Easy way to derive rotation matrix

It is very easy to derive rotation matrix.
This method is applied to general cases.

Derivation

Step1. Calculate rotation of unit vectors

Calculate rotation of unit vector of $x$ axis by $\theta$
\begin{align}
\left(
\begin{array}{c}
1\\
0
\end{array}
\right)
%
\overset{\text{Rotate by }\theta}{\longrightarrow}
%
\color{blue}{
\left(
\begin{array}{c}
\cos\theta\\
\sin\theta
\end{array}
\right)
}
\end{align}

Same way , rotate unit vector of $y$ axis
\begin{align}
\left(
\begin{array}{c}
0\\
1
\end{array}
\right)
%
\overset{\text{Rotate by }\theta}{\longrightarrow}
%
\color{red}{
\left(
\begin{array}{c}
-\sin\theta\\
\cos\theta
\end{array}
\right)
}
\end{align}

Step2. Arrange two vectors

Just arrange these two vectors and we got Rotation Matrix!
\begin{align}
\left(
\begin{array}{cc}
\color{blue}{\cos\theta}&\color{red}{-\sin\theta}\\
\color{blue}{\sin\theta}&\color{red}{\cos\theta}
\end{array}
\right)
\end{align}

Background: Why can we derive?

What we have to know is following simple property:

Useful property of matrix

Any matrix $M$ can be written as \begin{align} M=(Me_1,...,Me_n), \end{align} where \begin{align} e_i= \left( \begin{array}{c} 0 \\ \vdots \\ 1\\ \vdots \\ 0 \end{array} \right) (i~~~ \end{align} is a unit vector.

From this fact, we can derive any matrix by calculating transformation of each unit vectors.

Proof.
Let $(Me_i)_j$ be the $j$-th element of the vector $Me_i$.
Then $(Me_i)_j=\sum_{k=1}^n M_{jk}(e_i)_k=M_{ji}$.
Where, we use
\begin{align}
(e_i)_k=\delta_{ik}=
\begin{cases}
\,1& (k=i)\\
\,0& (k\neq i)
\end{cases}.
\end{align}
($\delta_{ik}$ is called Kronecker delta.)

Useful knowledge for calculation

Prefix

Name	Symbol	$10^n$
kilo	k	$10^3$
hecto	h	$10^2$
deca	da	$10^1$
		$10^0$
deci	d	$10^{-1}$
centi	c	$10^{-2}$

Pressure

\begin{align}
1 \mathrm{\,Pa}&=1 \mathrm{\,N/m^2}\\
\text{Atmospheric pressure}&=1013.25 \mathrm{\,hPa}= 0.101325\mathrm{\,MPa}
\end{align}

Others

Density of water:

$$\rho \sim 1 \mathrm{\,g/cm^3}=1000 \mathrm{\,kg/m^3}$$

Water pressure:

Let gravitational acceleration be $g\sim 10 \mathrm{\,m/s^2}$.
According to above calculation, water pressure at a depth of 1m is
$$10^4 \mathrm{\,(kg\cdot m/s^2)/m^2}=10^4\mathrm{\,N/m^2}=10\mathrm{\,k Pa}.$$

Contraction of completely anti-symmetric tensor

Contraction formula of the completely antisymmetric tensor (or Levi-Civita symbol) is indispensable for vector calculus.

If you think it is hard to remember the formula, it is wrong.
Actually, the derivation is very easy.

At the end of this article, you will be able to calculate contraction formula in your head!
$$\epsilon_{ijk}\epsilon^{ilm}=?$$

Preparations

Completely anti-symmetric tensor $\epsilon_{ijk}$ has following properties:

Anti symmetricity

\begin{align} \epsilon_{\color{red}{\LARGE i}\color{blue}{\LARGE j}k}=-\epsilon_{\color{blue}{\LARGE j}\color{red}{\LARGE i}k},\qquad \epsilon_{i\color{red}{\LARGE j}\color{blue}{\LARGE k}}=-\epsilon_{i\color{blue}{\LARGE k}\color{red}{\LARGE j}},\qquad \epsilon_{\color{red}{\LARGE i}j\color{blue}{\LARGE k}}=-\epsilon_{\color{blue}{\LARGE k}j\color{red}{\LARGE i}} \end{align}

From this property, we notice that if two indices are the same, completely anti-symmetric tensor is equal to zero.
For example, $\epsilon_{iik}=-\epsilon_{iik}$ then $2\epsilon_{iik}=0$.
We conclude that

\begin{align} \epsilon_{\color{red}{\LARGE ii}k}=\epsilon_{k\color{red}{\LARGE ii}}=\epsilon_{\color{red}{\LARGE i}k \color{red}{\LARGE i}}=0 \end{align}

Derivation (3 dimensions case)

Step. 1 Enumerate all the cases of $~\neq 0\Rightarrow$ Roughly determine the formula

Since $\epsilon_{ijk}$ is a completely anti-symmetric tensor, $\epsilon_{ijk}\epsilon^{ilm}\neq0$ only when

1. $j=l$, $k=m$,
2. $j=n$, $k=l$.

Then contraction formula can be represented as $$\epsilon_{ijk}\epsilon^{ilm}=a\cdot\delta_{jl}\delta_{km}+b\cdot\delta_{jm}\delta_{kl}$$ with two constants $a,b$.

Step. 2 Find unknown constants: Consider anti-symmetricity of indices

Since left-hand side is anti-symmetric with exchange of $j,k$ (or $l, m$), right-hand side must be the same*1: $$\epsilon_{ijk}\epsilon^{ilm}=a(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}).$$ Substituting $(j,k)=(l,m)=(2,3)$, we can get $a=1$. Now, we have got the following contraction formula!:

Contraction formula

$$\epsilon_{ijk}\epsilon^{ilm}=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}$$

Exercise

Calculate following case (contractions of two pair indices): $$\epsilon_{\color{red}{\large i}\color{blue}{\large j}k}\epsilon^{\color{red}{\large i}\color{blue}{\large j}m}=?$$ Do not use above formula. Calculate in your head (you can calculate faster!).

Higher dimensions case

The process is same as the three dimensions case.

*1:Or we can derive as follows: Since $\mathrm{LHS}=0$ when $j=k$ (or $l=m$), $a+b=0$.