MHB Is there a 2x2 matrix with real entries whose sine is [1, 2016; 0, 1]?

[Ackbach]

Here is this week's POTW:

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For any square matrix $A$, we can define $\sin(A)$ by the usual power series:
\[
\sin(A) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} A^{2n+1}.
\]
Prove or disprove: there exists a $2 \times 2$ matrix $A$ with real entries such that
\[
\sin(A) = \left( \begin{array}{cc} 1 & 2016 \\ 0 & 1 \end{array} \right).
\]

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[Ackbach]

Re: Problem Of The Week # 227 - Aug 03, 2016

This was Problem B-4 in the 1996 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

Suppose such a matrix $A$ exists. If the eigenvalues of $A$ (over
the complex numbers) are distinct, then there exists a complex
matrix $C$ such that $B=CAC^{-1}$ is diagonal. Consequently,
$\sin B$ is diagonal. But then $\sin A=C^{-1}(\sin B)C$ must
be diagonalizable, a contradiction. Hence the eigenvalues of $A$
are the same, and $A$ has a conjugate $B=CAC^{-1}$ over
the complex numbers of the form
\[
\left(
\begin{array}{cc}
x & y\\
0 & x
\end{array}
\right).
\]
A direct computation shows that
\[
\sin B = \left(
\begin{array}{cc}
\sin x & y\cdot \cos x\\
0 & \sin x
\end{array}
\right).
\]
Since $\sin A$ and $\sin B$ are conjugate, their eigenvalues
must be the same, and so we must have $\sin x=1$. This implies
$\cos x=0$, so that $\sin B$ is the identity matrix, as must be $\sin
A$, a contradiction.
Thus $A$ cannot exist.