# Why this matrix is a rotation matrix and how it work?

So I have the the column matrix $$X=\begin{pmatrix} a\\ b \end{pmatrix}$$ with $$a,b$$ real numbers and to this matrix I have the following vector associated: $$\vec{x}=a\cdot \vec{i}+b\cdot\vec{j}$$

I have the matrix: $$A(t)=\begin{pmatrix} \cos(t) & \sin(t)\\ -\sin(t) & \cos(t) \end{pmatrix}$$ with $$t$$ real number

$$B=\begin{pmatrix} 1\\ 1 \end{pmatrix}$$ , $$U=A(-\frac{\pi}{12})\cdot B$$ and $$V=A(\frac{\pi}{6})\cdot B$$

I need to find the cosinus of vectors $$\vec{u}$$ and $$\vec{v}$$ these vectors being associated to column matrix $$U$$ and $$V$$.The response is $$\frac{\sqrt{2}}{2}$$

So I just did calculations I it's a little bit of work but I find that the matrix A(t) is a rotation matrix with angle t and the cosinus between u and v is pi/4 but at school I didn't learn about rotation matrix.Can you explain me why the angle is pi/4? I mean, if you can, in simple terms how rotation matrix work.

Thanks!

• $A(t)$ acts on the vector $B$ by rotating it by $t$. So, $A(-\pi/12)$ rotates $B$ by $-\pi/12$, and $A(\pi/6)$ rotates $B$ by $\pi/6$. The difference between these two angles is $\pi/6 - (-\pi/12) = \pi/6 + \pi/12 = 3\pi/12 = \pi/4$. – Bungo Jul 7 '19 at 8:29
• I agree with @Bungo's comment/explanation. I just wanted to remark that $A(t)$ is a rotation by angle minus $t$ counterclockwise (i.e. $t$ clockwise). – Malkoun Jul 7 '19 at 9:03
• “How a rotation matrix works” has been answered in many places in many ways, for example math.stackexchange.com/questions/363652/… – David K Jul 7 '19 at 12:32
• So do you want to completely understand what makes that particular formulation of $A(t)$ be a rotation matrix, or do you just want to know how the angle between $u$ and $v$ comes to be $\pi/4$? Did Bungo's comment answer your question? If not, where did it stop making sense? Remember you can edit the question to make it clearer what you really need to know. – David K Jul 8 '19 at 1:56
• Bungo's comment was helpful.I understood.Thanks! – DaniVaja Jul 9 '19 at 8:44

The important thing is that every rotation $$\varphi$$ about the origin is a linear transformation, meaning that, for all vectors $$v, w$$ of the plane and scalar $$\lambda$$, we have $$\varphi(\lambda v) =\lambda\cdot\varphi(v) \\ \varphi(v+w) =\varphi(v) +\varphi(w)$$ These properties can be easily seen geometrically, in particular, drawing the second one says something like '$$\varphi$$ takes parallelograms to parallelograms'.
It is also clear that the composition of rotation by angle $$t$$ with rotation by angle $$s$$ is the rotation by $$s+t$$.
Now, by the basics of linear algebra, since $$\varphi$$ is linear, once a basis $$e_1,e_2$$ is fixed, it can be expressed as multiplication by the matrix whose columns are $$\varphi(e_1),\varphi(e_2)$$ (coordinated in basis $$e_1,e_2$$).
In the given example $$A(t)$$ is the matrix of rotation by angle $$-t$$, with respect to the standard basis $$i,j$$.
Based on the above, it should be straightforward that $$A(t)\cdot A(s)\ =\ A(t+s)$$ which boils down to the trigonometric addition formulas for $$\sin$$ and $$\cos$$ (and proves them at once).