Why Adjoint Laws Need Care
An adjoint turns an operator into another operator characterized by how it moves between the two slots of an inner product. Once adjoints are known to exist and be unique in the finite-dimensional inner product spaces considered here, their algebraic rules let us calculate adjoints of complicated expressions from the adjoints of their parts. The main source of care is the complex scalar field: with our convention that the inner product is linear in its first slot, taking the adjoint conjugates scalar coefficients.
Throughout, spaces are finite-dimensional inner product spaces over \(\mathbb{C}\). The inner product \(\langle\mathbf{x},\mathbf{y}\rangle\) is linear in \(\mathbf{x}\), conjugate-linear in \(\mathbf{y}\), and satisfies \(\langle\mathbf{x},\mathbf{y}\rangle=\overline{\langle\mathbf{y},\mathbf{x}\rangle}\). For a linear map \(T:U\to V\), its adjoint is a linear map \(T^*:V\to U\) satisfying the defining identity below. Existence and uniqueness of this map are established in the earlier material on adjoints.
The uniqueness in this definition can also be checked directly. If \(Q,R:V\to U\) both satisfy the defining identity, then \(\langle\mathbf{x},(Q-R)\mathbf{y}\rangle_U=0\) for every \(\mathbf{x}\in U\) and \(\mathbf{y}\in V\). For fixed \(\mathbf{y}\), choose \(\mathbf{x}=(Q-R)\mathbf{y}\). The result is \(\|(Q-R)\mathbf{y}\|^2=0\), so \((Q-R)\mathbf{y}=\mathbf{0}\). This holds for every \(\mathbf{y}\), hence \(Q=R\). We will use this uniqueness when a proposed map is shown to satisfy the defining identity.
The Algebraic Laws
For two maps to be added, they must have the same domain and codomain. For a composition \(ST\), the codomain of \(T\) must equal the domain of \(S\); here \(ST\) means \(S\circ T\), so \(T\) acts first. The adjoint reverses that order.
Proof. For the sum, take \(\mathbf{x}\in U\) and \(\mathbf{y}\in V\). Linearity in the first slot and the defining identities for \(S^*\) and \(T^*\) give \(\langle(S+T)\mathbf{x},\mathbf{y}\rangle_V=\langle S\mathbf{x},\mathbf{y}\rangle_V+\langle T\mathbf{x},\mathbf{y}\rangle_V=\langle\mathbf{x},(S^*+T^*)\mathbf{y}\rangle_U\). Uniqueness of the adjoint gives \((S+T)^*=S^*+T^*\).
For scalar multiplication, the same definition and the two-slot scalar rules give \(\langle(cT)\mathbf{x},\mathbf{y}\rangle_V=c\langle T\mathbf{x},\mathbf{y}\rangle_V=c\langle\mathbf{x},T^*\mathbf{y}\rangle_U=\langle\mathbf{x},\overline{c}\,T^*\mathbf{y}\rangle_U\). Thus \((cT)^*=\overline{c}\,T^*\) by uniqueness. For composition, take \(\mathbf{x}\in U\) and \(\mathbf{z}\in W\). Applying the defining identity first to \(S\), then to \(T\), yields \(\langle ST\mathbf{x},\mathbf{z}\rangle_W=\langle T\mathbf{x},S^*\mathbf{z}\rangle_V=\langle\mathbf{x},T^*S^*\mathbf{z}\rangle_U\). Therefore \((ST)^*=T^*S^*\).
Finally, apply the defining identity for \(T^*:V\to U\) to \(\mathbf{y}\in V\) and \(\mathbf{x}\in U\): \(\langle T^*\mathbf{y},\mathbf{x}\rangle_U=\langle\mathbf{y},T^{**}\mathbf{x}\rangle_V\). Conjugate symmetry and the defining identity for \(T\) also give \(\langle T^*\mathbf{y},\mathbf{x}\rangle_U=\overline{\langle\mathbf{x},T^*\mathbf{y}\rangle_U}=\overline{\langle T\mathbf{x},\mathbf{y}\rangle_V}=\langle\mathbf{y},T\mathbf{x}\rangle_V\). Consequently \(\langle\mathbf{y},T^{**}\mathbf{x}-T\mathbf{x}\rangle_V=0\) for every \(\mathbf{y}\in V\). Choosing \(\mathbf{y}=T^{**}\mathbf{x}-T\mathbf{x}\) shows that this difference has squared norm zero, so it is \(\mathbf{0}\). This holds for every \(\mathbf{x}\in U\), proving \(T^{**}=T\). \(\square\)
For matrices in orthonormal bases, the adjoint is the conjugate transpose: if \(T\) has matrix \(\mathbf{A}\), then \(T^*\) has matrix \(\mathbf{A}^*=\overline{\mathbf{A}}^{\top}\). Thus the operator laws become \((\mathbf{A}+\mathbf{B})^*=\mathbf{A}^*+\mathbf{B}^*\), \((c\mathbf{A})^*=\overline{c}\mathbf{A}^*\), and \((\mathbf{B}\mathbf{A})^*=\mathbf{A}^*\mathbf{B}^*\). The matrix formula here assumes orthonormal bases.
