1. ## Natural embedding

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

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

Thanks for any help.

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

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

What are $\displaystyle e_1$ and $\displaystyle e_2$? If they are identity elements of $\displaystyle A_1$ and $\displaystyle A_2$, then $\displaystyle Te_1 = e_2$ 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:
$\displaystyle Tx = x\oplus 0$
how can I show that
$\displaystyle Te_1=e_2$
where $\displaystyle e_1$ and $\displaystyle e_2$ are identity elements of algebras $\displaystyle A_1$ and $\displaystyle A_2$ respectively
and that
$\displaystyle \sigma(x) \in \sigma(Tx)$?

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

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

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

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