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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?

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  • $\begingroup$ 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/… $\endgroup$ – Cameron Williams Nov 20 at 17:31
  • $\begingroup$ I don't understand why $\sigma(x^{-1}y)\subseteq D(0, \|x\| \|y\|)$. $\endgroup$ – math112358 Nov 21 at 2:05
  • $\begingroup$ There's a slight typo in what I wrote. There should be a ${}^{-1}$ on the $\|x\|$. $\endgroup$ – Cameron Williams Nov 21 at 13:05
  • $\begingroup$ But $\sigma(xy)$ is not the substlet of $\sigma(x)\sigma(y)$. $\endgroup$ – math112358 Nov 21 at 16:13
  • $\begingroup$ 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. $\endgroup$ – Cameron Williams Nov 21 at 20:06
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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 $\sigma(z^{-1}w)=X$.

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  • $\begingroup$ Does the following inequality hold? $\sigma(xy)\subset \sigma(x)\sigma(y)$?When $\sigma(xy)=\sigma(x)\sigma(y)$? $\endgroup$ – math112358 Nov 21 at 7:56
  • $\begingroup$ 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 $. $\endgroup$ – Martin Argerami Nov 21 at 10:32

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