Examples of spectral decompositions

[Boromir]

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]

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.

 

[Boromir]

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]

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.

 

[blue_raver22]

for your question! Spectral decomposition is a fundamental concept in functional analysis and is used to decompose a linear operator into simpler components. Here are some examples of spectral decompositions for normal compact and non-compact operators on a Hilbert space:

1. Normal compact operator: Let T be a normal compact operator on a Hilbert space H. By the spectral theorem, we know that T has a spectral decomposition into a sum of projection operators, each corresponding to an eigenvalue of T. In other words, we can write T as T = ∑ λPλ, where λ are the eigenvalues of T and Pλ are the corresponding orthogonal projection operators. This decomposition allows us to understand the behavior of T in terms of its eigenvalues and eigenvectors.

2. Normal non-compact operator: For a normal non-compact operator on a Hilbert space, the spectral decomposition is slightly more complicated. In this case, the operator may have a continuous spectrum, which means it has an infinite number of eigenvalues. The spectral decomposition for such operators involves a measure-theoretic approach, where the operator is decomposed into a sum of integral operators with respect to a spectral measure.

3. Toeplitz operator: A Toeplitz operator is a type of normal non-compact operator that arises in the study of analytic functions. It is defined by a symbol function and can be decomposed into a sum of rank one operators, each corresponding to a point on the unit circle. This decomposition is known as the Szegő decomposition and is an important tool in the study of Toeplitz operators.

4. Compact self-adjoint operator: For a compact self-adjoint operator on a Hilbert space, the spectral decomposition is particularly simple. By the spectral theorem, we know that the operator can be written as T = ∑ λPλ, where λ are the eigenvalues of T and Pλ are the corresponding orthogonal projection operators. However, in this case, the sum is finite, as the operator only has a finite number of non-zero eigenvalues.

These are just a few examples of spectral decompositions for different types of operators on a Hilbert space. Each type of operator may have a different decomposition, depending on its properties and the techniques used to obtain it. I hope this provides some concrete examples of spectral decompositions and how they can be obtained for different types of operators.