Show that $$\exp \begin{pmatrix}
x & -y\\
y & x\end{pmatrix}= \exp(x) \begin{pmatrix}
\cos y& -\sin y\\
\sin y& \cos y\end{pmatrix}$$
for all $ x,y \in \mathbb{R} $.
My thought process is the following:
$$\left(\begin{array}{cc}
a & -b\\
b & a\end{array}\right) + \left(\begin{array}{cc}c & -d\\
d& c\end{array}\right) = \left(\begin{array}{cc}
a+c & -(b+d)\\
b+d & a+c
\end{array}\right)$$
and
$$\left(\begin{array}{cc}
a & -b\\
b & a\end{array}\right) \left(\begin{array}{cc}c & -d\\
d& c\end{array}\right) = \left(\begin{array}{cc}
ac-bd & -(ad+bc)\\
ad+bc & ac-bd
\end{array}\right).$$
Can I use this idea in my proof?
Best Answer
The argument of the exponential on the LHS $$\begin{pmatrix} x & -y\\ y & x\end{pmatrix} \;=\; x\,\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix} \;+\; y\,\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}$$ may be identified with the complex number $\,x+y\,i\,$ because the matrix summands on the RHS commute and $$\begin{pmatrix} 0 & -1\\ 1& 0\end{pmatrix}^2 \;=\; -\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}\,.$$ Under this correspondence we have $\:\begin{pmatrix} \cos y& -\sin y\\ \sin y& \cos y\end{pmatrix} \;\longleftrightarrow\;\cos y +i\sin y\,,\,$ and the identity under consideration is just Euler's identity $$e^{x+iy} \;=\; e^x\,(\cos y +i\sin y)\,.$$