Normality Compares an Operator with Its Adjoint
For a complex matrix, taking an adjoint involves both transposing and conjugating its entries. Normality asks whether the operator commutes with that adjoint. This condition includes important classes of operators, such as self-adjoint and unitary operators, but it is not the same as belonging to either class. Direct calculations with small matrices make the distinction precise.
Throughout, let \(V\) be a finite-dimensional complex inner product space, and let \(T:V\to V\) be a linear operator. The adjoint \(T^*\) is the operator characterized by \(\langle T\mathbf{u},\mathbf{v}\rangle=\langle\mathbf{u},T^*\mathbf{v}\rangle\) for all \(\mathbf{u},\mathbf{v}\in V\). In an orthonormal basis, if \(\mathbf{A}\) is the matrix of \(T\), the matrix of \(T^*\) is the conjugate transpose \(\mathbf{A}^*\): its entry in row \(i\), column \(j\) is the complex conjugate of the entry of \(\mathbf{A}\) in row \(j\), column \(i\).
The equality compares two operators, so it can be checked by computing both products entry by entry. A matrix is not normal merely because it has an adjoint, and the equation does not say that \(\mathbf{A}\) equals \(\mathbf{A}^*\). Those are different conditions. The examples below establish the basic cases by calculation.
Diagonal Matrices
Let \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\) be a diagonal complex matrix. Its adjoint is the diagonal matrix whose entries are the conjugates \(\overline{d_1},\dots,\overline{d_n}\). Since multiplying diagonal matrices multiplies corresponding diagonal entries, the two products agree.
Proof. For \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\), its adjoint is \(\mathbf{D}^*=\operatorname{diag}(\overline{d_1},\dots,\overline{d_n})\). Thus both products are the diagonal matrix with entries \(\overline{d_j}d_j=d_j\overline{d_j}=|d_j|^2\), for \(j=1,\dots,n\), so \(\mathbf{D}^*\mathbf{D}=\mathbf{D}\mathbf{D}^*\). Also, \(\mathbf{D}^*=\mathbf{D}\) holds exactly when \(\overline{d_j}=d_j\) for every \(j\), which is equivalent to every \(d_j\) being real. Finally, \(\mathbf{D}\) is unitary exactly when \(\mathbf{D}^*\mathbf{D}=\mathbf{I}_n\) and \(\mathbf{D}\mathbf{D}^*=\mathbf{I}_n\); both products are diagonal with entries \(|d_j|^2\), so these equalities hold exactly when \(|d_j|=1\) for every \(j\). \(\square\)
Worked Example: A Complex Diagonal Matrix
Test normality, self-adjointness, and unitarity for \(\mathbf{D}=\begin{bmatrix}1+i&0\\0&2i\end{bmatrix}\).
Conjugate each entry and transpose. Since this matrix is diagonal, transposing does not move any nonzero entries:
Compute each product separately. The first product is
Reversing the order gives
The products agree, so \(\mathbf{D}\) is normal. It is not self-adjoint because, for example, \(1+i\) is not real. It is not unitary because its diagonal entries do not both have absolute value \(1\): \(|1+i|^2=2\) and \(|2i|^2=4\).
Check: The two displayed products are both \(\operatorname{diag}(2,4)\), as predicted by the squared absolute values of the diagonal entries.
Self-Adjoint and Unitary Operators
A self-adjoint operator satisfies \(T^*=T\). Substituting this equality into the normality condition gives \(T^*T=T^2=TT^*\), so self-adjoint operators are normal. A unitary operator satisfies \(T^*T=TT^*=I\), which also makes it normal. These implications do not say that every normal operator is self-adjoint or unitary, as the diagonal example already demonstrates.
Proof. If \(T\) is self-adjoint, then \(T^*=T\). Therefore \(T^*T=TT\) and \(TT^*=TT\), so \(T^*T=TT^*\). If \(T\) is unitary, then by definition \(T^*T=I\) and \(TT^*=I\). The two products are equal, so \(T\) is normal. \(\square\)
Worked Example: A Self-Adjoint Matrix That Is Not Diagonal
Check directly that \(\mathbf{H}=\begin{bmatrix}2&i\\-i&1\end{bmatrix}\) is self-adjoint and normal.
