<s>[INST] Solve the matrix equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$ Solve the equation $\sin(X)=\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}$, where $X\in M_2(\Bbb C)$ and $a\in \Bbb C$.
I discussed the case whether $X$ is diagonalisable or not.
If $X$ is diagonalisable, we have $X\sim \begin{pmatrix}x & 0\\0 & y\end{pmatrix}$, thus $\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}\sim \begin{pmatrix}\sin(x) & 0\\0 & \sin(y)\end{pmatrix}$. There is no solution if $a\neq 0$.
If $X$ is not diagonalisable and then at least triangularizable, we have $X\sim \begin{pmatrix}x & y\\0 & x\end{pmatrix}$, thus $\begin{pmatrix}1 & a\\0 & 1\end{pmatrix}\sim \begin{pmatrix}\sin(x) & y\cos(x)\\0 & \sin(x)\end{pmatrix}$. There is no solution if $a\neq 0$.
Any error?
 [/INST] As you correctly infer, there is no diagonalizable solution $X$.
Let $A = \left(\begin{smallmatrix}1&a\\0&1 \end{smallmatrix} \right)$.
Notably, we compute
$$
\sin\pmatrix{\lambda&1\\0&\lambda} = \pmatrix{\sin\lambda &\cos\lambda \\0&\sin \lambda}=:M(\lambda)
$$
and so, $\sin(X) = A$ has a solution if and only if there exists a $\lambda$ such that $A \sim M(\lambda)$.
Now, since $A$ has eigenvalue $1$, it must be that $\lambda = (2k + 1)\pi$ for some $k \in \Bbb Z$.  However, this would mean that $\cos \lambda = 0$, which means that $M(\lambda) \sim I \nsim A$.  
So, there is indeed no solution.
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