Problem source

The exponential of a square matrix ${\bf A}$ is defined to be
$$
\exp ({\bf A}) = \sum_{r=0}^\infty {1\over r!} {\bf A}^r \,,
$$
where ${\bf A}^0={\bf I}$  and  $\bf I$ is the identity matrix. 
Let 
$$
{\bf M}=\left(\begin{array}{cc} 0 & -1 \\ 1 & \phantom{-} 0
\end{array}
\right) \,.
$$
Show that ${\bf M}^2=-{\bf I}$ and hence express $\exp({\theta {\bf M}})$ as a single  $2\times 2$ matrix,
where $\theta$ is a real number.
Explain the geometrical significance of $\exp({\theta {\bf M}})$.
Let
 $$
{\bf N}=\left(\begin{array}{rr} 0 & 1 \\ 0 & 0 \end{array}\right) \,.
$$
Express similarly  $\exp({s{\bf N}})$, where $s$ is  a real number, and explain the geometrical significance of $\exp({s{\bf N}})$.
For which values of $\theta$ does
$$
\exp({s{\bf N}})\; \exp({\theta {\bf M}})\,  = \,
\exp({\theta {\bf M}})\;\exp({s{\bf N}})
$$
for all $s$?
Interpret this fact geometrically.

Solution source

\begin{align*}
\mathbf{M}^2 &= \begin{pmatrix} 0 & - 1 \\ 1 & 0 \end{pmatrix}^2  \\
&= \begin{pmatrix} 0 \cdot 0 + (-1) \cdot 1  & 0 \cdot (-1) + (-1) \cdot 0 \\
1 \cdot 0 + 0 \cdot 1 & 1 \cdot (-1) + 0 \cdot 0 \end{pmatrix} \\
&= \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix} \\
&= - \mathbf{I}
\end{align*}

\begin{align*}
\exp(\theta \mathbf{M}) &= \sum_{r=0}^\infty \frac1{r!} (\theta \mathbf{M})^r \\
&= \sum_{r=0}^\infty \frac{1}{r!} \theta^r \mathbf{M}^r \\
&= \cos \theta \mathbf{I} + \sin \theta \mathbf{M} \\
&= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}
\end{align*}

This is a rotation of $\theta$ degrees about the origin.

\begin{align*}
&& \mathbf{N}^2 &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}^2 \\
&& &= \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \\
\Rightarrow && \exp(s\mathbf{N}) &= \sum_{r=0}^\infty \frac{1}{r!} (s\mathbf{N})^r \\
&&&= \mathbf{I} + s \mathbf{N} \\
&&&= \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix}
\end{align*}

This is a shear, leaving the $y$-axis invariant, sending $(1,1)$ to $(1+s, 1)$.

Suppose those matrices commute, for all $s$, ie
\begin{align*}
&& \begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1 &s \\ 0 & 1 \end{pmatrix} \\
\Rightarrow && \begin{pmatrix} \cos \theta - s \sin \theta & -\sin \theta + s \cos \theta \\
\sin \theta & \cos \theta \end{pmatrix} &= \begin{pmatrix} \cos \theta & s \cos \theta - \sin \theta \\
\sin \theta & s \sin \theta + \cos \theta \end{pmatrix} \\
\Rightarrow && \sin \theta &= 0 \\
\Rightarrow && \theta &=n \pi, n \in \mathbb{Z}
\end{align*}

Clearly it doesn't matter when we do nothing. If we are rotating by $\pi$ then it also doesn't matter which order we do it in as the stretch happens in both directions equally.