Top 5 2x2 matrices ranked

LinearAlgebra

• In your linear algebra classes, you are introduced to matrices for the first time, but you may not be away of some of these cool properties that can occur with simple matrices.

Rank 5) The Identity Matrix

$$\begin{array}{r}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$$

• Its the identity matrix. not much to say. its everywhere, and given any vector $\left[\begin{array}{c}a\\ b\end{array}\right]$ it will always give you back the same vector

Rank 4 (and Rank 3) Quaternion Matrices (Specifically, the group of order 8)

• Quaternions are one of the coolest small order group who elements have their own unique representation. You can construct every element in the group with only three matrices!

$$\begin{array}{r}\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right],\left[\begin{array}{cc}0& i\\ i& 0\end{array}\right],\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$$

Rank 2) Fibonacci Matix!

• When we examine this matrix $\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$ and transform the vector $\left[\begin{array}{c}0\\ 1\end{array}\right]$, you return to get vector $\left[\begin{array}{c}1\\ 1\end{array}\right]$. It doesn’t look all too impressive until we repeatedly compose this linear transformation with itself. lets see what happens to the following :

$$\begin{array}{r}\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\circ \left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]\end{array}$$ $$\begin{array}{r}\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]=\left[\begin{array}{c}2\\ 3\end{array}\right],\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}2\\ 3\end{array}\right]=\left[\begin{array}{c}3\\ 5\end{array}\right],...\end{array}$$

putting these vectors side by side, we notice something intersting:

$$\begin{array}{r}\left[\begin{array}{c}0\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 2\end{array}\right],\left[\begin{array}{c}2\\ 3\end{array}\right],\left[\begin{array}{c}3\\ 5\end{array}\right],...,\left[\begin{array}{c}{f}_n\\ f_{n+1}\end{array}\right]\end{array}$$

• We have constructed the Fibonacci sequence by only using matrices! Using this fact, we can compute any nth fibbocanni term using the various techniques in linear algebra!
• Our general formula becomes :

$$\begin{array}{r}\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\underset{\text{n times}}{\underbrace{\circ ...\circ }}\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}{f}_n\\ f_{n+1}\end{array}\right]\end{array}$$

• Here are more links if you are interested in learning more!

Fibonacci on Crack

Rank 1) Rotation Matrices and giving us easy trig identities

Using the rotation matrix

$$\begin{array}{r}{\left[\begin{array}{c}{R}_{\theta }\end{array}\right]}_B^B=\left[\begin{array}{cc}\cos (\theta )& -\sin (\theta )\\ \sin (\theta )& \cos (\theta )\end{array}\right]\end{array}$$

We can easily derive Trig identities instead of trying to rely on rote memorization. Here is the process on why it works:

• Rotation $a$ + Rotation $b$ = Rotation $a+b$, Hence you can conceptualize composition of these items as adding the rotation between the two angels. You apply rotation by angle $a$ 1st, then by angle $b$ second. This gives us that:

$$\begin{array}{r}{R}_a\circ R_b=R_{a+b}=R_{b+a}\end{array}$$

This naturally gives us the following identity:

$$\begin{array}{r}{\left[R_a\circ R_b\right]}_B^B=[R_a{]}_B^B[R_b{]}_B^B\end{array}$$

Giving us this beautiful result that

$$\begin{array}{r}{\left[\begin{array}{c}{R}_{a+b}\end{array}\right]}_B^B=[R_a{]}_B^B[R_b{]}_B^B\end{array}$$

What does this tell us? It gives us a quick way to compute cos(a + b) and sin(a + b) as

$$\begin{array}{r}\left[\begin{array}{cc}\cos (a+b)& -\sin (a+b)\\ \sin (a+b)& \cos (a+b)\end{array}\right]=\left[\begin{array}{cc}\cos (a)& -\sin (a)\\ \sin (a)& \cos (a)\end{array}\right]\left[\begin{array}{cc}\cos (b)& -\sin (b)\\ \sin (b)& \cos (b)\end{array}\right]\end{array}$$

Using the matrix multiplication techniques, we can easily see that

$$\begin{array}{rl}\cos (a+b)& =\cos (a)\cos (b)-\sin (a)\sin (b)\\ \sin (a+b)& =\sin (a)\cos (b)+\cos (a)\sin (b)\end{array}$$

• Using this same process, one could even find a triple angle formula by composing a third matrix with a third angle