Worked Matrix Calculations
Worked Example: A Sum and a Complex Scalar Multiple
Let \(\mathbf{A}=\begin{bmatrix}1+i&2\\i&1\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}i&1-i\\2&-i\end{bmatrix}\). Calculate the adjoint of their sum and compare it with the sum of their adjoints. Then calculate the adjoint of \((2-i)\mathbf{A}\).
Conjugate each entry and then transpose to get \(\mathbf{A}^*=\begin{bmatrix}1-i&-i\\2&1\end{bmatrix}\) and \(\mathbf{B}^*=\begin{bmatrix}-i&2\\1+i&i\end{bmatrix}\). Adding the original matrices gives \(\mathbf{A}+\mathbf{B}=\begin{bmatrix}1+2i&3-i\\2+i&1-i\end{bmatrix}\), so \((\mathbf{A}+\mathbf{B})^*=\begin{bmatrix}1-2i&2-i\\3+i&1+i\end{bmatrix}\). The separate adjoints give \(\mathbf{A}^*+\mathbf{B}^*=\begin{bmatrix}1-i-i&-i+2\\2+1+i&1+i\end{bmatrix}=\begin{bmatrix}1-2i&2-i\\3+i&1+i\end{bmatrix}\), as required.
Next, multiply each entry of \(\mathbf{A}\) by \(2-i\): \((2-i)\mathbf{A}=\begin{bmatrix}(2-i)(1+i)&(2-i)2\\(2-i)i&2-i\end{bmatrix}=\begin{bmatrix}3+i&4-2i\\1+2i&2-i\end{bmatrix}\). Its adjoint is \(\begin{bmatrix}3-i&1-2i\\4+2i&2+i\end{bmatrix}\). On the other hand, \(\overline{2-i}=2+i\), and \((2+i)\mathbf{A}^*=\begin{bmatrix}(2+i)(1-i)&(2+i)(-i)\\(2+i)2&2+i\end{bmatrix}=\begin{bmatrix}3-i&1-2i\\4+2i&2+i\end{bmatrix}\). The two results agree, verifying the scalar law in this example.
Worked Example: The Adjoint of a Composition
Let \(\mathbf{T}=\begin{bmatrix}1&i\\0&1-i\end{bmatrix}\) and \(\mathbf{S}=\begin{bmatrix}i&1\\1&0\end{bmatrix}\). Compute \((\mathbf{S}\mathbf{T})^*\) and \(\mathbf{T}^*\mathbf{S}^*\). This checks both the reversed order and the complex conjugation.
First multiply in the composition order: \(\mathbf{S}\mathbf{T}=\begin{bmatrix}i&1\\1&0\end{bmatrix}\begin{bmatrix}1&i\\0&1-i\end{bmatrix}=\begin{bmatrix}i&i^2+1-i\\1&i\end{bmatrix}=\begin{bmatrix}i&-i\\1&i\end{bmatrix}\). Taking the conjugate transpose gives \((\mathbf{S}\mathbf{T})^*=\begin{bmatrix}-i&1\\i&-i\end{bmatrix}\).
The individual adjoints are \(\mathbf{T}^*=\begin{bmatrix}1&0\\-i&1+i\end{bmatrix}\) and \(\mathbf{S}^*=\begin{bmatrix}-i&1\\1&0\end{bmatrix}\). Multiplying in the reversed order gives \(\mathbf{T}^*\mathbf{S}^*=\begin{bmatrix}1&0\\-i&1+i\end{bmatrix}\begin{bmatrix}-i&1\\1&0\end{bmatrix}=\begin{bmatrix}-i&1\\(-i)(-i)+(1+i)&-i\end{bmatrix}=\begin{bmatrix}-i&1\\i&-i\end{bmatrix}\). Here \((-i)(-i)+(1+i)=-1+1+i=i\). This agrees entry by entry with \((\mathbf{S}\mathbf{T})^*\).
Worked Example: A Rectangular Matrix and Its Double Adjoint
Consider the linear map \(\mathbf{A}:\mathbb{C}^3\to\mathbb{C}^2\) with matrix \(\mathbf{A}=\begin{bmatrix}1+i&0&2\\i&1-i&-1\end{bmatrix}\) in the standard orthonormal bases. Calculate \(\mathbf{A}^*\) and \(\mathbf{A}^{**}\), then check the defining inner-product identity on specific vectors.
