# investigating an operator

• January 15th 2012, 07:18 AM
mgarson
investigating an operator
Hello!

I have a problem that comes in two parts.

(1) $H$ is a Hilbert space with orthonormal basis $\{e_k\}$. We are given a positivie strictly increasing sequence $\{\alpha_k\}$with the following property $\lim_{k\rightarrow \infty}\frac{\alpha_k}{\alpha_{k+1}}=1$. Our task is to show that there exists a unique operator $M\in B(H)$ such that $Me_k=\frac{\alpha_k}{\alpha_{k+1}}e_{k+1}$.

(2) Determine $\|M\|$ and the spectrum of M (hint: use the eigenvalues of the adjoint $M^*$).

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My attempt

(1) For the first one I was thinking that I could just simply find the operator by using $Mx$ where $x\in H$. Here it goes,

$x\in H$ gives me $x=\sum_{k=1}^\infty e_k=\lim_{n\rightarrow \infty}\sum_{k=1}^n e_k$.
So,
$Mx=M\lim_{n\rightarrow \infty}\sum_{k=1}^n e_k=\lim_{n\rightarrow \infty}\sum_{k=1}^n Me_k = \lim_{n\rightarrow \infty}\sum_{k=1}^n \frac{\alpha_k}{\alpha_{k+1}}e_{k+1}=\sum_{ k=1}^\infty \frac{\alpha_k}{\alpha_{k+1}}e_{k+1}$
(the last equality by Parseval's formula and the second by continuity (boundedness) of $M$).

Well I now know what a bounded linear operator with the above stated property looks like. But is it unique? Futhermore, its "look" complicates things in the second part.

(2) I know the definition $\|M\|=\sup_{\|x\|=1}|Mx|$. But this doesn't seem all that easy to compute using what I know about my operator.

And what will the eigenvalues of $M^*$ tell me? I know that $\sigma (M)=\overline{\sigma (M^*)}$ (where I by $\sigma (M)$ denote the spectrum of $M$) and this does not help me unless I can use the eigenvalues of $M^*$ to find $\sigma(M^*)$. This would be easy if $M^*$ was compact, then I would know that $\sigma (M^*)= \{0\}\cup \{eigenvalues of M^*\}$.

Thanks! (Nerd)
• January 15th 2012, 12:19 PM
Opalg
Re: investigating an operator
The operator M is a weighted shift (check that link for some useful guidance). It shifts each basis vector $e_k$ to the next one $e_{k+1},$ multiplying it by the weight $\alpha_k/\alpha_{k+1}.$ Since its value at each basis vector is specified, it must be unique (you know its value at each finite linear combination of basis vectors and hence, by continuity, at every vector).

The adjoint operator is a backward weighted shift, taking each $e_k$ to a multiple of $e_{k-1}$ and sending $e_1$ to 0. The advantage of looking at the adjoint is that it has many eigenvalues, whereas M itself does not.

For convenience, write $\beta_k = \alpha_k/\alpha_{k+1}$, and let $B = \sup\{\beta_k:k\in\mathbb{N}\}$. If $x = \textstyle\sum \xi_ke_k$ then $Mx = \textstyle\sum \beta_k\xi_ke_{k+1}$, and

$\|Mx\|^2 = \sum|\beta_k\xi_k|^2\leqslant B^2\sum|\xi|^2 = B^2\|x\|^2.$

That shows that $\|M\|\leqslant B$. By looking at $\|Me_k\|$ for each k, you should be able to show the reverse inequality.
• January 16th 2012, 07:00 AM
mgarson
Re: investigating an operator
Quote:

Originally Posted by Opalg
The advantage of looking at the adjoint is that it has many eigenvalues, whereas M itself does not.

Mabye I'm missing something trivial, but how do the eigenvalues of the adjoint help me find the spectrum of it? I know that I have the spectum of a compact operator if I have the eigenvalues. Is my $M^*$ compact?
• January 16th 2012, 09:05 AM
Opalg
Re: investigating an operator
Quote:

Originally Posted by mgarson
Mabye I'm missing something trivial, but how do the eigenvalues of the adjoint help me find the spectrum of it? I know that I have the spectum of a compact operator if I have the eigenvalues. Is my $M^*$ compact?

The idea is that M* has so many eigenvalues that they force the spectrum to be as big as it could possibly be. To see how that might work, here is a slightly simplified example.

Let S be the unilateral shift operator, defined on the basis vectors by $Se_k = e_{k+1}$ (so S is like your operator M except that it does not have the weights $\alpha_k/\alpha_{k+1}$). Its adjoint S* is the backwards shift, defined by $S^*e_1=0$ and $S^*e_k = e_{k-1}$ for k>1.

Let $\lambda$ be a fixed complex number with $0<|\lambda|<1$ and let x be the vector given by $\textstyle x=\sum\lambda^ke_k.$ (Notice that that sum converges in H because the sum of the squares of the absolute values of the coefficients is $\textstyle\sum|\lambda|^{2k}<\infty.$) Then $S^*x = \textstyle \sum\lambda^ke_{k-1} = \lambda x.$

Thus $\lambda$ is an eigenvalue of S*, with eigenvector x. That holds for every nonzero $\lambda$ in the open unit ball. In fact, 0 is also an eigenvalue, because $S^*e_1=0.$ So the spectrum of S* contains the entire open unit ball. Since the spectrum is always closed, it contains the closed unit ball. But $\|S^*\| = 1$, and the absolute value of an element of the spectrum can never be greater than the norm of the operator. Conclusion: the spectrum of S* is the closed unit ball. Therefore the spectrum of S is also the closed unit ball (though S has no eigenvalues).

Notice that although S and S* are not compact, S* has a huge number of eigenvalues (uncountably many, which a compact operator never could have).

Finding the spectrum of the operator M* is a similar exercise, though you have to decide how to deal with the weights.