Putnam 1996, Problem B-4

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  
\[
\sin A = \begin{pmatrix} 1 & 1996 \\ 0 & 1 \end{pmatrix}.
\]

Solution:
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  
\[
\begin{pmatrix} x & y \\ 0 & x \end{pmatrix}.
\]  

A direct computation shows that  
\[
\sin B = \begin{pmatrix} \sin x & y \cdot \cos x \\ 0 & \sin x \end{pmatrix}.
\]  

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.  

Alternate solution (due to Craig Helfgott and Alex Popa): Define both \(\sin A\) and \(\cos A\) by the usual power series. Since \(A\) commutes with itself, the power series identity  
\[
\sin^2 A + \cos^2 A = I
\]  
holds. But if \(\sin A\) is the given matrix, then by the above identity, \(\cos^2 A\) must equal \(\begin{pmatrix} 0 & -2 \cdot 1996 \\ 0 & 0 \end{pmatrix}\) which is a nilpotent matrix. Thus \(\cos A\) is also nilpotent. However, the square of any \(2 \times 2\) nilpotent matrix must be zero (e.g., by the Cayley-Hamilton theorem). This is a contradiction.