Conjugating each entry and transposing gives \(\mathbf{A}^*=\begin{bmatrix}1-i&-i\\0&1+i\\2&-1\end{bmatrix}\). Conjugating and transposing again returns \(\mathbf{A}^{**}=\begin{bmatrix}1+i&0&2\\i&1-i&-1\end{bmatrix}=\mathbf{A}\). In particular, the entry \(i\) in row 2, column 1 becomes \(-i\) in row 1, column 2 of \(\mathbf{A}^*\), then returns to \(i\) in the second adjoint.
For a direct check, choose \(\mathbf{x}=(1,i,1)^\top\in\mathbb{C}^3\) and \(\mathbf{y}=(1-i,2)^\top\in\mathbb{C}^2\). Multiplication gives \(\mathbf{A}\mathbf{x}=(3+i,2i)^\top\): the second entry is \(i+(1-i)i-1=i+(1+i)-1=2i\). Since the inner product is linear in the first slot, \(\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle=(3+i)(1+i)+(2i)2=2+4i+4i=2+8i\). Also, \(\mathbf{A}^*\mathbf{y}=(-4i,\,2+2i,\,-2i)^\top\), so \(\langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle=1(4i)+i(2-2i)+1(2i)=4i+(2+2i)+2i=2+8i\). The two inner products agree, as the defining identity requires.
Common Mistake: Forgetting to Conjugate the Scalar
A frequent error is to write \((c\mathbf{A})^*=c\mathbf{A}^*\) for a complex scalar \(c\). For a concrete countercheck, take \(c=i\) and \(\mathbf{A}=\begin{bmatrix}1&i\\0&1\end{bmatrix}\). Then \(i\mathbf{A}=\begin{bmatrix}i&-1\\0&i\end{bmatrix}\), and therefore \((i\mathbf{A})^*=\begin{bmatrix}-i&0\\-1&-i\end{bmatrix}\). But \(\mathbf{A}^*=\begin{bmatrix}1&0\\-i&1\end{bmatrix}\), so the incorrect calculation gives \(i\mathbf{A}^*=\begin{bmatrix}i&0\\1&i\end{bmatrix}\), which is not the adjoint just computed. The correction is to conjugate the scalar: \(\overline{i}\mathbf{A}^*=-i\mathbf{A}^*=\begin{bmatrix}-i&0\\-1&-i\end{bmatrix}\). For real scalars the conjugate does not change the scalar, which can hide this error; for complex scalars, the conjugation is essential.
There are two further checks that help prevent mistakes. First, for a product, reverse the order as well as taking adjoints: the adjoint of \(ST\) is \(T^*S^*\), not \(S^*T^*\). Second, keep the map types visible. If \(T:U\to V\), then \(T^*:V\to U\), so a proposed identity must have matching domains and codomains. These checks are useful before any entries are calculated.
Using the Laws Efficiently
For a sum of several maps and scalar multiples, apply the sum and scalar laws term by term. For a product of several compatible maps, reverse the entire order. For example, for \(R:W\to X\), \(S:V\to W\), and \(T:U\to V\), repeated application gives \((RST)^*=T^*S^*R^*\). Indeed, \((RST)^*=T^*(RS)^*=T^*S^*R^*\), where the first equality applies the composition law to \(R S\) composed after \(T\), and the second applies it to \(RS\). The domains then line up from \(X\) back to \(U\).
These identities are not separate computational tricks: each follows from the same defining inner-product equation and the uniqueness of the adjoint. That perspective helps distinguish the roles of the laws. Additions remain in the same order, scalar coefficients are conjugated, compositions reverse order, and applying the adjoint twice returns to the original map. In a calculation, checking those four features before simplifying entries often catches errors early.
Check Your Understanding
Use the defining identity or the conjugate-transpose formula in orthonormal bases to justify each answer.
- If \(T:U\to V\) and \(c\in\mathbb{C}\), what is the adjoint of \(cT\)? Explain why the scalar changes.
- Let \(\mathbf{A}=\begin{bmatrix}1&i\\2&0\end{bmatrix}\) and \(\mathbf{B}=\begin{bmatrix}i&1\\0&1\end{bmatrix}\). Calculate \((\mathbf{A}+\mathbf{B})^*\) and \(\mathbf{A}^*+\mathbf{B}^*\).
- If \(T:U\to V\) and \(S:V\to W\), state the domain and codomain of \(T^*S^*\) and identify the adjoint law it illustrates.
- For \(\mathbf{C}=\begin{bmatrix}1+i&2\\0&i\end{bmatrix}\), calculate \((i\mathbf{C})^*\) and compare it with \(i\mathbf{C}^*\).
- Give the defining-inner-product argument that \(T^{**}=T\) for a map \(T:U\to V\).
