MHB Matrix Ops: R(x)v & R(x)w Rotate Counter-Clockwise

[brunette15]

I have the following matrix R(x) = [cos(x) -sin(x) ; sin(x) cos(x)]

Now consider the unit vectors v = [1;0] and w = [0,1].

Now if we compute R(x)v and R(x)w the vectors are supposed to rotate about the origin by the angle x in a counter clockwise direction. I am struggling to see how this works.

Can anyone please further explain this idea?

Thanks in advance!

 

[Prove It]

I have the following matrix R(x) = [cos(x) -sin(x) ; sin(x) cos(x)]

Now consider the unit vectors v = [1;0] and w = [0,1].

Now if we compute R(x)v and R(x)w the vectors are supposed to rotate about the origin by the angle x in a counter clockwise direction. I am struggling to see how this works.

Can anyone please further explain this idea?

Thanks in advance!

First of all, it helps if you can visualise a point in both Cartesian form (which it is usually written in its matrix form as) and its polar form, in other words, in terms of its distance from the origin (radius) and its direction (angle swept out).

Consider a point $\displaystyle \begin{align*} (x, y) \end{align*}$. It can be written in its polar form as $\displaystyle \begin{align*} \left( r\cos{ ( \theta ) }, r\sin{ (\theta )} \right) \end{align*}$. Suppose it is rotated by an angle of $\displaystyle \begin{align*} \alpha \end{align*}$ in the anticlockwise direction. Then the new point $\displaystyle \begin{align*} \left( x' , y' \right) \end{align*}$ has the same distance, but now its angle has $\displaystyle \begin{align*} \alpha \end{align*}$ added to it, thus $\displaystyle \begin{align*} \left( x' , y' \right) = \left( r\cos{ \left( \theta + \alpha \right) } , r\sin{ \left( \theta + \alpha \right) } \right) \end{align*}$. This doesn't really help us though, because we would like to be able to see a transformation in terms of the original x and y. Thankfully they simplify with the compound angle identities as

$\displaystyle \begin{align*} x' &= r\cos{ \left( \theta + \alpha \right) } \\ &= r \left[ \cos{ \left( \theta \right) } \cos{ \left( \alpha \right) } - \sin{ \left( \theta \right) } \sin{ \left( \alpha \right) } \right] \\ &= r\cos{ \left( \theta \right) } \cos{ \left( \alpha \right) } - r\sin{ \left( \theta \right) } \sin{ \left( \alpha \right) } \\ &= x\cos{ \left( \alpha \right) } - y\sin{ \left( \alpha \right) } \end{align*}$

and

$\displaystyle \begin{align*} y' &= r\sin{ \left( \theta + \alpha \right) } \\ &= r\left[ \sin{\left( \theta \right) } \cos{ \left( \alpha \right) } + \cos{ \left( \theta \right) } \sin{ \left( \alpha \right) } \right] \\ &= r \sin{ \left( \theta \right) } \cos{ \left( \alpha \right) } + r\cos{ \left( \theta \right) } \sin{ \left( \alpha \right) } \\ &= y\cos{ \left( \alpha \right) } + x\sin{ \left( \alpha \right) } \end{align*}$

Can you see how it would look in matrix form now?

 

[I like Serena]

I have the following matrix R(x) = [cos(x) -sin(x) ; sin(x) cos(x)]

Now consider the unit vectors v = [1;0] and w = [0,1].

Now if we compute R(x)v and R(x)w the vectors are supposed to rotate about the origin by the angle x in a counter clockwise direction. I am struggling to see how this works.

Can anyone please further explain this idea?

Thanks in advance!

Hey brunette15! (Smile)

See how v = [1;0], the first unit vector, "selects" the leftmost column of the matrix?
So the leftmost column has to be the image of the first unit vector.
Indeed, [cos(x); sin(x)] is the unit vector rotated by an angle of x.

Same for the second unit vector, w = [0;1], that must be mapped to [-sin(x) ; cos(x)].

In general, if you want to find any matrix, consider what the images of the unit vectors must be.
Put them beside each other in a matrix and presto! (Mmm)

 

[brunette15]

Thankyou!