The Adjoint as a Defining Identity
For a linear operator \(T\), the adjoint is another operator that transfers \(T\) from one slot of an inner product to the other. On a complex inner product space, this transfer involves complex conjugation. The goal here is to establish that the transfer is always possible in finite dimensions and that there is only one operator that performs it.
Throughout, \(V\) is a finite-dimensional vector space over \(\mathbb{C}\), and its inner product is linear in the first slot and conjugate-linear in the second. Thus \(\langle a\mathbf{x},\mathbf{y}\rangle=a\langle\mathbf{x},\mathbf{y}\rangle\) and \(\langle\mathbf{x},a\mathbf{y}\rangle=\overline{a}\langle\mathbf{x},\mathbf{y}\rangle\). We also use conjugate symmetry, \(\langle\mathbf{x},\mathbf{y}\rangle=\overline{\langle\mathbf{y},\mathbf{x}\rangle}\), and positive definiteness, which says that \(\langle\mathbf{x},\mathbf{x}\rangle=0\) exactly when \(\mathbf{x}=\mathbf{0}\).
The equality must hold for every pair of vectors, not merely for a chosen basis pair or a single test vector. Its order of arguments matters: with our convention, the first slot is linear and the second slot is conjugate-linear. We will construct an operator satisfying the identity, then show that no two different operators can satisfy it.
Orthonormal Coordinates in Finite Dimensions
An ordered list \(\mathcal{E}=(\mathbf{e}_1,\dots,\mathbf{e}_n)\) is an orthonormal basis when it is a basis and \(\langle\mathbf{e}_i,\mathbf{e}_j\rangle=1\) if \(i=j\), and \(0\) otherwise. Every finite-dimensional complex inner product space has such a basis. For completeness, this follows by orthogonalizing any basis.
Proof. We use induction on \(n=\dim V\). If \(n=0\), the empty list is an orthonormal basis. If \(n\geq1\), choose a basis \((\mathbf{v}_1,\dots,\mathbf{v}_n)\). Set \(\mathbf{u}_1=\mathbf{v}_1/\|\mathbf{v}_1\|\); this is defined because \(\mathbf{v}_1\ne\mathbf{0}\). Suppose \(\mathbf{u}_1,\dots,\mathbf{u}_{k-1}\) have been constructed as orthonormal vectors spanning the same subspace as \(\mathbf{v}_1,\dots,\mathbf{v}_{k-1}\). Define $$ \mathbf{w}_k=\mathbf{v}_k-\sum_{j=1}^{k-1}\langle\mathbf{v}_k,\mathbf{u}_j\rangle\mathbf{u}_j. $$ For \(1\leq\ell<k\), orthonormality gives \(\langle\mathbf{w}_k,\mathbf{u}_\ell\rangle=\langle\mathbf{v}_k,\mathbf{u}_\ell\rangle-\langle\mathbf{v}_k,\mathbf{u}_\ell\rangle=0\). Also, \(\mathbf{w}_k\ne\mathbf{0}\), since otherwise \(\mathbf{v}_k\) would be in the span of the preceding basis vectors, contradicting their linear independence. Therefore \(\mathbf{u}_k=\mathbf{w}_k/\|\mathbf{w}_k\|\) is defined, has norm \(1\), and is orthogonal to the preceding vectors. The vectors constructed through step \(k\) span the same space as \(\mathbf{v}_1,\dots,\mathbf{v}_k\), by the displayed definition of \(\mathbf{w}_k\). Induction produces an orthonormal basis of \(V\). \(\square\)
In an orthonormal basis \(\mathcal{E}\), write \([\mathbf{x}]_{\mathcal{E}}\) for the column of coordinates of \(\mathbf{x}\). If \([\mathbf{x}]_{\mathcal{E}}=\mathbf{x}_c\) and \([\mathbf{y}]_{\mathcal{E}}=\mathbf{y}_c\), then $$ \langle\mathbf{x},\mathbf{y}\rangle=\mathbf{x}_c^{\top}\overline{\mathbf{y}_c}. $$ Indeed, expanding both vectors in the basis leaves only terms with matching basis indices. This coordinate formula is the key to constructing the adjoint.
Constructing the Adjoint from a Matrix
Choose an orthonormal basis \(\mathcal{E}=(\mathbf{e}_1,\dots,\mathbf{e}_n)\) of \(V\). Write the matrix of \(T\) in this basis as \(\mathbf{A}=(a_{ij})\), meaning that \(T\mathbf{e}_j=\sum_{i=1}^n a_{ij}\mathbf{e}_i\). The conjugate transpose \(\mathbf{A}^*\) has entries \((\mathbf{A}^*)_{ij}=\overline{a_{ji}}\). We use this matrix to define an operator \(S\) by prescribing its action on the basis: $$ S\mathbf{e}_j=\sum_{i=1}^n\overline{a_{ji}}\mathbf{e}_i \qquad (1\leq j\leq n). $$ There is a unique linear operator with these values on a basis. Its matrix in \(\mathcal{E}\) is \(\mathbf{A}^*\).
