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How could I find the eigenvectors of the following matrix?

$$\left(\begin{matrix} \cos(\theta) & \sin(\theta)\\ \sin(\theta)& -\cos(\theta)\end{matrix}\right)$$

I found the eigenvalues $1$ and $-1$ but I'm struggling with calculating the eigenvectors.

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Once you have the eigenvalues $\lambda=\pm1$, you can proceed to find corresponding eigenvectors by the usual method of solving the equation $A\mathbf x=\lambda\mathbf x$, which amounts to finding the kernel of $A-\lambda I$. It might be more instructive, however, to approach this geometrically.

The matrix $A$ represents a reflection in some line through the origin. On that line, reflection is the identity map, so we expect to have $1$ as an eigenvalue with that line as its eigenspace. Reflection can be thought of as reversing the component of a vector that’s perpendicular to the mirror line, so the other eigenvalue must be $-1$, with eigenspace the orthogonal complement of that line. In $\mathbb R^2$, this will be the perpendicular line. (We already knew that we’d have orthogonal eigenvectors because $A$ is symmetric, but it’s nice to verify that a different way.) What this means in terms of the computations is that once we’ve found an eigenvector for one eigenvalue, we can immediately write down an eigenvector for the other. This matches the eigenvalues you found by, I presume, finding the roots of the characteristic polynomial, so, so far, so good.

Let $\mathbf u=(\cos\phi,\sin\phi)^T$ be a unit vector in the direction of the line of reflection. Then, by the above description of reflection we have $A\mathbf x=2(\mathbf x\cdot\mathbf u)\mathbf u-\mathbf x$ for all vectors $\mathbf x$. Rewriting this as $(I+A)\mathbf x=2(\mathbf x\cdot\mathbf u)\mathbf u$ and expanding both sides, we end up with the equations $$\begin{align}(1+\cos\theta)x+y\sin\theta&=2x\cos^2\phi+2y\cos\phi\sin\phi \\ x\sin\theta+(1-\cos\theta)y&=2x\cos\phi\sin\phi+2y\sin^2\phi. \end{align}$$ Equating coefficients, we have $\sin\theta=2\cos\phi\sin\phi=\sin2\phi$, so $\theta=2\phi$. This means that the mirror line makes an angle of $\frac\theta2$ with the $x$-axis and thus we have for eigenvectors $\left(\cos\frac\theta2,\sin\frac\theta2\right)^T$ and $\left(-\sin\frac\theta2,\cos\frac\theta2\right)^T$.

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  • $\begingroup$ Dear Sir, I do not understand why Ax=2(x.u)u-x? Please Can you illustrate. $\endgroup$ Jan 23 '19 at 13:36
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Let $$A = \left(\begin{matrix}\cos(\theta) & \sin(\theta)\\ \sin(\theta)& -\cos(\theta)\end{matrix}\right).$$

$\lambda = +1,-1$ as you have already mentioned.

So eigenvectors corresponding to $\lambda = 1$ is got by solving for $x$ in $$Ax = x.$$

So do the row echelon conversion (or any other method to solve the simultaneous equation) and we get, $$ x = \begin{bmatrix} \frac{-\sin(\theta)}{\cos(\theta)-1} \cdot c \\ c \end{bmatrix},$$ where $c \in \mathbb{R}$.

Similarly one gets eigenvectors span for $\lambda = -1$.

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  • $\begingroup$ If you do some trigonometric manipulations on this eigenvector, you can get a more obviously geometrically meaningful result. $\endgroup$
    – amd
    Feb 11 '17 at 21:04

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