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Let $A$ be a unital $C^*$-algebra and $w$ be a state (i.e a positive linear functional such that $\|w\|=w(1_A)=1$. I'm trying to prove the following:

a) if $a$ is selfadjoint and $w(a^2)=w(a)^2$ then for any $b\in A$ we have $w(ab)=w(a)w(b)=w(ba)$.

b) if $\|a\|\le 1$ and $T=\left( \begin{array}{cc} 1_A & a \\ a^* & 1_A \\ \end{array} \right) $ then $T$ is positive.

What I have done:

for part a) I proved that for any $a\in A$ we have $|w(a)|\le w(|a|^2)^\frac{1}{2}$ and I know that for any state we can define a semi inner product as follows $<a,b>=w(b^*a)$ but I cant get the result I want.

for part b) it is clear that $T$ is selft adjoint, so if $\sigma(T)\subset[0,\infty)$ then $T$ is positive, I'm not sure how to find the spectrum of $T$.

Thank you for your help.

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For a): as you recalled, we have Cauchy-Schwarz for $w(x^*y)$, i.e. $$ |w(x^*y)|^2\leq w(x^*x)w(y^*y)\quad\forall x,y\in A. $$ Now for every $b$ in $A$ $$ w(ab)-w(a)w(b)=w((a-w(a)1)b)=w(a'b)\quad\mbox{with}\quad a'=a-w(a)1. $$ Note that $a'$ is self-adjoint and $a'^*a'=a'^2=a^2-2w(a)a+w(a)^21$. Hence $w(a'^2)=w(a^2)-w(a)^2=0$ by assumption. Therefore, by Cauchy-Schwarz: $$ |w(a'b)|^2\leq w(a'^*a')w(b^*b)= w(a'^2)w(b^*b)=0\quad\Rightarrow\quad w(a'b)=w(ab)-w(a)w(b)=0$$ for every $b\in A$. It follows that $w(ab)=w(a)w(b)$, and we can show in a similar fashion that $w(ba)=w(b)w(a)$.

For b): observe the matrix identity $$ (I-T)^*(1-T)=\left(\matrix{0&-a\\-a^*&0}\right)\left(\matrix{0&-a\\-a^*&0}\right)=\left(\matrix{aa^*&0\\0&a^*a}\right). $$ It follows that $\|I-T\|^2=\|(I-T)^*(I-T)\|=\max\{\|aa^*\|,\|a^*a\|\}=\|a\|^2$.

If $\|a\|\leq 1$, we get $\|I-T\|\leq 1$, so $T=I-(I-T)$ is positive.

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  • $\begingroup$ Hi Julien, thank your for the hints, is it ture that a diagonal matrix is positive if the elements of the diagonal are positive, I ask this because of the last matrix in the decomposition, and is it possible to compute the eigenvalues some how and then decide? thank you again. $\endgroup$ – i.a.m Apr 14 '13 at 2:11
  • $\begingroup$ Oh, I see, perfect! :) $\endgroup$ – i.a.m Apr 14 '13 at 2:18
  • $\begingroup$ Perfect, any ideas about a)? $\endgroup$ – i.a.m Apr 14 '13 at 3:31
  • $\begingroup$ Don't be, thank you for all the help :) $\endgroup$ – i.a.m Apr 14 '13 at 3:43
  • $\begingroup$ Note that $w(a)$ is a real number and you have proved that $|w(a)|^2\leq w(|a|^2)$, and now the equality holds, and I believe this can happen only when $a=w(a)Id$. $\endgroup$ – ougao Apr 14 '13 at 4:10

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