5
$\begingroup$

If the eigenfunctions of a linear operator are known, is there a way to calculate the eigenfunctions of the corresponding adjoint operator based on the known eigenfunctions? In other words, what's the relation between the eigenfunctions of an operator and its adjoint?

Thanks!

$\endgroup$
1
  • $\begingroup$ Also, how about the eigenfunctions of the inverse operator if it's invertible? $\endgroup$
    – chaohuang
    Apr 16, 2012 at 18:23

2 Answers 2

5
$\begingroup$

In the finite-dimensional case, the eigenfunctions for the adjoint are the dual basis to the basis of eigenfunctions. That is, if $f_1,\ldots,f_n$ are the eigenfunctions for an operator $T$, then the eigenfunctions $g_1,\ldots,g_n$ for the adjoint are defined by the equations $$ \langle g_i,f_j\rangle = \delta_{ij}, $$ where $\langle-,-\rangle$ is the inner product on the space of functions.

If we write $g_i = a_{i1}f_1 + a_{i2}f_2 + \cdots + a_{in}f_n$, then we can use the above equations to solve for the coefficients $a_{ij}$. The solution is that the matrix of coefficients $a_{ij}$ is the inverse of the matrix whose entries are $\langle f_i,f_j\rangle$.

I imagine that similar statements are true for certain types of non-self-adjoint operators on a Hilbert space, but I'm not an expert on functional analysis.

$\endgroup$
4
$\begingroup$

Jim answered the finite-dimensional case very nicely; let me address the infinite-dimensional case.

If you don't specify some conditions on your operator, the answer is no. Consider $H=\ell^2(\mathbb{N})$, and let $T$ be the "reverse shift" operator, given by $$ T(a_1,a_2,\ldots)=(a_2,a_3,\ldots). $$ Then it is easy to see that every $\lambda\in\mathbb{C}$ with $|\lambda|<1$ is an eigenvalue with eigenvector $(\lambda,\lambda^2,\lambda^3,\ldots)$. More properly, one is free to choose the first coordinate, but that is irrelevant here.

Now, $T^*$ is the usual shift $$ T^*(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots) $$ and it has no eigenvalues (and so no eigenvectors).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .