# 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?

• It's not clear how you decided that there is no solution when $a \neq 0$ – Omnomnomnom Oct 10 '16 at 14:00
• @Omnomnomnom By the trace. $\sin(x)+\sin(y)=2\iff \sin(x)=\sin(y)=1$. – Aforest Oct 10 '16 at 14:14
• I meant in part 2... – Omnomnomnom Oct 10 '16 at 14:15
• For that first part, the quicker observation to make is that $\pmatrix{1&a\\0&1}$ is not diagonalizable, whereas $\sin(X)$ clearly must be – Omnomnomnom Oct 10 '16 at 14:16
• @Omnomnomnom $\sin(x)=1\implies \cos(x)=0$ – Aforest Oct 10 '16 at 14:19

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 $M(\lambda) \sim I \nsim A$.