Linear Algebra True/Flase Are these two statements true or false, if brief justification/counterexample could be given it would be appreciated.
$(1)$ $I_{V}$ is an identity operator on vector space $V$, $\dim V=n$ and $A$ is a matrix corresponding to $I_{V}$, then $A=I_{n}$.
$(2)$ If $A$ is a real $ 2\times 2$ matrix, without real eigenvalues then $A$ is a rotation matrix.  
My solution to $(2)$ is false as you can take a counterexample being the rotation matrix
  $\left[ {\begin{array}{cc}
   cos(\theta) & -sin(\theta) \\
   sin(\theta) & cos(\theta) \\
  \end{array} } \right]  $
would have eigenvalue of $\lambda = \frac{1}{cos(\theta)}$.  
Any help would be appreciated.
 A: $ (1) $ is easily seen to be true. First off, "an" identity operator is actually "the" (unique) identity operator, since any linear operator on $ V $ is uniquely determined by its action on a basis $ \{e_i\}_{i=1}^N $ of $ V $.
Now, the matrix representing $ I_V $ on any basis (for instance $ \{e_i\} $) will have as its $ i $-th column $ I_V(e_i) = e_i $, so that in fact $ A = I_n $.
$ (2) $: the matrix you bring as an example has eigenvalues $ \cos(\theta) \pm \mathrm{i}\sin(\theta) $, which are complex, unless $ \theta = k\pi $ for some $ k \in \mathbb Z $ — but then the matrix is simply the identity matrix, which fixes all points, or the matrix sending each point to its opposite, which still has the whole of $ \mathbb R^2 $ as eigenspace.
On the other hand, if an endomorphism in $ \mathbb R^2 $ doesn't have real eigenvalues, this doesn't imply it's a rotation. A trivial example is the composition of a rotation by an "expansion" (multiply by 2 your example matrix; the eigenvalues are simply scaled by a factor of 2 and thus remain complex). The resulting matrix is not a rotation because its determinant isn't 1.
More generally, any matrix
$$ \begin{pmatrix}a & b \\ c & d\end{pmatrix} $$
with $ (a+d)^2 + 4bc < 0 $ will have no real eigenvalues, and you can make more than just rotation matrices out of these parameters.