Proof. First suppose \(n>0\), choose an orthonormal basis as above, and let \(\mathbf{x}_c=(x_1,\dots,x_n)^{\top}\) and \(\mathbf{y}_c=(y_1,\dots,y_n)^{\top}\). The coordinates of \(T\mathbf{x}\) are \(\mathbf{A}\mathbf{x}_c\); the coordinates of \(S\mathbf{y}\) are \(\mathbf{A}^*\mathbf{y}_c\). Using the coordinate formula for the inner product, $$ \langle T\mathbf{x},\mathbf{y}\rangle =(\mathbf{A}\mathbf{x}_c)^{\top}\overline{\mathbf{y}_c} =\mathbf{x}_c^{\top}\mathbf{A}^{\top}\overline{\mathbf{y}_c}. $$ On the other hand, $$ \langle\mathbf{x},S\mathbf{y}\rangle =\mathbf{x}_c^{\top}\overline{\mathbf{A}^*\mathbf{y}_c} =\mathbf{x}_c^{\top}\mathbf{A}^{\top}\overline{\mathbf{y}_c}, $$ because \(\overline{\mathbf{A}^*}=\mathbf{A}^{\top}\). Thus \(S\) satisfies the defining identity for every \(\mathbf{x},\mathbf{y}\), so an adjoint exists. For uniqueness, suppose \(S\) and \(U\) both satisfy that identity. Fix any \(\mathbf{y}\in V\), and set \(\mathbf{z}=(S-U)\mathbf{y}\). Subtracting their two identities gives \(\langle\mathbf{x},\mathbf{z}\rangle=0\) for every \(\mathbf{x}\in V\). In particular, take \(\mathbf{x}=\mathbf{z}\). Then \(\langle\mathbf{z},\mathbf{z}\rangle=0\), so positive definiteness implies \(\mathbf{z}=\mathbf{0}\). This holds for every \(\mathbf{y}\), hence \(S=U\). If \(n=0\), the only vector in \(V\) is \(\mathbf{0}\) and the only map \(V\to V\) is the zero map. It satisfies the identity, and no other map exists. This also establishes the theorem in dimension zero. \(\square\)
Worked Examples in Complex Coordinates
Worked Example: Verifying an Adjoint in \(\mathbb{C}^2\)
Use the standard inner product \(\langle\mathbf{x},\mathbf{y}\rangle=x_1\overline{y_1}+x_2\overline{y_2}\). Let \(T\) have matrix $$ \mathbf{A}=\begin{bmatrix}1+i&2\\i&1-i\end{bmatrix}. $$ Construct its adjoint and verify the defining identity on a pair of vectors.
Taking the conjugate transpose gives $$ \mathbf{A}^*=\begin{bmatrix}1-i&-i\\2&1+i\end{bmatrix}. $$ Choose \(\mathbf{x}=(1+i,\,2-i)^{\top}\) and \(\mathbf{y}=(i,\,1)^{\top}\). First compute \(T\mathbf{x}\): $$ \mathbf{A}\mathbf{x} =\begin{bmatrix}(1+i)(1+i)+2(2-i)\\ i(1+i)+(1-i)(2-i)\end{bmatrix} =\begin{bmatrix}2i+4-2i\\(-1+i)+(1-3i)\end{bmatrix} =\begin{bmatrix}4\\-2i\end{bmatrix}. $$ Therefore \(\langle T\mathbf{x},\mathbf{y}\rangle=4\overline{i}+(-2i)\overline{1}=-4i-2i=-6i\).
Now compute the candidate adjoint applied to \(\mathbf{y}\): $$ \mathbf{A}^*\mathbf{y} =\begin{bmatrix}(1-i)i-i\\2i+1+i\end{bmatrix} =\begin{bmatrix}1\\1+3i\end{bmatrix}. $$ It follows that $$ \langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle =(1+i)\overline{1}+(2-i)\overline{(1+3i)} =(1+i)+(2-i)(1-3i) =(1+i)+(-1-7i) =-6i. $$ The two inner products agree, as required.
Worked Example: Constructing the Adjoint in a Different Orthonormal Basis
Let \(\mathcal{E}=(\mathbf{q}_1,\mathbf{q}_2)\) be an orthonormal basis of a complex inner product space, and define \(T\mathbf{q}_1=\mathbf{q}_2\), \(T\mathbf{q}_2=i\mathbf{q}_1\). Find \(T^*\), then check the identity for particular vectors.
