Show that f is diagonalizable The question is: Suppose that $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is linear, and that $f^{2015} = I$. Show that $f$ is diagonalizable. [hint:  If f is a multiple of the identity, it’s easy. Otherwise, it suffices to prove that f has two distinct eigenvalues.]
My proof so far:
Let $\vec{x}$ be a nonzero vector in $\mathbb{R}^2$. Then
\begin{align*}
f^{2015}\vec{x} & = I\vec{x}\\
f^{2015}\vec{x} - I\vec{x}&= 0\\
(f^{2015}-I)\vec{x} & = 0
\end{align*}
If we let $f = a^{\frac{1}{2015}}I$ where $a$ is a scalar, then it follows that $f^{2015} = I$. This implies that $f$ is already a diagonal matrix, and the repeated multiplication of a diagonal matrix is still diagonal, so it follows that $f^{2015}$ is diagonalizable. 
My questions: First, does this proof answer the question, or am I wrong?
Second, I'm not sure how to use the second part of the hint, if I am supposed to at all, as I cannot think of any ways to show $f$'s eigenvalues other than what I've done, but wouldn't $(f^{2015}-I)\vec{x} = 0$ imply that $f^{2015}$ only has eigenvalues of $\lambda_1 = \lambda_2 = 0$? What am I missing here? $f$ is in $\mathbb{R}^2$, so it can't have more than 1 eigenvalue, but based on what I've shown, 1 should be an eigenvalue already. Therefore, I shouldn't be able to have complex eigenvalues, but then what is the second one?  
 A: Choose any basis for $\mathbb{R}^2$ and represent $f$ as a matrix $A$.  If $A$ is not diagonalizable, then its Jordan canonical form is $\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)$, for some real $a$ (which is its repeated eigenvalue).
That is, there is some invertible $Q$ such that $A=Q\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)Q^{-1}$
and hence $$I=A^{2015}=(Q\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)Q^{-1})^{2015}=Q\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)^{2015}Q^{-1}$$
Multiplying on the left by $Q^{-1}$ and on the right by $Q$ we conclude that $I=\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)^{2015}$.  As it happens, it is easy to calculate this matrix; $$I=\left(\begin{smallmatrix}a&1\\0&a\end{smallmatrix}\right)^{2015}=\left(\begin{smallmatrix}a^{2015}&2015a^{2014}\\0&a^{2015}\end{smallmatrix}\right)$$
Since this equals the identity, we must have $0=2015a^{2014}$, so $a=0$.  But then $a^{2015}\neq 1$.  Hence this case cannot happen, and so $A$ must be diagonalizable.  Hence also $f$ is diagonalizable.
