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We are given the following problem:

Consider the matrix $$ A = \left[\begin{array}{rrr} \cos\theta & \sin\theta\\ \sin\theta & -\cos\theta \end{array}\right] $$, where $\theta \in \mathbb R$.

a) Show that $A$ has an eigenvector in $\mathbb{R^2}$ with eigenvalue $1$.

We start the problem by taking the difference of the original matrix with the product of lambda and the identity:

$$A-\lambda I = \left[\begin{array}{rrr} \cos\theta -\lambda & \sin\theta\\ \sin\theta & -\cos\theta -\lambda \end{array}\right] $$

We then find its determinant:

$$det(A-\lambda I) = (\cos\theta -\lambda)(-\cos\theta -\lambda) - \sin\theta^2$$

which gives us: $\lambda^2 - 1$ and eigenvalues of $\lambda_1 = 1, \lambda_2 = -1$

Using $\lambda_1=1$ we have:

$$B(1) = \left[\begin{array}{rrr} \cos\theta -1 & \sin\theta\\ \sin\theta & -\cos\theta -1 \end{array}\right] $$

How do I reduce it and answer the problem: Show that $A$ has an eigenvector in $\mathbb{R^2}$ with eigenvalue $1$?

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  • $\begingroup$ We already know it has an eigenvalue of 1 (above), but we have yet to find the eigenvector. I would surmise reducing the matrix B to row echelon form and solving the system, this will give us our vector for $\lambda_1$ $\endgroup$ – user4640007 May 17 '15 at 3:09
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You want to solve the equation

$$\left(\begin{array}{cc} \cos\theta - 1 & \sin\theta \\ \sin\theta & -\cos\theta -1\end{array}\right)\left(\begin{array}{cc} x \\ y\end{array}\right) = \left(\begin{array}{cc} 0 \\ 0\end{array}\right).$$

Since this is an eigenvalue equation, the two rows are linearly dependent. Let's work with the first row without loss of generality. Then

$$x(\cos\theta-1) + y\sin\theta = 0.$$

That is to say that

$$ x = \frac{\sin\theta}{1-\cos\theta}y.$$

Thus your eigenvector is nothing more than

$$\left(\begin{array}{cc} \dfrac{\sin\theta}{1-\cos\theta} \\ 1\end{array}\right).$$

If you wanted to work with the second equation, you could relate the two eigenvectors you get by using the Pythagorean identity.

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there is no need to compute the eigenvalues. the reason is that the matrix $A$ represents the reflection on the line $ y = \tan(\theta/2) x.$ therefore the eigenvalues are $1, -1.$ the corresponding eigenvectors are $\pmatrix{\cos(\theta/2)\\\sin(\theta/2)}, \pmatrix{-\sin(\theta/2)\\\cos(\theta/2)}.$

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