Find eigenvector of the linear operator

Question

Task is to find an eigenvector of the following linear operator:
$f \to \int^{x}_{-x} f(t)dt$ in the linear span $\langle cos(x), sin(x), ...,cos(mx),sin(mx)\rangle$.

I know how to find eigenvectors and have the general idea about what linear span and linear operators are, but still I feel like I'm a bit out of my depth with this problem.

Since we see integral in function, mapping transforms original vector to $\mathbb{R}$. But I keep wondering what is an original vector space, what's $x$ and $t$ here.

Please note, I probably would fail to understand the complete solution and it's not what I want. I would appreciate just some hints on how to approach this problem. Thanks!

Answer 1

Remember that the vector space you're working in is a space of functions. Let's call that space of functions $V$ ($V$ is the span of the functions you listed; those functions form a basis for V). Because the integral in the definition of the operator has $x$ in its bounds, the result of the integration is not just a real number, but a function of $x$. And in fact, the result lies in the span of the functions, i.e., in $V$. This means that the operator maps $V$ to $V$. (This is important if we're going to talk about eigenvalues and eigenvectors, which only make sense for linear maps from a vector space to itself.)

I would start by doing this: calculate $\int_{-x}^x \sin(kt) dt$ and $\int_{-x}^x \cos(kt) dt$. Your answers will be functions of $x$ (that depend on the constant $k$), and you should be able to write them in terms of the basis functions.

You'll probably be able to see some eigenvectors just by doing that. But if you're more comfortable working with matrices, I would write down the matrix for your operator with respect to the given basis. If this seems hard, you could take $m = 1$ first, so that your basis contains just two elements ($\sin x $ and $\cos x$) and the matrix for the operator is $2 \times 2$. Then see what happens when $m=2$, and so on, until you can understand the pattern. Once you have the matrix, call it $M$, you can do the usual $\det( M - \lambda I) = 0$ business to find the eigenvalues, or better yet, just look at the matrix and see what the eigenvalues are.

Answer 2

The operator is related to an indefinite integral, except it's symmetric. You can quickly see what it does:

$$\cos(x)\mapsto \int_{-x}^x\cos(t)dt=-\sin t|_{-x}^x=-2\sin x$$

What's immediately apparent is that an integral of an odd function over a symmetric interval is always zero. Therefore, the entire subspace $\sin x,\sin 2x,\cdots$ is an eigenspace with an eigenvalue equal to $0$. Additionally, all the cosines map to sines... so the operator is nilpotent (square of this operator is a null operator).

To sum up, all the sines are eigenvectors with eigenvalue of 0, and these are all of them (meaning that the operator is not diagonalizable, as the number of eigenvectors is half the dimension of the space).

Answer 3

Linear operator is just a linear transformation. That is a function $T$ that preserves constant multiplication and function addition. That is:

• $T(rf(x))=rT(f(x))$ and

• $T((f+g)(x))=T(f(x))+T(g(x))$

Linear span means all the linear combinations of said functions. In particular they are restricting the domain of your linear operator to that set:

$$ f(x) \in \left<\sin(x),\cos(x),\ldots,\sin(mx),\cos(mx)\right>\\ =\{a_1\sin (x)+b_1\cos(x)+a_2\sin(2x)+b_2\cos(2x)+\cdots+ a_m\sin(mx)+b_m\cos(mx) \} $$

So if you take a generic $f(x)$ in the domain of $f(x) \to \int_{-x}^{x}f(x)\ dx$ it has the form:

$$ f(x)=a_1\sin (x)+b_1\cos(x)+a_2\sin(2x)+b_2\cos(2x)+\cdots+ a_m\sin(mx)+b_m\cos(mx) $$ Note each term can be integrated by substitution: $$\int_{-x}^{x}a_k\sin(kx)\ dx=0$$

and $$\int_{-x}^{x}b_k\cos(kx)\ dx=\frac{-2b_k}{k}\sin(kx)$$

Suppose $f(x)\neq 0$ is an eigenvector then $\int_{-x}^xf(x)\ dx=\lambda f(x)$ for some constant $\lambda$

$$ \int_{-x}^{x}f(x)\ dx=-\frac{2b_1}{1}\sin(x)-\frac{2b_2}{2}\sin(2x)-\cdots- \frac{2b_m}{m}\sin(mx) $$

So

$$\lambda(a_1\sin (x)+b_1\cos(x)+a_2\sin(2x)+b_2\cos(2x)+\cdots+ a_m\sin(mx)+b_m\cos(mx))\\ =-\frac{2b_1}{1}\sin(x)-\frac{2b_2}{2}\sin(2x)-\cdots- \frac{2b_m}{m}\sin(mx)$$

Since the set $\{\sin(x),\cos(x),\sin(2x),\cos(2x),\ldots,\sin(mx),\cos(mx)\}$ is linearly independent, you can conclude all the sine terms have zero coefficients $a_i=0$ for all $i$ and similarly the coefficients of cosine $\lambda b_i$ are all zero but $f(x) \neq 0$ so $\lambda=0$ and the eigenvectors are made up of cosines only.