Adjoints in Complex Coordinates
For a complex matrix, transposing its entries is not enough to compute its adjoint: each entry must also be conjugated. That distinction affects every test for self-adjointness, unitarity, and normality. The most reliable hand calculation is therefore systematic: compute the conjugate transpose first, then calculate the products required by the definition, and compare every entry.
We work over \(\mathbb{C}\), with the standard inner product on column vectors defined to be linear in its first slot: \(\langle\mathbf{x},\mathbf{y}\rangle=\sum_{j=1}^n x_j\overline{y_j}\). For a scalar \(z=a+bi\), its complex conjugate is \(\overline{z}=a-bi\). For a matrix \(\mathbf{A}\), the conjugate transpose is obtained by transposing and conjugating each entry.
These definitions ask different questions. Self-adjointness compares the matrix directly with its adjoint. Unitarity requires both products to equal the identity. Normality asks only that the two products agree; their common value need not be the identity. A matrix can satisfy more than one of these properties, so test each requested property against its own definition.
Proof. For the product identity, the \((i,j)\)-entry of \((\mathbf{A}\mathbf{B})^{*}\) is the conjugate of the \((j,i)\)-entry of \(\mathbf{A}\mathbf{B}\), so it is \(\overline{\sum_{k=1}^{n}a_{jk}b_{ki}}=\sum_{k=1}^{n}\overline{b_{ki}}\,\overline{a_{jk}}\). The \((i,j)\)-entry of \(\mathbf{B}^{*}\mathbf{A}^{*}\) is exactly \(\sum_{k=1}^{n}(\mathbf{B}^{*})_{ik}(\mathbf{A}^{*})_{kj} =\sum_{k=1}^{n}\overline{b_{ki}}\,\overline{a_{jk}}\). Thus the entries agree. Applying conjugation and transposition twice returns every entry to its original position and value, proving \((\mathbf{A}^{*})^{*}=\mathbf{A}\). If \(\mathbf{A}^{*}=\mathbf{A}\), then \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\), so \(\mathbf{A}\) is normal. \(\square\)
The product identity explains a useful feature of adjoints: the order reverses. It also justifies the standard connection with inner products. If \(\mathbf{A}\) is an \(m\times n\) matrix, then for \(\mathbf{x}\in\mathbb{C}^{n}\) and \(\mathbf{y}\in\mathbb{C}^{m}\), \(\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle=\langle\mathbf{x},\mathbf{A}^{*}\mathbf{y}\rangle\). Indeed, expanding the left side gives \(\sum_i(\sum_j a_{ij}x_j)\overline{y_i}\), which equals \(\sum_jx_j\overline{\sum_i\overline{a_{ij}}y_i}\), the right side. This is why \(\mathbf{A}^{*}\) represents the adjoint map in standard complex coordinates.
A Reliable Hand-Calculation Procedure
For an \(n\times n\) matrix, use the same sequence whether the question asks about one property or several:
Transpose the matrix, then conjugate each entry. Keep the result visible so that conjugation is not silently lost in later products.
Calculate \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\) when testing normality or unitarity. Their order matters.
For normality, compare the two products with each other. For unitarity, check that each product is \(\mathbf{I}_n\). For self-adjointness, compare \(\mathbf{A}^{*}\) directly with \(\mathbf{A}\).
To check an asserted eigenpair \((\lambda,\mathbf{v})\), calculate \(\mathbf{A}\mathbf{v}\) and confirm it equals \(\lambda\mathbf{v}\), with \(\mathbf{v}\ne\mathbf{0}\). Solve \((\mathbf{A}-\lambda\mathbf{I}_n)\mathbf{x}=\mathbf{0}\) to describe the full eigenspace.
Worked Example: A Matrix That Is Not Normal
Let \(\mathbf{A}=\begin{bmatrix}1&i\\0&2\end{bmatrix}\). Compute its adjoint, test normality, and find the eigenspace for each of the eigenvalues \(1\) and \(2\).
