Spectrum of product of self-adjoint operators contained in $\mathbb{R}$

Question

Let $S,T$ be self-adjoint bounded operators on a complex Hilbert space. In this post, it is shown that $\sigma(ST)\subset\mathbb{R}$. The answerer uses that $\sigma(ST)\cup\{0\}=\sigma(TS)\cup\{0\}$ and that $\sigma(U)=\sigma(U^{*})^{*}$ for any operator $U$. I know that these results are indeed true. But how does he use these results to conclude that $\sigma(ST)\subset\mathbb{R}$?

If I play around with his arguments, I find that $$\sigma(ST)\cup\{0\}=\sigma(TS)\cup\{0\}=\sigma((TS)^{*})^{*}\cup\{0\}=\sigma(S^{*}T^{*})^{*}\cup\{0\}=\sigma(ST)^{*}\cup\{0\}.$$

But, for example, $\mathbb{C}^{*}=\mathbb{C}$ and of course $\mathbb{C}\not\subset\mathbb{R}$.

Martin Argerami · Accepted Answer · 2020-03-08T01:09:20.400

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Personally, I cannot make sense of that answer. More importantly, the assertion is not true if we only require $S,T$ to be selfadjoint. Consider $$ S=\begin{bmatrix} 1&0\\0&-1\end{bmatrix},\ \ \ \ T=\begin{bmatrix} 0&1\\1&0\end{bmatrix}. $$ Then $\sigma(ST)=\{i,-i\}$.

If you require, say $S\geq0$, then the assertion does work because $$ \sigma(ST)\cup\{0\}=\sigma(S^{1/2}TS^{1/2})\cup\{0\}\subset\mathbb R. $$

edited Mar 08 '20 at 01:09

answered Mar 07 '20 at 19:06

Martin Argerami

205,756

Thanks for your reply! Doesn't this only work for positive operators? (Taking square roots?) – Calculix Mar 07 '20 at 19:13
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No. The operator $-I$ is not positive and it has square root $i,I$. You can do that for any selfadjiont. – Martin Argerami Mar 07 '20 at 20:05
I agree with you example, but it leaves me a bit puzzled. In any C-algebra $A$, the set positive elements is precisely $A^{+}={a^{}a:a\in A}$ (see thm 2.2.5 in Murphy's book on C-algebras). But then this would imply that $T=Z^{}Z$ is automatically positive. – Calculix Mar 07 '20 at 21:41
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$(iI)^{2}=-I$, but $(iI)^{*}=-iI$. So $iI$ is a square root of $-I$, but it is not a self-adjoint square. – Calculix Mar 07 '20 at 21:49
You are right, my bad. The square of a selfadjoint is positive. – Martin Argerami Mar 08 '20 at 00:25
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And the assertion is not true. Please see the edit. – Martin Argerami Mar 08 '20 at 00:47

Spectrum of product of self-adjoint operators contained in $\mathbb{R}$

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