Rotation matrix vs regular matrix

  • Thread starter dmoney123
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  • #1
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Can you calculate eigenvalues and eigenvectors for rotation matrices the same way you would for a regular matrix?

If not, what has to be done differently?
 

Answers and Replies

  • #2
Deveno
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it don't see why not. the existence of real eigenvalues will depend on the angle of rotation (most angles will give complex eigenvalues).
 
  • #3
chiro
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The determinant of a rotation matrix should always be 1 (since it preserves length) so there should always be eigenvalues and eigenvectors that can be calculated given a rotation matrix.
 
  • #4
Deveno
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The determinant of a rotation matrix should always be 1 (since it preserves length) so there should always be eigenvalues and eigenvectors that can be calculated given a rotation matrix.
the characteristic polynomial of:

[tex]\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}[/tex]

is:

[tex]x^2 - (2\cos\theta)x + 1[/tex]

which has real solutions only when:

[tex]4\cos^2\theta - 4 \geq 0 \implies \cos\theta = \pm 1[/tex]

for angles that aren't integer multiples of pi, this will lead to complex eigenvalues.
 
  • #5
AlephZero
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the characteristic polynomial of:

[tex]\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}[/tex]

is:

[tex]x^2 - (2\cos\theta)x + 1[/tex]
... and the eigenvalues are [itex]\cos\theta \pm i \sin\theta[/itex]. Now, I wonder what that fact might have to do with "rotation"... :smile:
 
  • #6
chiro
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the characteristic polynomial of:

[tex]\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}[/tex]

is:

[tex]x^2 - (2\cos\theta)x + 1[/tex]

which has real solutions only when:

[tex]4\cos^2\theta - 4 \geq 0 \implies \cos\theta = \pm 1[/tex]

for angles that aren't integer multiples of pi, this will lead to complex eigenvalues.
What has that got to do with what I said?
 

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