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Examples of spectral decompositions

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Boromir

Banned
Feb 15, 2014
38
I would like examples of spectral decompositions and how they are obtained for normal compact operators and normal non-compact operators on a hilbert space.
I have googled it, but all I can find are proofs of the theorems; no concrete examples. I can't imagine it is that hard, for example for normal compact operator we only need the eigenvalues and eigenvectors of the operator.

Thanks
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,702
I would like examples of spectral decompositions and how they are obtained for normal compact operators and normal non-compact operators on a hilbert space.
The easiest example of a spectral resolution is the finite-dimensional case, where the spectral decomposition of a normal $n\times n$ matrix is equivalent to the fact that the matrix can be diagonalised. The spectral subspaces are then just the eigenspaces of the matrix.

A more complicated example is given by the resolvent of a selfadjoint differential operator. In that case, the spectral resolution is obtained by the techniques of Sturm–Liouville theory.
 
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Boromir

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Feb 15, 2014
38
The easiest example of a spectral resolution is the finite-dimensional case, where the spectral decomposition of a normal $n\times n$ matrix is equivalent to the fact that the matrix can be diagonalised. The spectral subspaces are then just the eigenspaces of the matrix.

A more complicated example is given by the resolvent of a selfadjoint differential operator. In that case, the spectral resolution is obtained by the techniques of Sturm–Liouville theory.
ok are compact normal operator more promising (easier) ground?
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
I would like examples of spectral decompositions and how they are obtained for normal compact operators and normal non-compact operators on a hilbert space.
I have googled it, but all I can find are proofs of the theorems; no concrete examples. I can't imagine it is that hard, for example for normal compact operator we only need the eigenvalues and eigenvectors of the operator.

Thanks
Based on what Opalg said consider the following example.

Let $X$ be the space of all smooth ($\mathcal{C}^{\infty}$) function that are $2\pi$-periodic. In other words, $X$ consists of all functions of all smooth $f:\mathbb{R}\to \mathbb{C}$ such that $f(x+2\pi) = f(x)$.

We define a (Herminitian) inner product $\left< \cdot , \cdot \right>: X\times X \to \mathbb{C}$ by,
$$ \left< f,g\right> = \int_{-\pi}^{\pi} f\cdot \overline{g} $$

Verify that $(X, \left< \cdot , \cdot \right>)$ is a compact space with an inner product. Define the following operator,
$$ L:X\to X \text{ by }L(f) = f'' $$
We argue that this operator is self-adjoint, by integration by parts,
$$ \left< L(f),g\right> = \int_{-\pi}^{\pi} f''\cdot \overline{g} = f'\overline{g}'\bigg|_{-\pi}^{\pi} - \int_{-\pi}^{\pi}f'\overline{g}' = -\int_{-\pi}^{\pi}f'\overline{g}' $$
Note, the evaluation at $\pm \pi$ cancels out because $f',g'$ are $2\pi$-periodic functions. Now by repeating integration by parts a second time we see,
$$ -\int_{-\pi}^{\pi}f'\overline{g}' = \int_{-\pi}^{\pi} f\overline{g}'' = \left< f,L(g)\right> $$
Now that $\left< L(f),g\right> = \left<f,L(g)\right>$ is self-adjoint.

It now follows by the theory of self-adjoint operators that the eigenvalues of $L$ are real. Let us suppose that $k$ is an eigenvalue of $k$ so $L(f) = kf$ for some non-zero $f$. Thus, we are led to the differencial equation $f'' = kf$.

Suppose that $k>0$ so can then write $k=a^2$ for real number $a$, the solution to the DE gives us that $f = c_1\sin(ax) + c_2\cos(ax)$. But $f$ must have period $2\pi$ and so it means that $a = n$ must be an integer. Therefore, $\sin(nx)$ and $\cos(nx)$ where $n\in \mathbb{Z}$ is a collection of eigenvectors of $L$. Since sine is odd and cosine is even we can describe these eigenvectors of $L$ by $E=\{1,\sin(nx),\cos(nx)\}$ where $n\geq 1$. Furthermore, every eigenvector with positive eigenvalue of $L$ is a linear combination of these ones.

The case $k\leq 0$ leads to no eigenvalues, it is an exercise to confirm this. Thus, our list above $E$ is a complete list of all eigenvalues of $L$, in the sense that every eigenvalue is a linear combination of the ones we gave above.

It is not true however that given any $f\in X$ we can express $f$ as a linear combination of $E$. But we can do it if we allow infinitely many of the $E$, more precisely, $f$ can be expressed as an infinite series (convergence in the inner-product sense) of those eigenvalues, this is exactly what Fourier analysis is about.