Find a $2 \times 2$ unitary matrix with eigenvalues $1, -1$ and the eigenvectors $$\begin{pmatrix}1\\ 0\end{pmatrix},\:\begin{pmatrix}0\\ 1\end{pmatrix}$$
My attempt:
Let $$A = \begin{pmatrix} a_1 & a_2\\ a_3 & a_4\end{pmatrix}$$ so that $$\det\begin{pmatrix}a_1-\lambda &a_2\\ a_3&a_4-\lambda \end{pmatrix} = 0$$ but it's look like a bit difficult to continue, maybe there is a simple way?
Best Answer
Let $A\in \mathbb{C}^{2\times 2}$ be a unitary matrix such that its eigenvalues are $1$ and $-1$ with eigenvectors $(1,0)$ and $(0,1)$. This means (by definition), that $A(1,0)^{T}=(1,0)^{T}$ and $A(0,1)^{T}=-(0,1)^{T}$ (here the $T$ just means transposing). Let's write
$$ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$
The equations above imply that
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} =\begin{pmatrix} a \\ c \end{pmatrix} =\begin{pmatrix} 1 \\ 0 \end{pmatrix} $$ and $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} =\begin{pmatrix} b \\ d \end{pmatrix} =\begin{pmatrix} 0 \\ -1 \end{pmatrix} $$
Now, componentwise, you get $a=1$, $c=0$, $b=0$, $d=1$, so that
$$ A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $$
Now, this matrix has eigenvalues $1$ and $-1$, and eigenvectors $(1,0)$ and $(0,1)$, as desired. It just remains to check that indeed $A$ is unitary. For example, you can argue that this is the case because the columns of $A$ form an orthonormal basis for $\mathbb{C}^{2}$, or you can simply compute $A^{*}A$ and check that it is the identity matrix.
Hope this helps!