## A doubt about the parts of the spectrum of tensor products

Let $\mathcal{H}$ be any complex Hilbert space of infinite dimensional. By an operator $T$ I mean a linear bounded transformation from $\mathcal{H}$ into $\mathcal{H}$, i.e, $T:\mathcal{H}\rightarrow\mathcal{H}$ is a linear transformation, such that $\left\|Tx\right\|\leq \beta \left\|x\right\|$, for some $\beta\geq 0$ in $\mathbb{C}$. The sets $\mathcal{N}(T)$ and $\mathcal{R}(T)$ denote the kernel and the range of $T$. Also, $\mathcal{B}[\mathcal{H}]$ is the set the of all operators defined in $\mathcal{H}$.

The spectrum of $T$ is the set $\sigma(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)\neq 0$ or $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $I\in\mathcal{B}[\mathcal{H}]$ is the identity operator (i.e., the spectrum of $T$ is the set of all $\lambda$ such that $(\lambda I - T)$ fail to have a bounded inverse on $\mathcal{R}(\lambda I - T)=\mathcal{H}$). A classical partition of the spectrum is:

i. Point Spectrum: $\sigma_{P}(T) = (\lambda \in \mathbb{C}: \mathcal{N}(\lambda I - T) \neq 0)$.

ii. Continuous Spectrum: $\sigma_{C}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0 , \mathcal{R}(\lambda I - T)^{-}=\mathcal{H}$ and $\mathcal{R}(\lambda I - T)\neq \mathcal{H} )$, where $\mathcal{R}(.)^{-}$ denotes the closure of $\mathcal{R}(.)$.

iii. Residual Spectrum: $\sigma_{R}(T) = (\lambda \in\mathbb{C}: \mathcal{N}(\lambda I - T)= 0$ and $\mathcal{R}(\lambda I - T)^{-}=\mathcal{H})$.

Let $\otimes$ denote the tensor product. Consider $T=(A\otimes B)\in\mathcal{B}[\mathcal{H}\otimes\mathcal{H}]$, where $A\in\mathcal{B}[\mathcal{H}]$ and $B\in\mathcal{B}[\mathcal{H}]$. It is well known that $\sigma(T) = \sigma(A\otimes B) = \sigma(A)\sigma(B)$. My questions are:

a. $\sigma_{C}(T) = \sigma_{C}(A)\otimes \sigma_{C}(B)$ ?

b. $\sigma_{R}(T) = \sigma_{R}(A)\otimes \sigma_{R}(B)$ ?

Thank you very much for your attention.

-
 Do you mean the Hilbert space tensor product (which is another Hilbert space, with Hilbert basis $\{e_i\otimes e_j\}$ if $\{e_j\}$ is a basis of $H$), or the Banach space tensor product (a Banach space isometric to the space of nuclear operators on $H$) ? – Pietro Majer Oct 28 at 17:35 I mean the Hilbert space tensor product. – portella Oct 28 at 18:11