# $\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}$ eigenvalue?

How do I compute the eigenvector of $\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}$?

$\begin{bmatrix}\cos\theta&\sin\theta\\\sin\theta&-\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\cos\theta+y\sin\theta\\x\sin\theta -y\cos\theta\end{bmatrix}$

The I got this system of equations: ${x\cos\theta+y\sin\theta=\lambda x\\x\sin\theta -y\cos\theta=\lambda y}\implies$

$\det\begin{bmatrix}\lambda-\cos\theta&-\sin\theta\\\sin\theta&-\lambda-\cos\theta\end{bmatrix}=0\implies -\lambda^2+\cos^2\theta+\sin^2\theta=0\implies\lambda=1$

Questions:

1) How do I compute $\lambda$?

2) Is it possible to get a simple value for the given matrix $A$?

• Your second entry of the product is wrong. Are you familiar with the fact that $\lambda$ is a root of $\det(A-\lambda I)$? Typically, all textbooks write it the line after the definition of "eigenvalue". – user228113 Aug 27 '17 at 11:17
• Which is also the reason why no one ever says "the eigenvalue of $A$". – user228113 Aug 27 '17 at 11:20
• When calculating eigenvalues with the determinant formula, the $\lambda$s always occur along the diagonal. The result will become a second degree polynomial in a polynomial of the complex exponential function. – mathreadler Aug 27 '17 at 11:29
• Geometrically, this is a counterclockwise rotation of $\theta$ radians composed with a reflection about the horizontal axis. So, except for specific values of $\theta$, e.g. $\theta = 0$, expect complex eigenvectors since rotations in $\mathbb{R}^2$ preserve nothing. – Kaj Hansen Aug 27 '17 at 11:31
• "How do I compute $\lambda$"? Solve a quadratic equation. – GEdgar Aug 27 '17 at 11:39

1. Just below the highlighted equation, you have $\sin \theta y$, which is wrong; the $y$ should be next to the $\cos \theta$ (preferably before it, to improve readability).

2. In the determinant just about "Questions", the $\lambda$ in the second line should be in the second column, not the first, so that the bottom row reads $-\sin \theta ~~~ \lambda + \cos \theta$.

If you continue from there, you have some hope of correctly computing $\lambda$.

• Please check my edit! – Pedro Gomes Aug 27 '17 at 11:31
• The second line of your last matrix has sine and cosine swapped. You really need to actually do the algebra right if you want to get anywhere. – John Hughes Aug 27 '17 at 11:34
• Could you please check the answer I provided? Thanks! – Pedro Gomes Aug 27 '17 at 12:30
• Why is there a $-\lambda$ in the lower right of your matrix? I reiterate: you have to do the algebra right; that's the first step. – John Hughes Aug 28 '17 at 14:45

\begin{align} & \det\left(\begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = (\cos\theta-\lambda)^2 + \sin^2\theta \\[10pt] = {} & \lambda^2 - 2\cos\theta + 1 = 0 \\[10pt] &\text{if and only if } \lambda = \cos\theta \pm i\sin\theta. \end{align}

• Check the characteristic equation again, please. The eigenvalues are $\lambda=\pm1$. The determinant of this matrix is $=-1$, which is also the product of the eigenvalues. Geometrically this linear transformation is an orthogonal reflection w.r.t. a line forming an angle $\theta/2$ with the $x$-axis. – Jyrki Lahtonen Sep 5 '17 at 6:05

To hand-calculate the eigenvalues of a $2\times2$ matrices, I believe the best way is not to expand $\det(\lambda I-A)$, but to write down the polynomial $\lambda^2-\operatorname{tr}(A)\lambda + \det A$ directly. This is always more efficient and less error prone.

In your case, clearly $A$ has trace zero and its determinant is $-1$. Hence the characteristic polynomial is $\lambda^2-1$ and the eigenvalues are $\pm1$.