# spectrum of elements in $C^*$ algebra

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?

• Well we know that $\sigma(x) \subseteq D(0, \|x\|)$ so $\sigma(x^{-1}y)\subseteq D(0, \|x\| \|y\|)$ which is the best we can do, really. Cf my post here: math.stackexchange.com/questions/972529/… – Cameron Williams Nov 20 at 17:31
• I don't understand why $\sigma(x^{-1}y)\subseteq D(0, \|x\| \|y\|)$. – math112358 Nov 21 at 2:05
• There's a slight typo in what I wrote. There should be a ${}^{-1}$ on the $\|x\|$. – Cameron Williams Nov 21 at 13:05
• But $\sigma(xy)$ is not the substlet of $\sigma(x)\sigma(y)$. – math112358 Nov 21 at 16:13
• I didn't say it is. $D(z, r)$ represents the disc centered at $z$ with radius $r$. I'm not multiplying two spectra. I'm just giving a bound. Basically if $z\in \sigma(x^{-1}y)$, then $|z| \le \|x^{-1}\| \|y\|$ is the best you can do in general. – Cameron Williams Nov 21 at 20:06

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 $$\sigma(z^{-1}w)=X$$.
• Does the following inequality hold? $\sigma(xy)\subset \sigma(x)\sigma(y)$?When $\sigma(xy)=\sigma(x)\sigma(y)$? – math112358 Nov 21 at 7:56
• No. Try $x=\begin{bmatrix}0&1\\0&0\end{bmatrix}$, $y=\begin{bmatrix}0&0\\1&0\end{bmatrix}$. The equality almost never occurs, it's not true even when $y=x$. – Martin Argerami Nov 21 at 10:32