On the sine of a two by two matrix with integer entries Let $A \in M(2,\mathbb Z)$ ; Let 
$\sin A :=\sum _{n=0}^\infty \dfrac {(-1)^n}{(2n+1)!}A^{2n+1}$ (This series is known to converge absolutely in the Banach space $\mathcal L(\mathbb R^2)$ under Operator norm ) . 
If $\sin A \in M(2,\mathbb Z)$ , then is it true that $A^2=O$ ? 
 A: Note - Haven't received that many downvotes for a single answer for a while. 
I suspect people downvote this because it uses something too advanced for
a problem that looks trivial. In fact, if the eigenvalues $\lambda_1, \lambda_2$ 
of $A$ are both real (say $A$ is real symmetric), then this problem is indeed trivial. This is because when $\lambda_1, \lambda_2$ are real,
$$|\det\sin(A)| = |\sin(\lambda_1)\sin(\lambda_2)| \le 1 \quad\text{ and }\quad |{\rm tr} \sin(A)| = |\sin(\lambda_1) + \sin(\lambda_2)| \le 2$$
and the fact $\pi$ is irrational is enough to force $\lambda_1 = \lambda_2 = 0$.
The problem is $\lambda_1, \lambda_2$ need not be real....

Original answer
This answer uses a little bit of transcendental number theory. It might be an overkill for this particular problem. However, it confirm whenever both $A$ and $\sin A$ are integer matrices, then $A^2 = 0$.
Let $\lambda_1, \lambda_2$ be eigenvalues of $A$.
When $A \in M(2,\mathbb{Z})$, its characteristic polynomial
$\chi_A(\lambda) \stackrel{def}{=}\det(\lambda I - A)$ will be a polynomial with integer coefficients. Since $\lambda_1, \lambda_2$ are roots of this polynomial, they will be algebraic numbers.
Notice $\sin A$ has $\sin(\lambda_1)$ and $\sin(\lambda_2)$ as its eigenvalues.
If $\sin A \in M(2,\mathbb{Z})$, then by same argument, $\sin\lambda_1$ and $\sin\lambda_2$ will be algebraic numbers.
By Lindermann Weierstrass theorem.
If $\alpha$ is any non-zero algebraic number, then $\sin\alpha$ will be transcendental. In order for all four numbers $\lambda_1, \lambda_2, \sin\lambda_1, \sin\lambda_2$ to be algebraic, we need $\lambda_1 = \lambda_2 = 0$. This leaves us with two possibilities:


*

*$A$ is diagonalizable and hence is equals to the zero matrix.

*$A$ is similar to a Jordan block of order $2$ with zero diagonal.
In this case, $A$ will be nilpotent and satisfies $A^2 = 0$.


In both cases, if both $A, \sin A \in M(2,\mathbb{Z})$, then $A^2 = 0$.
