I don't get his "modern" proof. Specifically the step where he says "it's easy to see geometrically that these matrices differ by a rotation" seems to be doing a lot of heavy lifting. The first matrix transforms e1 to (a,-b), the second scales e1 to (c,0). If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
You're assuming that we know that the length of vector (a, -b) is a²+b². We don't know that.
We start by assuming that the position vector (a, -b) has length c. This implies that we can rotate that vector until it becomes the position vector (c, 0).
As you note, we can create the two vectors above from (1, 0) using linear transformation matrices [(a, b), (-b, a)] and [(c, 0), (0, c)]
So we could create the position vector (c, 0) by starting at (1, 0), applying the linear transformation [(a, b), (-b, a)], then applying a rotation to bring it back to the e1 axis.
Thus for some rotation matrix R,
R × [(a, b), (-b, a)] = [(c, 0), (0, c)]
The determinant of a rotation matrix is 1, so the determinant of the left side is 1×(a²+b²), while the determinant of the right side is c², which is how we end up with a²+b²=c².
Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already.
> Now the only thing which I'm not sure of is whether there's a way to show that the determinant of a rotation matrix is 1 without assuming the Pythagorean identity already
You can define the determinant that way. Now the question is why the cross multiplication formula for determinant accurately computes the area.
You can prove that via decomposition into right triangles https://youtu.be/_OiMiQGKvvc?si=TyEge1_0W4rb648b
Or you can go in reverse from the coordinate formula, to prove that the area is correctly predicted by the determinant.
Yep - I'm just not sure if any of those proofs implicitly assume Pythagoras, and haven't thought through them properly.
I was initially going to say we know that det R = 1 by using the trigonometric identity cos²x+sin²x=1, but then found out that all the proofs of it seem to assume Pythagoras, and in fact, the identity is called the Pythagorean trigonometric identity.
Draw it on graph paper:
Set B as the origin, and let BC (the 'a' side), be on the the positive side of the x-axis. Let AC be on the positive side of the y-axis.
The left matrix is a clockwise rotation and scaling. This is clearly seen if you draw the transformation applied to the two axis basis vectors. (The scaling factor isn't obvious yet.)
Then the left matrix varies (1,0) to the side AB, which has magnitude c. Z and carries (0,1) to an perpendicular line of the (importantly) same magnitude,
So it's a rotation and a scaling by c.
The right matrix obviously is a scaling by c.
> If you can see that you obtain one of these vectors by rotating the other, then you've shown that their lengths are equal (i.e. a²+b²=c²), which is what we want to show in the first place.
Yes, that's the point!