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 $1$?

  • $\begingroup$ 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 $\lambda_1$ $\endgroup$ May 17, 2015 at 3:09

2 Answers 2


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.


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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