Matrices and power series For any square matrix $A$, we can define $\sin A$ using the formal power series as follows:
\begin{equation}
\sin A=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}A^{2n+1}
\end{equation}
Prove or disprove: there exists a $2\times2$ real matrix $A$ such that
\begin{equation}
\sin A=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}
 A: The relation $\, \sin(A)^2+\cos(A)^2=I$ also hold for matrices
So we would have to find a matrix for cosine such that $C^2=\begin{pmatrix}0&-4040\\0&0\end{pmatrix}$ which is not possible.
A: Notice that
\begin{equation}
B=\begin{bmatrix}1 & 2020 \\0& 1\end{bmatrix}
\end{equation}
is not diagonalizable. If it was $1$ would be the only eigenvalue and $B$ would be equal to the identity matrix.
Therefore if $\sin A = B$, $A$ isn't diagonalizable either. The (complex) Jordan normal form of $A$ would be
\begin{equation}
\overline A=\begin{bmatrix}\lambda & 1 \\0& \lambda\end{bmatrix}
\end{equation}
And
\begin{equation}
\sin \overline A=\begin{bmatrix}\sin \lambda & \cos \lambda \\0& \sin \lambda\end{bmatrix}
\end{equation}
Hence $\lambda$ belongs to $\pi/2 + 2\pi \mathbb Z$  and
\begin{equation}
\sin \overline A=\begin{bmatrix}\sin \lambda & 0 \\0& \sin \lambda\end{bmatrix} = \begin{bmatrix}1 & 0 \\0& 1\end{bmatrix}
\end{equation}
is diagonalizable. A contradiction.
A: Does this help!?:
$$A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \implies B=A^n=\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} \implies \sin A=\begin{bmatrix} \sin 1 & \cos 1 \\ 0 & \sin 1 \end{bmatrix}.$$
Use $$\sin A= \sum_{n=0}^{\infty}(-1)^n \frac{A^{2n+1}}{(2n+1)!}$$
