<s>[INST] Orthonormalbasis of a isometry $2 \times 2$ matrix I am a bit stuck in this easy question.
Let $A\in \mathbb R^{2 \times2}$ be a isometry so $A^{-1} = \bar{A}^T$. Show that $\left\{\begin{pmatrix} a \\ c \end{pmatrix},\begin{pmatrix} b \\ d \end{pmatrix}\right\}$ is a orthonormal basis. 
$A = \begin{pmatrix} a&b \\ c&d \end{pmatrix} \Longrightarrow \bar{A}^T = \begin{pmatrix} a&c \\ b&d \end{pmatrix}$
$A \cdot \bar{A}^T = \begin{pmatrix} a^2+b^2 & ac+bd \\ ac+bd&c^2+d^2 \end{pmatrix} = \begin{pmatrix} 1&0 \\ 0&1 \end{pmatrix} $
I tried to use Gram-Schmidt but too many variables were left so I couldn't prove it.
Now i also know that is has to be
$\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} b \\ d \end{pmatrix}\right\rangle =0 \Longrightarrow ab+cd = 0 $ and $\left\langle\begin{pmatrix} a \\ c \end{pmatrix}, \begin{pmatrix} a \\ c \end{pmatrix}\right\rangle =1 \Longrightarrow a^2+c^2 = 1$ but I also can't prove that because I only have $a^2+b^2 = 1 = c^2 + d^2$ and $ac+bd = 0$. 
 [/INST] Hint: Write 
$$
v_1 = \pmatrix{a\\c} \qquad v_2 = \pmatrix{b\\d}
$$
Verify that for $A = \pmatrix{a&b\\c&d}$, we have
$$
A^TA = \pmatrix{
\langle v_1,v_1 \rangle & \langle v_2,v_1 \rangle\\
\langle v_1,v_2 \rangle & \langle v_2,v_2 \rangle}
$$
Since $A^{-1} = {A^T}$, we have ${A^T}A = I$.  
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