Transpose first and conjugate the entries \(i\) and \(-i\) as needed:
Now multiply in each order. For \(\mathbf{A}^{*}\mathbf{A}\), the rows of \(\mathbf{A}^{*}\) are \((1,0)\) and \((-i,2)\), while the columns of \(\mathbf{A}\) are \((1,0)^{\top}\) and \((i,2)^{\top}\). For \(\mathbf{A}\mathbf{A}^{*}\), use the columns of \(\mathbf{A}^{*}\), \((1,-i)^{\top}\) and \((0,2)^{\top}\):
The products differ, for example in their \((1,1)\)-entries, so \(\mathbf{A}\) is not normal. It is not self-adjoint either, since \(\mathbf{A}^{*}\ne\mathbf{A}\).
For \(\lambda=1\), solve \((\mathbf{A}-\mathbf{I}_2)\mathbf{x}=\mathbf{0}\). Writing \(\mathbf{x}=(x,y)^{\top}\), the equations are \(iy=0\) and \(y=0\), so \(y=0\) and \(x\) is free. Hence \(E_1=\operatorname{span}\{(1,0)^{\top}\}\). For \(\lambda=2\), the equations from \((\mathbf{A}-2\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) reduce to \(-x+iy=0\), so \(x=iy\) and \(E_2=\operatorname{span}\{(i,1)^{\top}\}\).
Check: \(\mathbf{A}(1,0)^{\top}=(1,0)^{\top}=1(1,0)^{\top}\), and \(\mathbf{A}(i,1)^{\top}=(2i,2)^{\top}=2(i,1)^{\top}\). Both spanning vectors are nonzero and satisfy the eigenvector equation.
Unitary and Self-Adjoint Examples
Worked Example: Unitary but Not Self-Adjoint
Consider \(\mathbf{B}=\begin{bmatrix}0&i\\i&0\end{bmatrix}\). Compute \(\mathbf{B}^{*}\), classify \(\mathbf{B}\), and find the eigenspaces for \(\lambda=i\) and \(\lambda=-i\).
Transposing leaves the matrix unchanged, and conjugation changes each \(i\) to \(-i\), giving
Since \(\mathbf{B}^{*}\ne\mathbf{B}\), it is not self-adjoint. The products are
Both products equal \(\mathbf{I}_2\), so \(\mathbf{B}\) is unitary and, because the products agree, normal. To find \(E_i\), solve \((\mathbf{B}-i\mathbf{I}_2)\mathbf{x}=\mathbf{0}\). Its first equation is \(-ix+iy=0\), so \(y=x\); the second equation gives the same condition. Thus \(E_i=\operatorname{span}\{(1,1)^{\top}\}\). For \(\lambda=-i\), the first equation from \(\mathbf{B}+i\mathbf{I}_2\) is \(ix+iy=0\), so \(y=-x\), giving \(E_{-i}=\operatorname{span}\{(1,-1)^{\top}\}\).
Check: \(\mathbf{B}(1,1)^{\top}=(i,i)^{\top}=i(1,1)^{\top}\), and \(\mathbf{B}(1,-1)^{\top}=(-i,i)^{\top}=-i(1,-1)^{\top}\). The displayed products also verify unitarity directly.
Worked Example: Self-Adjoint but Not Unitary
Let \(\mathbf{C}=\begin{bmatrix}2&i\\-i&2\end{bmatrix}\). Test self-adjointness, normality, and unitarity, then determine the eigenspaces for \(\lambda=3\) and \(\lambda=1\).
Conjugating the transpose gives \(\mathbf{C}^{*}=\begin{bmatrix}2&i\\-i&2\end{bmatrix}=\mathbf{C}\), so \(\mathbf{C}\) is self-adjoint. Consequently, it is normal; the products confirm this and also test unitarity:
This common product is not \(\mathbf{I}_2\), so \(\mathbf{C}\) is not unitary. For \(\lambda=3\), the first equation of \((\mathbf{C}-3\mathbf{I}_2)\mathbf{x}=\mathbf{0}\) is \(-x+iy=0\), so \(x=iy\); the second equation, \(-ix-y=0\), is then satisfied. Thus \(E_3=\operatorname{span}\{(i,1)^{\top}\}\). For \(\lambda=1\), the first equation is \(x+iy=0\), so \(x=-iy\); the second equation is then satisfied as well. Hence \(E_1=\operatorname{span}\{(-i,1)^{\top}\}\).
