Properties of the Conjugate Transpose

We've already seen that $(A + B)^T = A^T + B^T$, which holds for real or complex matrices.

Also recall that $A^\dagger= (A^T)^* = (A^*)^T$ is the conjugate transpose of $A$ i.e. the result of transposing the matrix then taking the complex conjugate of each entry (or vice-versa). 

Now, suppose that we have two complex matrices $A$ and $B$ of the same dimension (so we can add them).

Then $(A + B)^\dagger = A^\dagger + B^\dagger$.

The working is essentially the same as we saw for $(A + B)^T = A^T + B^T$ except we need to apply the complex conjugate as well as the transpose.

To see this, let’s consider an example with the following matrices:

$$A = \begin{bmatrix} i & 0 & 1 - i \\ 2 & 4 & -1 \end{bmatrix}, \quad B = \begin{bmatrix} -6i & 1 & 2 \\ 3i & -i & 1 + i \end{bmatrix}$$

Then:

$$\begin{align*} (A + B)^\dagger &= \left( \begin{bmatrix} i & 0 & 1 - i \\ 2 & 4 & -1 \end{bmatrix} + \begin{bmatrix} -6i & 1 & 2 \\ 3i & -i & 1 + i \end{bmatrix} \right)^\dagger \\ &= \begin{bmatrix} -5i & 1 & 3 - i \\ 2 + 3i & 4 - i & i \end{bmatrix}^\dagger \\ &= \begin{bmatrix} -5i & 2 + 3i \\ 1 & 4 - i \\ 3 - i & i \end{bmatrix}^* \quad \text{(taking transpose)}\\ &= \begin{bmatrix} 5i & 2 - 3i \\ 1 & 4 + i \\ 3 + i & -i \end{bmatrix} \quad \text{(taking conjugate)} \end{align*}$$

Note that $A^\dagger = \begin{bmatrix} -i & 2 \\ 0 & 4 \\ 1 + i & -1 \end{bmatrix}$ and $B^\dagger = \begin{bmatrix} 6i & -3i \\ 1 & i \\ 2 & 1 - i \end{bmatrix}$ and so:

$$...$$