The columns of the matrix of \(T\) are the coordinates of \(T\mathbf{q}_1\) and \(T\mathbf{q}_2\), so $$ [T]_{\mathcal{E}}=\begin{bmatrix}0&i\\1&0\end{bmatrix}, \qquad [T^*]_{\mathcal{E}}=\begin{bmatrix}0&1\\-i&0\end{bmatrix}. $$ Thus \(T^*\mathbf{q}_1=-i\mathbf{q}_2\) and \(T^*\mathbf{q}_2=\mathbf{q}_1\). Take \(\mathbf{x}=2\mathbf{q}_1+(1-i)\mathbf{q}_2\) and \(\mathbf{y}=i\mathbf{q}_1+\mathbf{q}_2\). Then $$ T\mathbf{x}=2\mathbf{q}_2+(1-i)i\mathbf{q}_1 =(1+i)\mathbf{q}_1+2\mathbf{q}_2, \qquad T^*\mathbf{y}=i(-i)\mathbf{q}_2+\mathbf{q}_1 =\mathbf{q}_1+\mathbf{q}_2. $$ Using orthonormality and conjugating the coefficients in the second slot, $$ \langle T\mathbf{x},\mathbf{y}\rangle =(1+i)\overline{i}+2\overline{1} =(1+i)(-i)+2=3-i, $$ while $$ \langle\mathbf{x},T^*\mathbf{y}\rangle =2\overline{1}+(1-i)\overline{1} =3-i. $$ The construction works in this orthonormal basis just as it does in standard coordinates.
Uniqueness in Practice and a Common Pitfall
The uniqueness proof gives a useful way to detect an incorrect candidate: find vectors for which the defining identity fails. In particular, for a complex matrix, merely transposing without conjugating generally gives the wrong operator. Conjugation is required because scalars in the second slot of the inner product are conjugated.
Worked Example: Why the Ordinary Transpose Can Fail
On \(\mathbb{C}^2\) with the standard inner product, let $$ \mathbf{A}=\begin{bmatrix}1&i\\0&2\end{bmatrix}. $$ The correct candidate is \(\mathbf{A}^*=\begin{bmatrix}1&0\\-i&2\end{bmatrix}\), not the ordinary transpose \(\mathbf{A}^{\top}=\begin{bmatrix}1&0\\i&2\end{bmatrix}\). Test the ordinary transpose using \(\mathbf{x}=\mathbf{e}_2\) and \(\mathbf{y}=\mathbf{e}_1\). We have \(T\mathbf{e}_2=i\mathbf{e}_1+2\mathbf{e}_2\), so $$ \langle T\mathbf{e}_2,\mathbf{e}_1\rangle=i. $$ But \(\mathbf{A}^{\top}\mathbf{e}_1=\mathbf{e}_1+i\mathbf{e}_2\), and therefore $$ \langle\mathbf{e}_2,\mathbf{A}^{\top}\mathbf{e}_1\rangle =\overline{i}=-i. $$ Since \(i\ne-i\), the transpose fails the defining identity. With the conjugate transpose, \(\mathbf{A}^*\mathbf{e}_1=\mathbf{e}_1-i\mathbf{e}_2\), giving $$ \langle\mathbf{e}_2,\mathbf{A}^*\mathbf{e}_1\rangle =\overline{-i}=i =\langle T\mathbf{e}_2,\mathbf{e}_1\rangle. $$ The conjugation changes the result to the value required by the identity.
For a proof that an operator is the adjoint, it is not enough to observe that its matrix looks plausible. One must establish the identity for all pairs of vectors. The coordinate calculation in the existence proof does exactly that: it begins with arbitrary coordinate columns and shows that both inner products reduce to the same expression. Conversely, the uniqueness argument depends on positive definiteness: if the difference between two candidates sends \(\mathbf{y}\) to \(\mathbf{z}\), testing the identity with \(\mathbf{x}=\mathbf{z}\) forces \(\mathbf{z}=\mathbf{0}\).
The result is specific to finite-dimensional spaces over \(\mathbb{C}\) equipped with an inner product. Finite dimensionality supplies an orthonormal basis, which makes the construction possible; the inner product’s positive definiteness supplies uniqueness. The conjugate transpose formula is valid when the matrix is written in an orthonormal basis. This qualification matters: the same coordinate rule should not be applied to an arbitrary basis without accounting for the inner product in those coordinates.
Check Your Understanding
Use the linear-first convention throughout. When checking a proposed adjoint, evaluate the inner products with the second argument conjugated.
- For \(\mathbf{A}=\begin{bmatrix}2&i\\1&1-i\end{bmatrix}\) in standard coordinates on \(\mathbb{C}^2\), write the matrix of the adjoint.
- Explain why \(\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{x}^{\top}\overline{\mathbf{y}}\) in orthonormal coordinates.
- For the operator in the second worked example, find \(T^*\mathbf{q}_1\) and \(T^*\mathbf{q}_2\) from its matrix.
- In the uniqueness proof, why is it valid to choose \(\mathbf{x}=(S-U)\mathbf{y}\)? What property of the inner product then gives \((S-U)\mathbf{y}=\mathbf{0}\)?
- Give a reason that an ordinary transpose need not satisfy the adjoint identity for a complex matrix.
