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# Eigenvalues of rotation (TIFR 2014 problem 9)

Question:

Let $$A(\theta)=\begin{pmatrix} cos(\theta) & sin(\theta) \\ -sin(\theta) & cos(\theta) \end{pmatrix}$$

where $$\theta \in (0,2\pi)$$. Mark the correct statement below:

A. $$A(\theta)$$ has eigenvectors in $$\mathbb{R}^2$$ for all $$\theta \in (0,2\pi)$$

B.  $$A(\theta)$$ does not have eigenvectors in $$\mathbb{R}^2$$ for any $$\theta \in (0,2\pi)$$

C.  $$A(\theta)$$ has eigenvectors in $$\mathbb{R}^2$$ for exactly one values of  $$\theta \in (0,2\pi)$$

D. $$A(\theta)$$ has eigenvectors in $$\mathbb{R}^2$$ for exactly 2 values of $$\theta \in (0,2\pi)$$

Discussion:

If we try to write the linear operator on $$\mathbb{R}^2$$ which is the clockwise rotation by angle $$\theta$$ then we see that with respect to standard basis, the matrix is $$A(\theta)$$. This is quite easy to see, for $$(1,0) \to (cos(\theta),-sin(\theta))$$, and $$…$$ (draw the picture if you are not convinced about this).

What does having an eigenvector mean in this context? Well, as always, it means that $$Av=\lambda v$$ for some $$v \neq 0$$ in $$\mathbb{R}^2$$. But geometrically, this means that upon applying the transformation (in this case rotation) we get the resulting vector in the spanning space of $$v$$, that is the resulting vector must be in the line passing through $$v$$ and the origin.

Geometrically, after rotation, the vector is in the same line is possible only when either the angle of rotation is $$0$$ or the angle of rotation is $$\pi$$. That is either we did not move at all or we basically reflected about origin.

$$\theta \neq 0$$, so we are left with $$\theta=\pi$$. Therefore, we know option C is true.

If you do not trust your geometric sense at all, then you would want to look at the characteristic polynomial.

We have $$det(A(\theta))=cos^2(\theta)+sin^2(\theta)=1$$ and $$trace(A(\theta))=2cos(\theta)$$.

Therefore, the characteristic polynomial is $$x^2-2cos(\theta)x+1$$. This has a real root if the discriminant is positive. That is if $$4cos^2(\theta)-4 \ge 0$$ i.e, $$cos^2(\theta) \ge 1$$. We know that the maximum value of cos is 1 so $$cos(\theta)=1$$ or $$cos(\theta)=-1$$. Since $$\theta \ne 0$$ we are forced to conclude $$\theta = \pi$$ and we have the answer once more.

October 23, 2017