Rotation matrix formula derivation?

Question

I would like to understand where to following formula for a 2-D rotation matrix is derived from:

$R =\frac{1}{a^2+b^2} \left(\begin{array}{cc} a^2-b^2 & 2ab\\ -2ab & a^2-b^2 \end{array}\right)$

I believe the answer is geometric and if it is, a picture would be nice.

@LordSharktheUnknown I would not like to bring sin and cos into this if possible. — Tony, Jun 23 '17 at 19:14
@LordSharktheUnknown Well, $a$ may equal values greater than $1$ ... — Michael Hoppe, Jun 23 '17 at 19:39

score 1 · Accepted Answer · answered Jun 23 '17 at 19:24

1

Consider the unit vector

$$\vec u=\left(\frac a{\sqrt{a^2+b^2}},\frac b{\sqrt{a^2+b^2}}\right).$$

The rotation that applies $(1,0)$ to $\vec u$ and $(0,1)$ to a unit vector orthogonal to $\vec u$ is described by the matrix

$$R=\frac1{\sqrt{a^2+b^2}}\begin{bmatrix}a&b\\-b&a\end{bmatrix}.$$

Applying the rotation twice, i.e. squaring the matrix gives

$$R^2=\frac1{a^2+b^2}\begin{bmatrix}a^2-b^2&2ab\\-2ab&a^2-b^2\end{bmatrix}.$$

answered Jun 23 '17 at 19:24

2

Anyway, you shouldn't disregard the remark of Lord Shark the Unknown. – Jun 23 '17 at 19:27
isn't the matrix applying the rotation to the "opposite" direction now? Seems like the minus on the element that is set to b should be on the RH upper corner, not LH lower corner. – Tony Jun 26 '17 at 03:00
@Tony: a matter of convention. – Jul 03 '17 at 08:27

1 Answers1