- #1

- 32

- 1

If not, what has to be done differently?

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- Thread starter dmoney123
- Start date

- #1

- 32

- 1

If not, what has to be done differently?

- #2

Deveno

Science Advisor

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- #3

chiro

Science Advisor

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- #4

Deveno

Science Advisor

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- 6

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

Science Advisor

Homework Helper

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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"...

- #6

chiro

Science Advisor

- 4,797

- 133

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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