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\title{MATH 208 - HW \# 3 - Corrections}
\author{Nathan Marianovsky \\ October 20, 2017}

\begin{document}

\maketitle

\begin{problem}
Prove that the image of $f$ from the previous problem is $SO(3,\mathbb{R}) \subset gl(3,\mathbb{R})$.
\end{problem}

\vspace{.5cm}

\begin{solution}
\hspace{.1cm}
\begin{itemize}

\item
The following tells us that $\mathfrak{Im}(f) \subseteq SO(3,\mathbb{R})$:
\begin{itemize}

\item
$\forall$ $A \in so(3,\mathbb{R})$ we have $\det(A) = 0$ and $\det(e^A) = e^{\text{tr}(A)} = e^0 = 1$.

\item
If $P = e^A$ for an $A \in so(3,\mathbb{R})$, then $P^{-1} = e^{-A} = e^{A^T} = (e^A)^T = P^T$ implying $e^A \in SO(3,\mathbb{R})$. 

\end{itemize}

\vspace{.5cm}
\item
We also check that $f$ is injective by determining the kernel, $\ker(f) = \{A \in so(3,\mathbb{R}) \hspace{.1cm} | \hspace{.1cm} f(A) = I\}$, of the group homomorphism:
\begin{equation*}
\begin{split}
I &= f(A) \\
I &= I + \frac{\sin(\theta)}{\theta}A + \frac{\cos(\theta)-1}{\theta^2}A^2 \\
0 &= A\Big( \frac{\sin(\theta)}{\theta}I + \frac{\cos(\theta)-1}{\theta^2}A \Big) \\
A &= 0, \frac{\theta\sin(\theta)}{1-\cos(\theta)}I
\end{split}
\end{equation*}
Since no multiple of the identity is skew-symmetric it must be that $A = 0$ showing that the kernel is trivial. 

\vspace{.5cm}
\item
Now we handle surjectivity:
\begin{itemize}

\item
To accomplish this we are going to need Rodrigues' Rotation Formula which states given a vector $v \in \mathbb{R}^3$ and a unit vector $\hat{k} \in \mathbb{R}^3$ describing an axis of rotation about which $v$ rotates by an angle $\theta$, according to the right hand rule the resulting vector after rotation is defined as:
\begin{equation*}
v' = \cos(\theta)v + \sin(\theta)(\hat{k} \times v) + (1 - \cos(\theta))(\hat{k} \cdot v)\hat{k}
\end{equation*}
\begin{itemize}

\item
Let us quickly derive this formula. To begin take the vector we want to rotate and split it into components relative to the axis $\hat{k}$:
\begin{equation*}
v = v_\perp + v_\parallel
\end{equation*}
where the parallel component is nothing more than:
\begin{equation*}
v_\parallel = (v \cdot \hat{k})\hat{k}
\end{equation*}
and the perpendicular component is:
\begin{equation*}
v_\perp = v - v_\parallel = v - (\hat{k} \cdot v)\hat{k}
\end{equation*}
We now want to use the known identity:
\begin{equation*}
(a \cdot c)b - (a \cdot b)c = a \times (b \times c)
\end{equation*}
to rewrite the perpendicular component as:
\begin{equation*}
v_\perp = -\hat{k} \times (\hat{k} \times v)
\end{equation*}
To continue we note that since we are rotating about the axis $\hat{k}$, the component parallel to it will not change under the rotation. Therefore, we have:
\begin{equation*}
v_\parallel' = v_\parallel
\end{equation*}
The perpendicular component will transform according to:
\begin{equation*}
\begin{split}
\| v_\perp' \| &= \| v_\perp \| \\
v_\perp' = \cos(\theta)v_\perp &+ \sin(\theta)(\hat{k} \times v_\perp)
\end{split}
\end{equation*}
which can be simplified because:
\begin{equation*}
\hat{k} \times v_\perp = \hat{k} \times (v - v_\parallel) = \hat{k} \times v - \hat{k} \times v_\parallel = \hat{k} \times v
\end{equation*}
giving:
\begin{equation*}
v_\perp' = \cos(\theta)v_\perp + \sin(\theta)(\hat{k} \times v)
\end{equation*}
We know that the above transformation preserves the norm because $v_\perp$ and $\hat{k} \times v$ have the same length. Now we can write down the explicit form of the rotated vector as:
\begin{equation*}
\begin{split}
v' &=  v_\parallel' + v_\perp' \\
&= v_\parallel + \cos(\theta)v_\perp + \sin(\theta)(\hat{k} \times v) \\
&= v_\parallel + \cos(\theta)(v - v_\parallel) + \sin(\theta)(\hat{k} \times v) \\
&= \cos(\theta)v + (1 - \cos(\theta))v_\parallel + \sin(\theta)(\hat{k} \times v) \\
&= \cos(\theta)v + \sin(\theta)(\hat{k} \times v) + (1 - \cos(\theta))(\hat{k} \cdot v)\hat{k}
\end{split}
\end{equation*}

\end{itemize}

\item
With the rotation formula in hand, we want to transform it into matrix mode. So we define the following:
\begin{equation*}
K = \begin{pmatrix}
0 & -k_z & k_y \\
k_z & 0 & -k_x \\
-k_y & k_x & 0
\end{pmatrix} \hspace{.3cm} \text{and} \hspace{.3cm} K^2 = \begin{pmatrix}
-k_y^2-k_z^2 & k_xk_y & k_xk_z \\
k_xk_y & -k_x^2-k_z^2 & k_yk_z \\
k_xk_z & k_yk_z & -k_x^2-k_y^2
\end{pmatrix}
\end{equation*}
where:
\begin{equation*}
Kv = \hat{k} \times v \hspace{.3cm} \text{and} \hspace{.3cm} K^2v+v = (\hat{k} \cdot v)\hat{k}
\end{equation*}
allows us to rewrite the rotation formula as:
\begin{equation*}
\begin{split}
v' &= \cos(\theta)v + \sin(\theta)(\hat{k} \times v) + (1-\cos(\theta))(\hat{k} \cdot v)\hat{k} \\
&= \cos(\theta)v + \sin(\theta)Kv + (1-\cos(\theta))(K^2v+v) \\
&= v + \sin(\theta)Kv + (1-\cos(\theta))K^2v
\end{split}
\end{equation*}
This allows us to say that any $R \in SO(3,\mathbb{R})$ can be written down as:
\begin{equation*}
R = I + \sin(\theta)K + (1-\cos(\theta))K^2
\end{equation*}

\item
If we take $R$ as above then we can always find a $K \in so(3,\mathbb{R})$ s.t. $e^K = R$ with $k_1^2+k_2^2+k_3^2 = 1$. This basically shows surjectivity, however we would like to get rid of the restriction on the components of $K$. In full generality we had:
\begin{equation*}
e^A = I + \frac{\sin(\theta)}{\theta}A + \frac{(1-\cos(\theta))}{\theta^2}A^2
\end{equation*}
Therefore, we can say that given a rotation matrix, $R$, we can always find an $A \in so(3,\mathbb{R})$ s.t.:
\begin{equation*}
e^{\theta A} = R = I + \sin(\theta)A + (1-\cos(\theta))A^2
\end{equation*}
which lifts the restriction on the components summing to one. Thus, $\mathfrak{Im}(f) \cong SO(3,\mathbb{R})$. 

\end{itemize}

\end{itemize}
\end{solution}

\end{document}