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Let $\mathbf H$ be an infinite dimensional Hilbert space.

I want to find an example of a $2\times 2$ real symmetric positive definite matrix $M$ and a positive definite bounded operator $A : \mathbf H \times \mathbf H\to \mathbf H \times \mathbf H$ such that the spectrum of $A$ is discrete but the spectrum of $MA$ is not discrete (could be an interval for example).

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  • $\begingroup$ $A=1$, and then take any $M$ with non-discrete spectrum. (Here $\sigma(A)=\{ 1\}$ is discrete as a set, though actually the spectrum is not discrete in the more natural sense of empty essential spectrum. But then a bounded self-adjoint operator always has non-empty essential spectrum.) $\endgroup$
    – Christian Remling
    Commented Jul 23, 2023 at 19:10
  • $\begingroup$ Excuse me. $M$ is assumed to be a $2\times 2$ symmetric positive definite matrix so it has discrete spectrum which consists of its eigenvalues. $\endgroup$
    – SAKLY
    Commented Jul 23, 2023 at 19:29
  • $\begingroup$ What definition of "discrete spectrum" do you use? $\endgroup$
    – Jochen Glueck
    Commented Jul 23, 2023 at 22:10
  • $\begingroup$ @JochenGlueck: $\sigma_d$ = isolated eigenvalues of finite multiplicity, I have the advantage of never having seen any other definition. $\endgroup$
    – Christian Remling
    Commented Jul 23, 2023 at 22:26
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    $\begingroup$ @JochenGlueck Yes, if possible, we can restrict ourselves to this particular case for the operator $A$. $\endgroup$
    – SAKLY
    Commented Jul 24, 2023 at 17:14

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Here's a way to construct such an example:

For each integer $n \ge 1$ consider the matrices $A_n, M_n \in \mathbb{C}^{2 \times 2}$ given by $$ M_n = e_1 e_1^T = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \qquad \text{and} \qquad A_n = f_n f_n^T , $$ where $e_1 \in \mathbb{C}^2$ if the first canonical unit vector and where $f_n \in \mathbb{C}^2$ is a vector of Euclidean norm $1$ that satisfies $e_1^T f_n = \sqrt{1-\frac{1}{n}}$. Both $M_n$ and $A_n$ are orthogonal projections. The matrix $$ M_n A_n = e_1 (e_1^T f_n) f_n^T $$ has spectrum $\{0, (e_1^T f_n)^2\} = \{0, 1-\frac{1}{n}\}$.

Now consider the Hilbert space $\ell^2(\mathbb{N}; \mathbb{C}^2)$, which can be identified with $H \times H$ for $H := \ell^2(\mathbb{N}; \mathbb{C})$ by splitting each copy of $\mathbb{C}^2$ into its first and second component. On $\ell^2(\mathbb{N}; \mathbb{C}^2) = H \times H$ we define operators $A$ and $M$ by multiplying each sequence with the operator sequence $(A_n)$ and $(M_n)$, respectively.

Both operators $A$ and $M$ are orthogonal projections; in particular, they are positive semidefinite, self-adjoint, and have spectrum $\{0,1\}$. The operator $M$ can be written as the matrix $$ \begin{pmatrix} \operatorname{id} & 0 \\ 0 & 0 \end{pmatrix} $$ with respect to the decomposition $H \times H$. However, the spectrum of $MA$ contains all the numbers $1 - \frac{1}{n}$ and thus accumulates at $1$.

If you want positive definite rather than positive semidefinite operators, you can modiy the construction of $A_n$ and $M_n$ a bit to have the sequences $(A_n)$ and $(M_n)$ bounded below by a fixed multiple of the identity matrix (one just needs to explicitly compute the spectra of $M_nA_n$ then, which is a bit more computational work since those will be rank-$2$ rather than rank-$1$ operators.

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