First take the conjugate transpose. Conjugating changes \(i\) to \(-i\), and transposing then swaps the off-diagonal entries:
Thus \(\mathbf{H}\) is self-adjoint. To check normality by the defining equation, compute the common product \(\mathbf{H}^2\):
Since \(\mathbf{H}^*=\mathbf{H}\), the product \(\mathbf{H}\mathbf{H}^*\) is the same multiplication and equals \(\begin{bmatrix}5&3i\\-3i&2\end{bmatrix}\). Hence the two products agree.
Check: The entries follow from \(i(-i)=1\), \((-i)i=1\), and \(i^2=-1\); both products have diagonal entries \(5,2\) and off-diagonal entries \(3i,-3i\).
Worked Example: A Unitary Matrix That Is Not Self-Adjoint
Let \(\mathbf{U}=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\). Compute both products to verify unitarity and normality, then compare \(\mathbf{U}^*\) with \(\mathbf{U}\).
The entries are real, so the conjugation changes nothing; transposition gives
Multiplying in the first order gives
In the reverse order,
Both products are the identity, so \(\mathbf{U}\) is unitary and normal. But \(\mathbf{U}^*\ne\mathbf{U}\), so it is not self-adjoint.
Check: The first product has entries \(0\cdot0+1\cdot1=1\), \(0\cdot(-1)+1\cdot0=0\), \((-1)\cdot0+0\cdot1=0\), and \((-1)(-1)+0\cdot0=1\). The reverse product gives the same four entries.
A Direct Test for Nonnormality
To show that a matrix is not normal, it is enough to find one entry where \(\mathbf{A}^*\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^*\) differ. The order matters even for a small triangular matrix.
Worked Example: An Upper-Triangular Nonnormal Matrix
Test \(\mathbf{A}=\begin{bmatrix}1&1\\0&1\end{bmatrix}\) for normality.
The entries are real, so the adjoint is the transpose:
Compute the first product:
Compute the product in the opposite order:
The top-left entries are \(1\) and \(2\), so the products are unequal. Therefore \(\mathbf{A}\) is not normal.
Check: The two products have different diagonal entries, so no equality of the full matrices is possible.
Common Pitfall: Transpose Is Not Always the Adjoint
For a real matrix, the adjoint is the transpose. For a complex matrix, it is the conjugate transpose; omitting conjugation can change the answer. For example, consider the self-adjoint matrix \(\mathbf{H}=\begin{bmatrix}2&i\\-i&1\end{bmatrix}\) above. A mistaken calculation uses only the transpose and gets
That calculation incorrectly suggests that \(\mathbf{H}\) is not self-adjoint. The correction is to conjugate first, then transpose: \(\mathbf{H}^*=\begin{bmatrix}2&i\\-i&1\end{bmatrix}=\mathbf{H}\). The same distinction applies when testing normality: the products must use \(\mathbf{A}^*\), not merely \(\mathbf{A}^{\top}\).
Normality is a condition about commutation with the adjoint, not about being equal to the adjoint. The examples show several possibilities: a diagonal matrix can be normal without being self-adjoint or unitary; a self-adjoint matrix is normal; a unitary matrix is normal but need not be self-adjoint; and an upper-triangular matrix can fail normality. These direct tests provide a reliable starting point before using more structural results about normal operators.
Check Your Understanding
For each question, use the definition and show the relevant adjoint or matrix products.
- For \(\mathbf{D}=\begin{bmatrix}3&0\\0&-i\end{bmatrix}\), determine whether it is normal, self-adjoint, and unitary.
- Compute \(\mathbf{A}^*\) for \(\mathbf{A}=\begin{bmatrix}1&i\\2&0\end{bmatrix}\).
- Verify directly that \(\begin{bmatrix}0&i\\i&0\end{bmatrix}\) is normal. Is it self-adjoint?
- For \(\mathbf{B}=\begin{bmatrix}1&0\\1&1\end{bmatrix}\), compute \(\mathbf{B}^*\mathbf{B}\) and \(\mathbf{B}\mathbf{B}^*\), then decide whether \(\mathbf{B}\) is normal.
- State why \(\mathbf{A}^{\top}\) alone does not generally give the adjoint of a complex matrix \(\mathbf{A}\).
