Show that $J$ is diagonalizable and find an eigenbasis

Question

Let $V$ be the complex vector space with basis $B= (1, \cos(x), \sin(x))$. Define the linear operator $J: V \to V$ by

$$J([f])(x) = \int_0^\pi f(x-t) \,dt$$

for all $f$ contained in $V$. Show that $J$ is diagonalizable and find and eigenbasis.

Answer 1

I don't know what you've learned so far, but I'm guessing that you know how to deal with ordinary matrices. Given that $1$, $\cos x$, and $\sin x$ form a basis of $V$, you can represent any element $a+b\cos x+c\sin x$ of $V$ as the vector $\begin{pmatrix}a & b & c\end{pmatrix}^T$. To identify the matrix representing $J$, we can see what it does to the basis vectors $1$, $\cos x$, and $\sin x$: $$J[1](x) = \int_0^\pi1\;dt=\pi$$ $$J[\cos x](x)=\int_0^\pi\cos(x-t)\;dt=2\sin x$$ $$J[\sin x](x) = \int_0^\pi \sin(x-t)\;dt = -2\cos x$$ Evidently, $J$ can be represented by the matrix: $$J=\begin{pmatrix}\pi & 0 & 0\\0 & 0 & -2\\ 0 & 2 & 0\end{pmatrix}$$ Solving the equation $\det(J-\lambda I)=0$ yields the eigenvalues $\lambda = \pm 2i, \pi$. Using standard methods for finding eigenvectors of matrices, you should find that the following are eigenvalue/eigenfunction pairs: $$\lambda_1 = \pi, \qquad f_1=\begin{pmatrix}1 \\0 \\0\end{pmatrix}=1$$ $$\lambda_2 = 2i, \qquad f_2 = \begin{pmatrix}0 \\ 1 \\ -i\end{pmatrix} = \cos x - i \sin x$$ $$\lambda_3=-2i,\qquad f_3=\begin{pmatrix}0 \\1\\i\end{pmatrix} = \cos x + i \sin x$$ and I'm sure you can take it from here...