Find $\xi$ such that matrix is unitary and find $\lambda_i$ such that matrix hermitian i) Find $\xi \in \mathbb C$ such that $$\frac12 \begin{pmatrix} \xi & 0 & 1+i \\ 0 & 2 & 0 \\ 1-i & 0 & -1+i \end{pmatrix}$$ is unitary.
ii) Let $A := \textrm{diag}(\lambda_1,..,\lambda_n)$. Find $\lambda_i$ such that $A$ is hermitian.
 A: A complex square matrix $A\in\mathcal{M}_{n}(\mathbb C)$ is unitary if, by definition $A\bar{A}^t=\bar{A}^tA=\mathbb I_n$. Let's write the matrix product:
\begin{align*}
A\bar{A}^t=&
\frac12 \begin{pmatrix} \xi & 0 & 1+i \\ 0 & 2 & 0 \\ 1-i & 0 & -1+i \end{pmatrix}
\frac12 \begin{pmatrix} \bar{\xi} & 0 & 1+i \\ 0 & 2 & 0 \\ 1-i & 0 & -(1+i) \end{pmatrix}\\
=&\frac14 \begin{pmatrix} |\xi|^2+2 & 0 & (1+i)[\xi-(i+i)] \\ 0 & 4 & 0 \\ (1-i)[\bar{\xi}-(1-i)] & 0 & 4 \end{pmatrix}\\
\end{align*}
Hence, we impose $A\bar{A}^t=\mathbb I_n$ that leads to solve the following system:
$|\xi|^2+2=4$
$\xi-(i+i)=0$
$\bar{\xi}-(1-i)=0$
which has clearly a unique solution: $\xi=1+i$.
Moreover, computing $\bar{A}^tA$, we obtain a matrix that is in general different from $A\bar{A}^t$, but if $\xi=1+i$ then we reach the equality, as the following computation shows:
\begin{align*}
\bar{A}^tA=&
\frac12 \begin{pmatrix} \bar{\xi} & 0 & 1+i \\ 0 & 2 & 0 \\ 1-i & 0 & -(1+i) \end{pmatrix}
\frac12 \begin{pmatrix} \xi & 0 & 1+i \\ 0 & 2 & 0 \\ 1-i & 0 & -1+i \end{pmatrix}\\
=&\frac14 \begin{pmatrix} |\xi|^2+2 & 0 &\bar{\xi}(1+i)-2 \\ 0 & 4 & 0 \\ \xi(1-i)-2 & 0 & 4 \end{pmatrix}\\
\end{align*}
Hence $\xi=1+i$ allows indeed to get a unitary matrix.
For the second point: a matrix $A$ is hermitian when it coincides with its conjugate transpose, i.e. when $\bar{A}^t=A$. Now if you have a diagonal matrix, it conincides with its transpose. Hence the unique check to do is that the conjugate conicides with the original. But it happens if and only if EVERY elements of the matrix coincides with its conjugate, and this happens only when the elements are in $\mathbb R$. Hence a diagonal matrix $A=diag(\lambda_1,\dots,\lambda_n)$ is hermitian if and only of $\lambda_i\in\mathbb R$, for every $i=1,\dots,n$.
A: i) Call your matrix $m$. We need to find $\xi$ such that $m^* m = I$ (here $^*$ denotes the conjugate transpose). Writing out the matrix product, this gives 3 equations:
$$ -i/2 + (1+i)\xi/4 = 0,\\ i/2 + (1-i)\overline{\xi}/4 = 0, \\ 1/2 + \xi \overline{\xi}/4 = 1. $$
The first equation yields $\xi = 1+i$. This happens to also satisfy the other equations, so this is the unique solution.
ii) $A$ is hermitian iff it is equal to its own conjugate transpose. Hence this is satisfied exactly if all $\lambda_i$ satisfy $\overline{\lambda_i} = \lambda_i$, i.e. are real.
