The Adjoint Identity on \(\mathbb{C}^n\)
An adjoint transfers a linear operator from one side of an inner product to the other. Over \(\mathbb{C}\), this transfer must account for complex conjugation as well as the change from rows to columns. The exact placement of that conjugation depends on the inner-product convention, so we begin by fixing the convention used here: the inner product is linear in its first argument and conjugate-linear in its second.
Vectors in \(\mathbb{C}^n\) are columns. For \(\mathbf{x},\mathbf{y}\in\mathbb{C}^n\), the standard inner product is \(\langle\mathbf{x},\mathbf{y}\rangle=\sum_{j=1}^n x_j\overline{y_j}\). For a matrix \(\mathbf{A}\), its conjugate transpose \(\mathbf{A}^*\) is obtained by transposing the matrix and then taking the complex conjugate of every entry. These two operations are both important.
In this tutorial we work with the standard inner product and standard coordinates. If \(\mathbf{A}\in\mathbb{C}^{n\times n}\) represents \(T\), then \(T\mathbf{x}=\mathbf{A}\mathbf{x}\). The matrix formula will show that \(\mathbf{A}^*\) represents an operator satisfying the adjoint identity. In particular, this gives a direct method for computing and checking adjoints in \(\mathbb{C}^n\).
Why the Conjugate Transpose Appears
First express the standard inner product using matrix multiplication. The conjugate transpose of a column \(\mathbf{y}\) is the row \(\mathbf{y}^*\), so \(\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{y}^*\mathbf{x}\). This order agrees with the convention that the first argument is linear: multiplying \(\mathbf{x}\) by a scalar \(c\) multiplies the inner product by \(c\), while multiplying \(\mathbf{y}\) by \(c\) multiplies it by \(\overline{c}\).
Proof. Assume \(\mathbf{x},\mathbf{y}\in\mathbb{C}^n\) are arbitrary. By the matrix form of the inner product, \(\langle T\mathbf{x},\mathbf{y}\rangle=\mathbf{y}^*(\mathbf{A}\mathbf{x})\). Associativity of matrix multiplication gives \(\mathbf{y}^*\mathbf{A}\mathbf{x}\). On the other hand, the conjugate transpose satisfies \((\mathbf{A}^*\mathbf{y})^*=\mathbf{y}^*(\mathbf{A}^*)^*=\mathbf{y}^*\mathbf{A}\), because taking the conjugate transpose twice returns the original matrix and reverses the order of a product. Therefore \(\langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle=(\mathbf{A}^*\mathbf{y})^*\mathbf{x}=\mathbf{y}^*\mathbf{A}\mathbf{x}\). Both inner products are equal for every \(\mathbf{x},\mathbf{y}\), which proves the claimed identity. \(\square\)
The computation uses exactly the defining inner-product convention. In the formula \(\mathbf{y}^*\mathbf{A}\mathbf{x}\), the matrix that must act on \(\mathbf{y}\) in the second slot is \(\mathbf{A}^*\), since taking the conjugate transpose of \(\mathbf{A}^*\mathbf{y}\) produces \(\mathbf{y}^*\mathbf{A}\). The formula is not obtained by transposing alone: a complex scalar in the second argument is conjugated.
Write \(T(\mathbf{x})=\mathbf{A}\mathbf{x}\) using standard coordinates.
Form \(\mathbf{A}^*=\overline{\mathbf{A}}^{\top}\): entry \(a_{ij}\) becomes \(\overline{a_{ij}}\) in position \((j,i)\).
For chosen vectors, compute both \(\langle\mathbf{A}\mathbf{x},\mathbf{y}\rangle\) and \(\langle\mathbf{x},\mathbf{A}^*\mathbf{y}\rangle\), conjugating the entries in the second argument.
Computing and Checking the Formula
Worked Example: A Full Check for a \(2\times2\) Matrix
Let \(\mathbf{A}=\begin{bmatrix}1+i&2\\ i&1-i\end{bmatrix}\). Compute its adjoint matrix, then check the adjoint identity for \(\mathbf{x}=\begin{bmatrix}1+i\\2-i\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}i\\1-i\end{bmatrix}\).
Conjugating each entry gives \(\begin{bmatrix}1-i&2\\-i&1+i\end{bmatrix}\). Transposing this result gives
First apply \(\mathbf{A}\) to \(\mathbf{x}\). Its first entry is \((1+i)(1+i)+2(2-i)=2i+4-2i=4\). Its second entry is \(i(1+i)+(1-i)(2-i)=(i-1)+(1-3i)=-2i\). Thus \(\mathbf{A}\mathbf{x}=\begin{bmatrix}4\\-2i\end{bmatrix}\), and
Now apply \(\mathbf{A}^*\) to \(\mathbf{y}\). The first entry is \((1-i)i+(-i)(1-i)=(1+i)+(-1-i)=0\). The second entry is \(2i+(1+i)(1-i)=2i+2=2+2i\). Therefore \(\mathbf{A}^*\mathbf{y}=\begin{bmatrix}0\\2+2i\end{bmatrix}\), and
The two inner products are equal, as required. This check confirms both the computed matrix and the placement of the conjugates in the inner product.
Worked Example: A Second Complex Matrix
Let \(\mathbf{B}=\begin{bmatrix}i&2\\1-i&-i\end{bmatrix}\), \(\mathbf{x}=\begin{bmatrix}1\\i\end{bmatrix}\), and \(\mathbf{y}=\begin{bmatrix}1-i\\2\end{bmatrix}\). Find \(\mathbf{B}^*\) and verify the defining identity.
