Conjugate Transpose of a Matrix

The conjugate transpose of a matrix can be expressed in several equivalent ways:

• In mathematics, it is commonly denoted by $ A^* $.
• In physics and quantum mechanics, the notation $ A^{\dagger} $ is preferred, read aloud as “A dagger.”

This operation plays a central role in the study of unitary matrices \( U^{\dagger}U = I \) and Hermitian matrices \( H^{\dagger} = H \). 

Note. The conjugate transpose, sometimes called the "Hermitian adjoint," should not be confused with the classical adjugate matrix, which is the transpose of the cofactor matrix used to compute the inverse: $$ A^{-1} = \frac{1}{\det(A)}\,\mathrm{adj}(A) $$ These two notions are completely different.

Worked Example

Consider the complex matrix:

$$ U = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} $$

First, compute its transpose by swapping rows and columns:

$$ U^T = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix} $$

In this case, the matrix remains unchanged because it is diagonal: all nonzero elements lie on the main diagonal.

Next, take the complex conjugate of each entry in the matrix.

Note. The complex conjugate of a number is obtained by changing the sign of its imaginary part. For example, $ \overline{a + bi} = a - bi $, and $ \overline{i} = -i $.   For real numbers, the conjugate has no effect because the imaginary part is zero. For instance, $ \overline{1} = 1 $. Therefore, if a matrix contains only real numbers, its conjugate transpose is simply the transpose: $$ A^{\dagger} = A^T $$ In such cases, conjugation does not alter the matrix.

In our matrix \( U \), the only complex entry is the imaginary unit \( i \), whose conjugate is \( -i \).

$$ U^{\dagger} = \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} $$

This is the conjugate transpose of \( U \).

Properties of the Conjugate Transpose

The conjugate transpose behaves much like the ordinary transpose, except that each entry is also conjugated. Below are the main properties and their meanings.

• Double Conjugation
  Applying the conjugate transpose twice restores the original matrix: $$ (A^{\dagger})^{\dagger} = A $$

  Example:
  $$ A = \begin{pmatrix} i & 1 \\ 0 & 2i \end{pmatrix} $$ To find the conjugate transpose, swap rows and columns and flip the sign of the imaginary parts: $$ A^{\dagger} = \begin{pmatrix} -i & 0 \\ 1 & -2i \end{pmatrix} $$ Repeating the operation - taking the conjugate transpose of \( A^{\dagger} \) - swaps rows and columns again and restores the original signs: $$ (A^{\dagger})^{\dagger} = \begin{pmatrix} i & 1 \\ 0 & 2i \end{pmatrix} = A $$ The first dagger transforms the matrix, and the second undoes the transformation. Thus, the double conjugate transpose always returns the original matrix.

• Matrix Product
  When taking the transpose (or conjugate transpose) of a product, the order of the factors reverses. Complex conjugation then applies to each element: $$ (AB)^{\dagger} = B^{\dagger} A^{\dagger} $$

  Example:
  $$ A = \begin{pmatrix} i & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & 1 \\ 0 & i \end{pmatrix} $$ Compute the product: $$ AB = \begin{pmatrix} i & i \\ 0 & i \end{pmatrix}. $$ Now take the conjugate transpose of \( AB \): $$ (AB)^{\dagger} = \begin{pmatrix} -i & 0 \\ -i & -i \end{pmatrix}. $$ Separately, we have: $$ A^{\dagger} = \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix}, \quad B^{\dagger} = \begin{pmatrix} 1 & 0 \\ 1 & -i \end{pmatrix}. $$ Their product is: $$ B^{\dagger}A^{\dagger} = \begin{pmatrix} 1 & 0 \\ 1 & -i \end{pmatrix} \begin{pmatrix} -i & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} -i & 0 \\ -i & -i \end{pmatrix}. $$ Since $$ (AB)^{\dagger} = B^{\dagger}A^{\dagger}, $$ the property is verified.

• Matrix Sum
  The conjugate transpose acts element by element, so matrix addition is preserved: $$ (A + B)^{\dagger} = A^{\dagger} + B^{\dagger} $$

  Example:
  $$ A = \begin{pmatrix} i & 2 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} 1 & i \\ 3i & 0 \end{pmatrix} $$ Add the two matrices entry by entry: $$ A + B = \begin{pmatrix} i + 1 & 2 + i \\ 3i + 0 & 1 + 0 \end{pmatrix} = \begin{pmatrix} 1 + i & 2 + i \\ 3i & 1 \end{pmatrix} $$ Now take the conjugate transpose: $$ (A + B)^{\dagger} = \begin{pmatrix} 1 - i & -3i \\ 2 - i & 1 \end{pmatrix} $$ Separately, $$ A^{\dagger} = \begin{pmatrix} -i & 0 \\ 2 & 1 \end{pmatrix}, \quad B^{\dagger} = \begin{pmatrix} 1 & -3i \\ -i & 0 \end{pmatrix}. $$ Their sum is: $$ A^{\dagger} + B^{\dagger} = \begin{pmatrix} 1 - i & -3i \\ 2 - i & 1 \end{pmatrix}. $$ Since $$ (A + B)^{\dagger} = A^{\dagger} + B^{\dagger}, $$ the property holds true.

• Multiplication by a Complex Scalar
  When a matrix is multiplied by a complex number \( c \), the conjugate transpose also conjugates the scalar: $$ (cA)^{\dagger} = \overline{c}\,A^{\dagger} $$

  Example:
  $$ c = 2i, \quad A = \begin{pmatrix} 1 & i \\ 0 & 2 \end{pmatrix} \Rightarrow cA = \begin{pmatrix} 2i & -2 \\ 0 & 4i \end{pmatrix} $$ Now compute the conjugate transpose of \( cA \): $$ (cA)^{\dagger} = \begin{pmatrix} -2i & 0 \\ -2 & -4i \end{pmatrix} $$ On the other hand, since \( \overline{c} = -2i \), $$ \overline{c}\,A^{\dagger} = (-2i) \begin{pmatrix} 1 & 0 \\ -i & 2 \end{pmatrix} = \begin{pmatrix} -2i & 0 \\ -2 & -4i \end{pmatrix} $$ Therefore, $$ (cA)^{\dagger} = \overline{c}\,A^{\dagger}. $$

These properties make the conjugate transpose a fundamental tool in linear algebra, especially in the study of unitary and Hermitian matrices and in the analysis of linear operators on complex vector spaces.

And so forth.