Strang's Linear Algebra: rotation & eigenvectors

The rotation matrix R is a fairly basic matrix from linear algebra. When multiplied times a vector, it rotates the vector through the angle θ: We've seen it before (here).

R = [ cos(θ) -sin(θ) ] [ sin(θ) cos(θ) ]

In Strang section 6.2 (first challenge problem), we are asked to prove the reasonable supposition:

R(θ)n = R(nθ)

That is, n sequential applications of the R vector should give the same result as a single rotation through an angle of nθ.

To calculate Rⁿ we'll need the eigenvalues and eigenvectors of R. The interest of this problem comes from consideration of the following question:

what sort of vector could have the same direction after rotation by an arbitrary angle θ?

Use the standard formula:

det (A - λI) = 0 [ cos(θ) - λ -sin(θ) ] [ sin(θ) cos(θ) - λ ] [cos(θ) - λ]2 + sin2(θ) = 0 λ2 - 2 λ cos(θ) + cos2(θ) + sin2(θ) = 0 λ2 - 2 λ cos(θ) + 1 = 0

We need the quadratic equation to find λ:

[2 cos(θ) +/- sqrt[4 cos2(θ) - 4] ] / 2 [2 cos(θ) +/- 2 sqrt[cos2(θ) - 1] ] / 2 cos(θ) +/- sqrt[cos2(θ) - 1] cos(θ) +/- sqrt[-sin2(θ)] cos(θ) +/- i sin(θ) λ1 = cos(θ) + i sin(θ) λ2 = cos(θ) - i sin(θ)

Strang gives the eigenvectors as

x1 = (1,i) x2 = (i,1)

However, I think this is a misprint since R x₁ gives:

[ cos(θ) -sin(θ) ] [ 1 ] = [ cos(θ) - i sin(θ) ] [ sin(θ) cos(θ) ] [ i ] [ sin(θ) + i cos(θ) ]

but

λ1 x1 = cos(θ) + i sin(θ) [ 1 ] [ i ] = [ cos(θ) + i sin(θ) ] [ - sin(θ) + i cos(θ) ]

We could solve this:

(A - λI1) x1 = 0 (A - λI2) x2 = 0 For x1 [ -i sin(θ) -sin(θ) ] [ ? ] = [ 0 ] [ sin(θ) -i sin(θ) ] [ ? ] [ 0 ]

Without writing it out, it looks like we just need to try a change of sign:

x1 = (-1,i) A x1 = λ1 x1 [ cos(θ) -sin(θ) ] [ -1 ] = [ - cos(θ) - i sin(θ) ] [ sin(θ) cos(θ) ] [ i ] [ - sin(θ) + i cos(θ) ]

and λ₁ x₁ gives the same result:

λ1 x1 = cos(θ) + i sin(θ) [ -1 ] = [ - cos(θ) - i sin(θ) ] [ i ] [ - sin(θ) + i cos(θ) ]

Similarly, x₂ should be (i,-1) because R x₂ equals:

[ cos(θ) -sin(θ) ] [ i ] = [ sin(θ) + i cos(θ) ] [ sin(θ) cos(θ) ] [ -1 ] [ - cos(θ) + i sin(θ) ]

and

λ2 x2 = cos(θ) - i sin(θ) [ i ] [ -1 ] = [ sin(θ) + i cos(θ) ] [ - cos(θ) + i sin(θ) ]

Everything checks.

Now that we have the eigenvalues and eigenvectors, we need to diagonalize R. We construct the columns of S directly from the eiegenvectors:

S = [ -1 i ] [ i -1 ]

Λ is just

[ λ1 0 ] [ 0 λ2 ]

To get S^-1 let's try the trick for 2 x 2's:

S-1 = 1/2 [ -1 -i ] [ -i -1 ]

and multipy to confirm we do get I:

1/2 [ -1 -i ] [ -i -1 ] [ -1 i ] [ 1 0 ] [ i -1 ] [ 0 1 ]

Let's also confirm that S Λ S^-1 = R. Part one is:

[ cos(θ) + i sin(θ) 0 ] [ 0 cos(θ) - i sin(θ) ] [ -1 i ] [ -cos(θ) - i sin(θ) sin(θ) + i cos(θ) ] [ i -1 ] [ -sin(θ) + i cos(θ) -cos(θ) + i sin(θ) ]

Step 2:

1/2 [ -1 -i ] [ -i -1 ] [ -cos(θ) - i sin(θ) sin(θ) + i cos(θ) ] [ a b ] [ -sin(θ) + i cos(θ) -cos(θ) + i sin(θ) ] [ c d ] a = 1/2 [ cos(θ) + i sin(θ) - i sin(θ) + cos(θ) ] = cos(θ) b = 1/2 [ i cos(θ) - sin(θ) - sin(θ) - i cos(θ) ] = -sin(θ) c = 1/2 [ sin(θ) - i cos(θ) + i cos(θ) + sin(θ) ] = sin(θ) d = 1/2 [ i sin(θ) + cos(θ) + cos(θ) - i sin(θ) ] = cos(θ)

And it checks.

To do the exponentiation we just need to calculate Λⁿ. So what is:

λ1n = [ cos(θ) + i sin(θ) ]n λ2n = [ cos(θ) - i sin(θ) ]n

Strang drops another hint by reminding us of Euler's formula:

eiθ = cos(θ) + i sin(θ) (eiθ)n = einθ

which is the whole point! But what about:

λ2n = [ cos(θ) - i sin(θ) ]n

If we compare -θ with θ then:

cos(-θ) = cos(θ) sin(-θ) = -sin(θ) cos(θ) - i sin(θ) = cos(-θ) + i sin(-θ) = e-iθ (e-i&theta)n = e-in&theta

Now all we have to do is convert back to sine and cosine:

λ1n = cos(nθ) + i sin(nθ) λ2n = cos(nθ) - i sin(nθ)

The big computation we did above (to check that S Λ S^-1 = R, is exactly the same with the substitution of nθ for θ.

R(θ)n = [ cos(nθ) -sin(nθ) ] [ sin(nθ) cos(nθ) ]

We're done!