Exercise 1

\[ \begin{flalign*} \mathbf{L} &= a \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + b \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix} + c \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + d \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} 0 & a \\ a & 0 \end{pmatrix} + \begin{pmatrix} 0 & -b\mathrm{i} \\ b\mathrm{i} & 0 \end{pmatrix} + \begin{pmatrix} c & 0 \\ 0 & -c \end{pmatrix} + \begin{pmatrix} d & 0 \\ 0 & d \end{pmatrix} \\ &= \begin{pmatrix} c + d & a - \mathrm{i}b \\ a + \mathrm{i}b & -c +d \end{pmatrix} \end{flalign*} \]

The off-diagonal elements are complex conjugates. The real numbers \(c\) and \(d\) can always be choosen to evaluate to any given diagonal elements.