# 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.

• You have computed $\lambda$ incorrectly. Commented Apr 11, 2014 at 16:36
• Hint for $(1)$: It is not hard to write down the matrix that represents the identity transformation with respect to any basis. Just do it and the answer will be clear ... Commented Apr 11, 2014 at 16:48
• Hint for $(2)$: If $A$ has non-real eigenvalues, what are the possible values of $\det(A)$? What are the possible values of $\det(A)$ if $A$ is a rotation matrix? Commented Apr 11, 2014 at 16:49

$(1)$ 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.