In my research problem, I have to estimate the spectral radius of the following operator $\chi A$ where $\chi$ is a scalar function taking values 0 or 1 and $A$ is an operator. I can compute explicitly all the eigenvalues of $A$ and $A^*$ and $A^*A$. The problem is that $A$ is not an selfadjoint nor normal operator and using the norm $\|A\|$ is too rough a approximation.
My question is that is there any stricter estimation of the spectral radius $\rho(\chi A)$? Or can we say that
$\rho(\chi A)\leq\rho(A)$
Thanks for your help.
P.S: In my physical problem, $\chi(x)$ (taking values of 0 or 1) is a function of coordinate $x$ and $A$ is a tensorial convolution operator (e.g Green function). For example $\chi A$ acting on $v$ yields
$(\chi A) v = \chi(x) [A* v(x)] = \chi(x)$$\int$ $A_{ijkl}(x-x')v_{lk}(x') dx'$
Edit: Recently I found a related thread on this subject Singular-value inequalities There is an inequality on singular values
$\sum s_j(RS) \leq \sum s_j(R)s_j(S)$.
However I still wonder how to apply this theorem if the number of singular values of $R$, $S$ and $RS$ are different
I presume this is on a Hilbert space, so that your "scalar function" $\chi$ is a self-adjoint projection (which by the spectral theorem is unitarily equivalent to multiplication by a scalar function with values $\{0,1\}$ on $L^2$ of some measure). No, we can't say much about $\rho(\chi A)$ compared to $\rho(A)$. Consider the $2 \times 2$ case
$$ A = \pmatrix{a & b\cr c & d\cr},\ \chi = \pmatrix{1 & 0\cr 0 & 0\cr}$$
$$\chi A = \pmatrix{a & b\cr 0 & 0\cr}, \ \text{eigenvalues}\ 0,a $$
so $\chi A$ has spectral radius $|a|$. But for any $a$, the spectral radius of $A$ could be arbitrarily large (e.g. if $b = 0$ and $d$ is large) or could be $0$ (e.g. if $b = -a^2$, $c = 1$, $d = -a$).