Assume $R_{3\times3}$ is a rotation matrix.
Question
Is it true that there exists two quaternions representing this same rotation matrix $R_{3\times3}$ ?
Hint : $\theta = \arccos\left( \frac{\mathrm{trace}(R) - 1}{2} \right)$,$\theta$ can be +ve or negative. So it will affect $q=\cos(\theta/2)+u\sin(\theta/2)$ on sin part. And may give two solutions
Assume if we have $q=(x,y,z,w)=(\hat{v },w) $ represents a quaternion . Then we have a 3D rotation matrix given by this quaternion as
$R_{\theta}=\begin{bmatrix} 1 - 2y^2 - 2z^2&2xy + 2wz & 2xz - 2wy\\ 2xy - 2wz&1 - 2x^2 - 2z^2 &2yz + 2wx \\ 2xz + 2wy & 2yz - 2wx & 1 - 2x^2 - 2y^2\end{bmatrix}\tag1 $
3D rotation matrix.
If (1) is true then, can we say , there must exist another quaternion $q'=(x^{'},y^{'},z^{'},w^{'}) $ whose 3D rotation matrix is given as
$ R_{T}= \begin{bmatrix} 1 - 2{y'}^2 - 2{z'}^2&2{x'}{y'} + 2{w'}{z'} & 2{x'}{z'} - 2{w'}{y'}\\ 2{x'}{y'} - 2{w'}{z'}&1 - 2{x'}^2 - 2{z'}^2 &2{y'}{z'} + 2{w'}{x'} \\ 2{x'}{z'} + 2{w'}{y'} & 2{y'}{z'} - 2{w'}{x'} & 1 - 2{x'}^2 - 2{y'}^2 \end{bmatrix}\tag2$
and $R_{\theta}=R_{T}$ ?
A rotation is defined giving a versor $\mathbf{u}$ (axis of rotation) and an angle $\theta$. With quaternions in exponential notation a rotation of a vector $\mathbf{v}$ is given by $$ R_{\mathbf{u},2\theta}(\mathbf{v})= e^{-\mathbf{u}\theta}\mathbf{v}e^{\mathbf{u}\theta} $$ where $e^{\mathbf{u}\theta}=\cos \theta + \mathbf{u} \sin \theta$, so , as @A.E said, the two quaternions $e^{\mathbf{u}\theta}$ and $e^{(-\mathbf{u})(-\theta)}$ correspond to the same rotation. Strictly speaking, if we identify rotations $mod(2\pi)$, there are infinitely many quaternions corresponding to the some rotation, since $\cos\theta$ and $\sin \theta$ are $2\pi$-periodic functions. Using exponential notation you can put the rotation matrices in a more expressive form and you'll have a simple answer to the second question.