Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where 
$ A =
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}
$, then $A =\begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatrix}
    a & c \\
    b & d
  \end{bmatrix}
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}$ = $\begin{bmatrix}
    a^2 + c^2 & ab + cd \\
    ab + cd & b^2 + d^2
  \end{bmatrix}$ = $\begin{bmatrix}
    1 & 0 \\
    0 & 1
  \end{bmatrix}$
. Then $a^2 + c^2 = 1$ implies $a = \cos\theta$, and $c = \sin\theta$ or $c = - \sin\theta$ using the unit circle . 
Similarly $ d = \cos\theta$, and $b = \sin\theta$ or $b = -\sin\theta$.
So know I am stuck in showing how $b = -\sin\theta$ has to be chosen and $c = \sin\theta$.
Can someone please help? Thank you!
 A: If you had $A =\begin{bmatrix} \cos\theta & \sin\theta \\  \sin\theta & \cos\theta \end{bmatrix}$ or $A =\begin{bmatrix} \cos\theta & -\sin\theta \\  -\sin\theta & \cos\theta \end{bmatrix}$ instead of $A =\begin{bmatrix} \cos\theta & -\sin\theta \\  \sin\theta & \cos\theta \end{bmatrix}$, would it then be true that $A^T A = I \vphantom{\frac{\int^\int}\int} $?
A: Show if $A^TA = I$ and $\det A = 1$ where 
$ A =
  \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}
$, then $A =\begin{bmatrix}
    \cos\theta & -\sin\theta \\
    \sin\theta & \cos\theta
  \end{bmatrix}$.
MY ATTEMPT:
Since it is an orthogonal matrix So modulus of It's eigenvalue is equal to 1
Also since  $det(A)=1$ So Multiplication of eigenvalues=1 
Let eigenvalues of $A$ is $a+ib$ and $c+id$. So 
we have:
$a+ib=\frac{1}{c+id}$ with $|a+ib|=1$ and $|c+id|=1$
 So we have $a+ib=\frac{c-id}{c^2+d^2}$ =$c+id$ [since ${c^2+d^2}=1$]
That is $a+ib=c+i\times(-d)$
$\Rightarrow a=c , b=-d$ 
for some $\theta$,     $a+ib=(\cos\theta +i\sin\theta)$
[ r=1 since modulus of eigenvalue is 1]
So we have $a=c=\cos\theta$  and $b=-d=\sin\theta$
Therefore eigenvalues of this matrix are: $\cos\theta+i\sin\theta$ and $\cos\theta-i\sin\theta$.
That means \begin{bmatrix}
    a & b \\
    c & d
  \end{bmatrix}
is similar to
$D=\begin{bmatrix}
   \cos\theta+i\sin\theta & 0 \\
    0 & \cos\theta-i\sin\theta
  \end{bmatrix} $
A: You have that there is some $\theta$ for which you can write
$$a = \cos\theta \qquad\text{and}\qquad c = \sin\theta$$
Now, note that the vanishing elements of the matrix product give the equation 
$$a b + c d = 0 \tag{1}$$
and that the assumed relation $\det A = 1$ gives
$$a d - b c = 1 \tag{2}$$
Simply solve $(1)$ and $(2)$ for $b$ and $d$ in terms of $a$ and $c$.
A: from $A^TA = I$, we konw $(a, c)$ and $(b, d)$ is orthogonal, for some $\theta$, we know $(a, c) = \sqrt{(a^2 + c^2)}(\cos\theta, \sin\theta) = (\cos\theta, \sin\theta)$, the unit vector orthogonal to $(a, c)$ is $(-\sin\theta, \cos\theta)$, so we know, for some scalar $k$, $(b, d) = k(-\sin\theta, \cos\theta)$. $k$ is either $1$ or $-1$. since $\det(A) = 1$, 
here we used $\sqrt{a^2 + c^2} = 1$, and $\sqrt{b^2 + d^2} = 1$
Let us calculate
$$\det(A) = \det \begin{bmatrix} \cos\theta & \sin\theta \\  -k\sin\theta & k\cos\theta \end{bmatrix} = k(\cos^2\theta + \sin^2\theta) = k = 1$$
so 
$$ A = \begin{bmatrix} \cos\theta & \sin\theta \\  -\sin\theta & \cos\theta \end{bmatrix}$$
