In Sheldon Axler's Linear Algebra on page 76 he writes:
The rotation of a nonzero vector in $R^2$ is obviously never equals a scalar multiple of itself.
Why is rotation by 180 degrees not the same as multiplication by $-1$?
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1Note: In the second edition, the quote occurs on page 78, not page 76. (It might be different in the first edition.) – Daniel Hast Aug 13 '13 at 07:24
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1You're right about the rotation by 180 degrees. Not having that textbook at hand makes it difficult to judge, but presumably the context makes it plain that this is no cause for concern. – Jyrki Lahtonen Aug 13 '13 at 07:27
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The context of that sentence is:
For a more complicated example, consider the operator $T\in\mathcal L(\mathbf F^2)$ defined by $$T(w,x)=(-x,w).$$ If $\mathbf F=\mathbf R$, then this operator has a nice geometric interpretation: $T$ is just a counterclockwise rotation by $90^\circ$ about the origin in $\mathbf R^2$. An operator has an eigenvalue if and only if there exists a nonzero vector in its domain that gets sent by the operator to a scalar multiple of itself. The rotation of a nonzero vector in $\mathbf R^2$ obviously never equals a scalar multiple of itself. Conclusion: if $\mathbf F=\mathbf R$, the operator $T$ defined by 5.4 has no eigenvalues. However, if $\mathbf F=\mathbf C$, the story changes. To find the eigenvalues of $T$, [...]
I think it's reasonable to conclude that the sentence in question is referring specifically to $T$ by "the rotation", not a general rotation by any angle.
That said, you're right! Rotation by 180 degrees in the plane is the same as multiplication by $-1$. For that matter, rotation by 360 degrees is the same as multiplication by $1$.
Marc van Leeuwen
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Chris Culter
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1@Frosty I wouldn't go that far. Like most sentences, the evaluation of this sentence depends on the context. – Chris Culter Apr 08 '14 at 18:48