# 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 $\pi/6 - (-\pi/12) = \pi/6 + \pi/12 = 3\pi/12 = \pi/4$.
– user169852
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). 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/… 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. Jul 8 '19 at 1:56
• Bungo's comment was helpful.I understood.Thanks! 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).