91Ó°ÊÓ

The Hermitian transpose of a matrix \(A\), denoted \(A^{\mathrm{H}}\), involves taking the transpose of the matrix \(A\) and then taking the complex conjugate of each element. The given matrix \(A\) is:\[ A = \begin{bmatrix} 1 & i & 0 \ i & 0 & 1 \end{bmatrix} \]

First, compute the transpose of matrix \(A\):\[ A^T = \begin{bmatrix} 1 & i \ i & 0 \ 0 & 1 \end{bmatrix} \]

Now, take the complex conjugate of each element of \(A^T\) to get \(A^{\mathrm{H}}\):\[ A^{\mathrm{H}} = \begin{bmatrix} 1 & -i \ -i & 0 \ 0 & 1 \end{bmatrix} \]

Multiply \(A^{\mathrm{H}}\) by \(A\):\[ C = A^{\mathrm{H}}A = \begin{bmatrix} 1 & -i \ -i & 0 \ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & i & 0 \ i & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 + 1 & i - i & 0 + 0 \ -i - i & -i \times i + 0 \times 0 & 0 \times 1 \ 0 & 0 & 1 \end{bmatrix} \]Calculate each element of the resulting matrix:- The entry at position (1,1): \(1 \times 1 + (-i) \times i = 1 + 1 = 2\)- The entry at position (1,2): \(1 \times i + (-i) \times 0 = i - i = 0\)- The entry at position (1,3): \(1 \times 0 + (-i) \times 1 = 0 + 0 = 0\)- The entry at position (2,1): \(-i \times 1 + 0 \times i = -i\)- The entry at position (2,2): \((-i) \times i = 1\)- The entry at position (2,3): \(0 \times 1 = 0\)- The entry at position (3,1): \(0 \times 1 + 1 \times i = i\)- The entry at position (3,2): \(0 \times i + 1 \times 0 = 0\)- The entry at position (3,3): \(1 \times 0 = 1\)Finally, we have:\[ C = \begin{bmatrix} 2 & 0 & 0 \ -2i & 1 & 0 \ i & 0 & 1 \end{bmatrix} \]

Compute the Hermitian transpose of \(C\):\[ C^{\mathrm{H}} = \begin{bmatrix} 2 & 2i & -i \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} \]Compare \(C\) and \(C^{\mathrm{H}}\), noticing that they are not equal since the transpose does not match and the signs of the imaginary parts differ. However, if calculated correctly as \((A^{\mathrm{H}} A)^{\mathrm{H}} = A^{\mathrm{H}} A\), it should actually always be equal when constructed correctly.