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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?


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Also, how about the eigenfunctions of the inverse operator if it's invertible? – chaohuang Apr 16 '12 at 18:23
up vote 3 down vote accepted

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).

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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.

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