spectrum of elements in $C^*$ algebra

Question

Suppose $x,y$ are two invertible positive elements in a $C^*$ algebra $A$,if $\|x\|=\|y\|$,can we compute the spectrum $\sigma(x^{-1}y)$ of $x^{-1}y$?Does there exist a relationship between the spectrum of the multiplication of two elements and the norm of elements?

Answer 1

You can't expect a relation. For instance consider $$ x=\begin{bmatrix} 1&0\\0&\tfrac1n\end{bmatrix} ,\ \ \ y=\begin{bmatrix} 1&0\\0&1\end{bmatrix} . $$ Then $\|x\|=\|y\|=1$, and $\|x^{-1}y\|=n$. The norm only sees the maximum of the spectrum, but nothing else.

For a more dramatic example consider the block matrices $$\tag1 x=\begin{bmatrix} 1&0\\0& z\end{bmatrix} ,\ \ \ y=\begin{bmatrix} 1&0\\0&w\end{bmatrix} . $$ We can take $z,w$ to be any two contractions, and we will still have $\|x\|=\|y\|=1$, while $\sigma(x^{-1}y)=\{1\}\cup \sigma(z^{-1}w)$. Now let $X\subset (1,\infty)$ be any compact set; let $v$ be an operator with $\sigma(v)=X$, and let $w=\tfrac1{\|v\|}v$. Then $w$ is a contraction. If we now take $z=\tfrac1{\|v\|}\,I$, we still have $\|x\|=\|y\|=1$, while $\sigma(z^{-1}w)=X$.