Conjugating and transposing gives \(\mathbf{B}^*=\begin{bmatrix}-i&1+i\\2&i\end{bmatrix}\). Compute the first side: \(\mathbf{B}\mathbf{x}=\begin{bmatrix}i+2i\\(1-i)+(-i)i\end{bmatrix}=\begin{bmatrix}3i\\2-i\end{bmatrix}\). Since \(\overline{1-i}=1+i\),
For the other side, \(\mathbf{B}^*\mathbf{y}=\begin{bmatrix}(-i)(1-i)+(1+i)2\\2(1-i)+2i\end{bmatrix}=\begin{bmatrix}1+i\\2\end{bmatrix}\). Hence
Both sides agree. Notice that the \(i\) in the second entry of \(\mathbf{x}\) is not conjugated, because it occurs in the first argument; the second argument is the one conjugated in this convention.
Worked Example: Detecting the Need for Conjugation
Consider \(\mathbf{C}=\begin{bmatrix}1&i\\0&1-i\end{bmatrix}\). Compare its adjoint with its ordinary transpose, and use \(\mathbf{x}=\begin{bmatrix}0\\1\end{bmatrix}\), \(\mathbf{y}=\begin{bmatrix}1\\0\end{bmatrix}\) to check which matrix satisfies the adjoint identity for this pair.
The correct adjoint is \(\mathbf{C}^*=\begin{bmatrix}1&0\\-i&1+i\end{bmatrix}\). The product \(\mathbf{C}\mathbf{x}\) is the second column of \(\mathbf{C}\), so \(\mathbf{C}\mathbf{x}=\begin{bmatrix}i\\1-i\end{bmatrix}\). It follows that \(\langle\mathbf{C}\mathbf{x},\mathbf{y}\rangle=i\overline{1}+ (1-i)\overline{0}=i\).
Next, \(\mathbf{C}^*\mathbf{y}=\begin{bmatrix}1\\-i\end{bmatrix}\), so \(\langle\mathbf{x},\mathbf{C}^*\mathbf{y}\rangle=0\overline{1}+1\overline{-i}=i\). The identity holds. If we use the ordinary transpose instead, \(\mathbf{C}^{\top}\mathbf{y}=\begin{bmatrix}1\\i\end{bmatrix}\), giving \(\langle\mathbf{x},\mathbf{C}^{\top}\mathbf{y}\rangle=\overline{i}=-i\), which is not \(i\). This particular pair exposes the error made by omitting conjugation.
Common Pitfall: Transposing Without Conjugating
A frequent mistake is to use \(\mathbf{A}^{\top}\) as the adjoint matrix over \(\mathbb{C}\). The wrong calculation in the last example is concrete: it gives \(\langle\mathbf{x},\mathbf{C}^{\top}\mathbf{y}\rangle=-i\), while the required left side is \(i\). The correction is to conjugate every matrix entry as well as transpose, producing \(\mathbf{C}^*=\begin{bmatrix}1&0\\-i&1+i\end{bmatrix}\) and the matching value \(i\).
For real matrices, conjugating entries has no effect, so \(\mathbf{A}^*=\mathbf{A}^{\top}\). That real case can make it tempting to forget conjugation in the complex setting. A useful check is to inspect the entries: if the matrix contains nonreal entries, compute the conjugate transpose explicitly rather than assuming that the ordinary transpose suffices.
Also keep the inner-product convention visible. Here \(\langle\mathbf{x},\mathbf{y}\rangle=\mathbf{y}^*\mathbf{x}\), not \(\mathbf{x}^*\mathbf{y}\). Swapping those expressions changes which argument is linear. The matrix formula above follows from the stated convention and should be used together with it.
What the Formula Lets You Do
The formula turns an abstract identity into a reliable calculation: take the matrix of the operator, conjugate its entries, and transpose. The examples show the important verification pattern. First calculate the vector obtained by applying the original matrix, then take its inner product with the second vector. Separately apply the conjugate-transpose matrix to the second vector and take the inner product with the first. Agreement checks the defining identity for those vectors.
A check on selected vectors is a useful way to catch arithmetic or convention errors, but it is not by itself a proof that two operators satisfy the identity for every pair. The theorem proves the identity for all vectors by matrix algebra. In practical work, the formula is especially valuable when entries are complex: it preserves the order and conjugation required by the inner product without relying on visual patterns in the matrix.
Check Your Understanding
For each computation, use the inner product linear in the first argument and show enough steps to track every conjugate.
- Find the adjoint matrix of \(\begin{bmatrix}2&i\\1+i&-3i\end{bmatrix}\).
- For \(\mathbf{D}=\begin{bmatrix}1&-i\\2i&1+i\end{bmatrix}\), compute \(\mathbf{D}^*\) and check the adjoint identity for \(\mathbf{x}=\begin{bmatrix}1\\i\end{bmatrix}\) and \(\mathbf{y}=\begin{bmatrix}i\\1\end{bmatrix}\).
- Explain why the adjoint matrix of a real matrix is its transpose.
- For the matrix in question 1, identify the specific operation that would be missing if one used only the transpose.
- Write \(\langle\mathbf{x},\mathbf{y}\rangle\) as a matrix product and state which argument is conjugated under this convention.
