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# TIFR 2014 Problem 9 Solution - Eigenvalues of Rotation TIFR 2014 Problem 9 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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## Problem:

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.

## Helpdesk

• What is this topic: Linear Algebra
• What are some of the associated concept: Eigenvectors,Characteristic Polynomial
• Book Suggestions: Linear Algebra done Right by Sheldon Axler

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