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).



Spectrum of a product of a symmetric positive definite matrix and a positive definite operator
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.


