does not.
Common mistake: treating transpose as conjugate transpose. For the unitary matrix above, a wrong calculation replaces \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_2\) with \(\mathbf{U}^{\top}\mathbf{U}=\begin{bmatrix}0&i\\i&0\end{bmatrix}\), then concludes that \(\mathbf{U}\) is not unitary. The correction is to conjugate the entries as well as transpose them: \(\mathbf{U}^{*}=\tfrac{1}{\sqrt{2}}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}\), which gives \(\mathbf{U}^{*}\mathbf{U}=\mathbf{I}_2\).
A compact comparison helps keep the conditions separate.
| Condition | Equation to check | Eigenvalue consequence |
|---|---|---|
| Normal | \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\) | An eigenvector for \(\lambda\) is an eigenvector of \(\mathbf{A}^{*}\) for \(\overline{\lambda}\) |
| Unitary | \(\mathbf{U}^{*}\mathbf{U}=\mathbf{U}\mathbf{U}^{*}=\mathbf{I}\) | Every eigenvalue has modulus \(1\) |
| Diagonalizable | \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) for some invertible \(\mathbf{P}\) | Does not, by itself, imply normality |
The practical lesson is to keep the definitions in view. Compute the correct adjoint first, use the two defining products to test normality, and invoke a modulus conclusion only when its required hypothesis has been established.
Check Your Understanding
For each question, write the relevant defining equation before computing.
- Find \(\mathbf{A}^{*}\) for \(\mathbf{A}=\begin{bmatrix}1&i\\2&0\end{bmatrix}\).
- For \(\mathbf{B}=\begin{bmatrix}0&1\\i&0\end{bmatrix}\), compute \((i\mathbf{B})^{*}\) and compare it with \(i\mathbf{B}^{*}\).
- Explain why every eigenvalue of a unitary matrix has modulus one, identifying where the nonzero-eigenvector condition is used.
- Is \(\operatorname{diag}(3,-i)\) normal? Do its eigenvalues all have modulus one? Verify both answers.
- Give a reason, using the two products in the definition, that the diagonalizable matrix in the worked example is not normal.
