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Let $E$ be a complex Hilbert space

Let $A,B\in \mathcal{L}(E)^+$. Assume that there exists $z\in \mathbb{C}^*$ such that $AB=zBA$. Why $$AB=BA\;?$$

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If $AB=0$ or $BA=0$, you are done. Otherwise, from $BA=|z|^2BA$, you obtain that $|z|=1$.

We have $$\{0\}\cup\sigma(BA)=\{0\}\cup\sigma(AB)=\{0\}\cup\sigma(zBA) =\{0\}\cup z\,\sigma(BA). $$ So $$\tag1\lambda\in\sigma(BA)\iff z\lambda\in\sigma(BA).$$ We also have, using that $\{0\}\cup\sigma(TS)=\{0\}\cup\sigma(ST)$,
$$\tag2 \{0\}\cup\sigma(BA)=\{0\}\cup\sigma(B^{1/2}(B^{1/2}A))=\{0\}\cup\sigma(B^{1/2}AB^{1/2})\subset[0,\infty). $$ Note that $\sigma(BA)\ne\{0\}$. Because if $\sigma(BA)=\{0\}$, then (2) gives $\sigma(B^{1/2}AB^{1/2})=\{0\}$; but this is a positive operator, so we would get $B^{1/2}AB^{1/2}=0$, which is the same as $(A^{1/2}B^{1/2})^*A^{1/2}B^{1/2}=0$, so $A^{1/2}B^{1/2}=0$ which in turn gives $AB=0$, and then $BA=0$.

Now, using $(2)$, take $\lambda>0$ with $\lambda\in \sigma(BA)$. Then, from $(1)$, $z\lambda\in(0,\infty)$, which implies $z\in (0,\infty)$. So $z$ is a positive real number with $|z|=1$, i.e. $z=1$.

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  • $\begingroup$ Please why you have shown that $\sigma(BA)\ne\{0\}$?. Because I think that even if $BA\neq 0$ which has a nontrivial kernel then 0 will be in its spectrum. $\endgroup$ – Student May 13 '18 at 13:50
  • $\begingroup$ $\sigma(BA)\ne\{0\}$ does not preclude $0\in\sigma(BA)$. $\endgroup$ – Martin Argerami May 13 '18 at 14:11
  • $\begingroup$ But why you have shown that $\sigma(BA)\ne\{0\}$? Thank you very much for your help. $\endgroup$ – Student May 13 '18 at 14:15
  • $\begingroup$ Because if I cannot use a nonzero $\lambda$, I cannot gather much information from $z\lambda$. $\endgroup$ – Martin Argerami May 13 '18 at 14:27

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