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anonymous

  • one year ago

The matrix P that multiplies (x, y, z) to give (z, x, y) is also a rotation matrix. Find P and P^3. The rotation axis a = (1, 1, 1) doesn't move, it equals Pa. What is the angle of rotation from v = (2, 3, -5) to Pv = (-5, 2, 3)?

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  1. anonymous
    • one year ago
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    So far, I've found \[P = \left[\begin{matrix}0 & 0&1 \\1&0&0\\0&1&0\end{matrix}\right], P ^{3} = I\]

  2. JoshDanziger23
    • one year ago
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    fugisawa, If you're confident that P is a rotation matrix (as evidenced by the fact that it is orthogonal with determinant 1), then you can almost immediately deduce from P<>I and P^3=I that the rotation angle theta must be 2 pi/3. This is confirmed by the eigenvalues, 1 and exp(+/- 2 pi /3), and for even greater certainty you can see that v is perpendicular to the axis (v.a=0) allowing you to deduce the angle of rotation simply by taking the angle between v and Pv, ie, cos(theta) = (Pv).v/(||Pv|| ||v||)=-1/2 (of course this is only valid for vectors perpendicular to the axis). If you need further convincing that P is a rotation matrix then for any point v not on the axis you can find the vector perpendicular to the axis connecting it to v, and show that after multiplying v by P the new perpendicular is the same length as the old one, that it meets the axis at the same location and that angle between the old perpendicular and the new perpendicular is the same regardless of the position of v. This isn't hard to do. Hope this helps! Josh

  3. JoshDanziger23
    • one year ago
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    Sorry, for exp(+/- 2 pi/3) read exp(+/- 2 i pi/3)

  4. anonymous
    • one year ago
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    It helped indeed. Thank you. However, this exercise was taken from the book Linear Algebra and Its Applications, by professor Strang, from the first chapter, Matrices an Gaussian Elimination, in which orthogonality, determinants, and eigenvalues and eigenvectors are not yet discussed. The only mention about a rotation matrix is in an exercise that gives the formula for the x-y euclidean plane, and that's it. \[\left[\begin{matrix}\cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{matrix}\right]\]

  5. JoshDanziger23
    • one year ago
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    Here's how I would think about this without using orthogonality, determinants or eigenvalues: We hypothesize that matrix P produces a rotation of theta around an axis running along vector a=(1,1,1) going through the origin. We're given a point v=(x,y,z) and asked to determine the rotation around a. First let's construct the vector r perpendicular to the hypothesized axis of rotation (let's just call it "the axis") and connecting it to v. We know v=ca+r (where c is some scalar) and r.a=0. Taking the dot product of the first equation with a gives v.a=ca.a+r.a; given r.a=0 this implies c=(v.a)/(a.a) and hence (from the first equation) r = v-ca = v-a(v.a)/(a.a). In the same way we can find the vector s perpendicular to the axis connecting it to w=Pv: introducing a scalar d analogous to c we find d=(w.a)/(a.a) and s=w-a(w.a)/(a.a). Now if P is indeed a pure rotation around the axis we would expect for any point v not on the axis that (1) the perpendiculars r and s run from the same point on the axis; (2) they have the same length; and (3) the angle between them is the same regardless of the choice of v. This means (1) c=d, (2) ||r||=||s||, and (3) cos t = r.s/(||r|| ||s||) is constant. With a=(1,1,1), v=(x,y,z) and w=(z,x,y), we have, first, c=(v.a)/(a.a)=(x+y+z)/3 and d=(w.a)/(a.a)=(x+y+z)/3, so c=d is demonstrated. Second, r=(x-c,y-c,z-c) and s=(z-c,x-c,y-c), so ||r||=||s|| is easily demonstrated. Finally cos(theta) = r.s/(||r|| ||s||) = [(x-c)(z-c) + (y-c)(x-c) + (z-c)(y-c)] / [(x-c)^2 +(y-c)^2 + (z-c)^2] = -0.5 (after simplification), so theta = 120 degrees or 2 pi/3. v=(2,3,-5) is a special case where c=d=0, so r=v and s=w.

  6. anonymous
    • one year ago
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    Thank you. That's it! I changed books to Introduction to linear algebra and there it was, the formula for the rotation of vector, just as you posted above.

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