1. ## Natural embedding

Hi,
This is the next problem in that theme. How can I show the following:

If we take natural embedding
$T:A_1 \rightarrow A_1 \oplus A_1$
then it satisfy $Txy=TxTy$ but neither
$\sigma(Tx) \in \sigma(x)$
nor
$Te_1=e_2$
However we have $\sigma(x) \in \sigma(Tx)$.

Thanks for any help.

2. Originally Posted by Arczi1984
How can I show the following:

If we take natural embedding
$T:A_1 \rightarrow A_1 \oplus A_1$
then it satisfy $Txy=TxTy$ but neither
$\sigma(Tx) \in \sigma(x)$
nor
$Te_1=e_2$
However we have $\sigma(x) \in \sigma(Tx)$.
To me, the natural embedding of $A_1$ in $A_1 \oplus A_1$ would be the map $Tx = x\oplus x$. But in that case $\sigma(Tx) = \sigma(x)$. Perhaps the question intends the "natural embedding" to be the map $Tx = x\oplus 0$? In that case, 0 is always in the spectrum of Tx, but it need not be in the spectrum of x.

What are $e_1$ and $e_2$? If they are identity elements of $A_1$ and $A_2$, then $Te_1 = e_2$ if T is the first of the above maps, but not if T is the second map.

3. In that paper it is called 'natural embedding' but they didn't give any formula. When I was trying to solve this problem I took the second case witch You are mentioned.

How can I show that o need not be in the spectrum of x (in this case)?
If we take this natural embedding:
$Tx = x\oplus 0$
how can I show that
$Te_1=e_2$
where $e_1$ and $e_2$ are identity elements of algebras $A_1$ and $A_2$ respectively
and that
$\sigma(x) \in \sigma(Tx)$?

4. You start with $A = C[0,1]$ and $x(t) = t+1$, then $0 \not\in \sigma(x)$ but $0 \in \sigma(Tx)$. Also $Te_1 \ne (e_1,e_1) = e_2$, where $e_1 is the constant function x(t) = 1, \forall t \in [0,1]$

5. I saw two mistakes (I'm sorry for that).

1. Of course it shoul be: $\sigma(Tx) \subset \sigma(x)$ and $\sigma(x) \subset \sigma(Tx)$

2. $T:A_1 \rightarrow A_1 \oplus A_2$ instead of $T:A_1 \rightarrow A_1 \oplus A_1$