Check: \(\mathbf{C}(i,1)^{\top}=(3i,3)^{\top}=3(i,1)^{\top}\), and \(\mathbf{C}(-i,1)^{\top}=(-i,1)^{\top}=1(-i,1)^{\top}\). Neither vector is zero. The common product above confirms normality but disproves unitarity.
A Quick Check That Does Not Prove Normality
Normality imposes a useful necessary condition on the row and column lengths. The diagonal entry \((\mathbf{A}^{*}\mathbf{A})_{jj}\) is the sum of squared absolute values in column \(j\), while \((\mathbf{A}\mathbf{A}^{*})_{jj}\) is the sum in row \(j\). Therefore equality of the products forces corresponding row and column lengths to agree. This is a preliminary check, not a substitute for comparing all entries.
Proof. Normality means \(\mathbf{A}^{*}\mathbf{A}=\mathbf{A}\mathbf{A}^{*}\), so corresponding diagonal entries are equal. By matrix multiplication, \((\mathbf{A}^{*}\mathbf{A})_{jj}=\sum_{k=1}^{n}\overline{a_{kj}}a_{kj}=\sum_{k=1}^{n}|a_{kj}|^2\), the squared length of column \(j\). Also, \((\mathbf{A}\mathbf{A}^{*})_{jj}=\sum_{k=1}^{n}a_{jk}\overline{a_{jk}}=\sum_{k=1}^{n}|a_{jk}|^2\), the squared length of row \(j\). This proves the necessary equality. To show it is not sufficient, take \(\mathbf{D}=\begin{bmatrix}i&1\\1&0\end{bmatrix}\). Its first row and first column both have squared length \(2\); its second row and second column both have squared length \(1\). Yet \(\mathbf{D}^{*}\mathbf{D}=\begin{bmatrix}2&-i\\i&1\end{bmatrix}\) and \(\mathbf{D}\mathbf{D}^{*}=\begin{bmatrix}2&i\\-i&1\end{bmatrix}\), which are unequal. Thus matching row and column lengths does not ensure normality. \(\square\)
Common Mistakes
Transposing without conjugating. For \(\mathbf{B}=\begin{bmatrix}0&i\\i&0\end{bmatrix}\), a wrong calculation is \(\mathbf{B}^{*}=\mathbf{B}^{\top}=\mathbf{B}\), which would incorrectly label \(\mathbf{B}\) self-adjoint. The correction is to conjugate after transposing: \(\mathbf{B}^{*}=\begin{bmatrix}0&-i\\-i&0\end{bmatrix}\ne\mathbf{B}\). The error changes the classification even though the matrix happens to be symmetric under ordinary transposition.
Comparing only one product with the identity. The definition of unitarity requires \(\mathbf{A}^{*}\mathbf{A}=\mathbf{I}_n\) and \(\mathbf{A}\mathbf{A}^{*}=\mathbf{I}_n\). Do not replace this test with normality: normality requires the two products to match, not to equal the identity. Likewise, a product that is not the identity shows a matrix is not unitary, but does not by itself show that the matrix is not normal.
Reversing the product order without changing the factors. The adjoint of a product is \((\mathbf{A}\mathbf{B})^{*}=\mathbf{B}^{*}\mathbf{A}^{*}\), not \(\mathbf{A}^{*}\mathbf{B}^{*}\). For direct tests of normality, calculate both \(\mathbf{A}^{*}\mathbf{A}\) and \(\mathbf{A}\mathbf{A}^{*}\) as written and compare their entries.
Check Your Understanding
Use the definitions to classify each matrix, show the products needed for your conclusion, and verify any claimed eigenvectors by multiplication.
- Compute the adjoint of \(\begin{bmatrix}1&2i\\-i&3\end{bmatrix}\). Is the matrix self-adjoint?
- For \(\begin{bmatrix}0&-1\\1&0\end{bmatrix}\), compute both products with its adjoint and determine whether it is unitary and normal.
- Show that \(\begin{bmatrix}1&i\\0&1\end{bmatrix}\) is not normal by finding one entry where its two products differ.
- For \(\begin{bmatrix}2&0\\0&-1\end{bmatrix}\), find \(E_2\) and \(E_{-1}\), and check each eigenspace description from the defining equation.
- Explain why equal corresponding row and column lengths are necessary, but not sufficient, for normality.
