What can we say about the spectrum of the difference of positive elements?











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Let $mathcal{A}$ be a unital $C^*$-algebra with $a,binmathcal{A}_+$ such that $|a|<|b|$. Does it follow that $a-bnotinmathcal{A}_+$?



I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.



Let $mathcal{A}$ be a unital $C^*$-algebra with $ainmathcal{A}$ and $binmathcal{A}_+$ such that $|a|<|b|$. If $C^*(a,b)$ is abelian, then $a-bnotinmathcal{A}_+$?



Proof: Since $binmathcal{A}_+$ we have $|b|=r(b)$, so we find $xinsigma(b)$ with $x>|a|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)leq|a|-h(b)<0$. Since $h(a-b)insigma(a-b)$, we get $a-bnotinmathcal{A}_+$.










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    up vote
    2
    down vote

    favorite












    Let $mathcal{A}$ be a unital $C^*$-algebra with $a,binmathcal{A}_+$ such that $|a|<|b|$. Does it follow that $a-bnotinmathcal{A}_+$?



    I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.



    Let $mathcal{A}$ be a unital $C^*$-algebra with $ainmathcal{A}$ and $binmathcal{A}_+$ such that $|a|<|b|$. If $C^*(a,b)$ is abelian, then $a-bnotinmathcal{A}_+$?



    Proof: Since $binmathcal{A}_+$ we have $|b|=r(b)$, so we find $xinsigma(b)$ with $x>|a|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)leq|a|-h(b)<0$. Since $h(a-b)insigma(a-b)$, we get $a-bnotinmathcal{A}_+$.










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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Let $mathcal{A}$ be a unital $C^*$-algebra with $a,binmathcal{A}_+$ such that $|a|<|b|$. Does it follow that $a-bnotinmathcal{A}_+$?



      I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.



      Let $mathcal{A}$ be a unital $C^*$-algebra with $ainmathcal{A}$ and $binmathcal{A}_+$ such that $|a|<|b|$. If $C^*(a,b)$ is abelian, then $a-bnotinmathcal{A}_+$?



      Proof: Since $binmathcal{A}_+$ we have $|b|=r(b)$, so we find $xinsigma(b)$ with $x>|a|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)leq|a|-h(b)<0$. Since $h(a-b)insigma(a-b)$, we get $a-bnotinmathcal{A}_+$.










      share|cite|improve this question















      Let $mathcal{A}$ be a unital $C^*$-algebra with $a,binmathcal{A}_+$ such that $|a|<|b|$. Does it follow that $a-bnotinmathcal{A}_+$?



      I can find very little about the spectrum of the sum or difference of two elements. But I believe there should at least be some useful observations to be made for positive elements. For example, I did prove the following.



      Let $mathcal{A}$ be a unital $C^*$-algebra with $ainmathcal{A}$ and $binmathcal{A}_+$ such that $|a|<|b|$. If $C^*(a,b)$ is abelian, then $a-bnotinmathcal{A}_+$?



      Proof: Since $binmathcal{A}_+$ we have $|b|=r(b)$, so we find $xinsigma(b)$ with $x>|a|$. Then there exists a homomorphism $h$ on $C^*(a,b)$ such that $h(b)=x$. We find $h(a-b)=h(a)-h(b)leq|a|-h(b)<0$. Since $h(a-b)insigma(a-b)$, we get $a-bnotinmathcal{A}_+$.







      spectral-theory c-star-algebras






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      edited Nov 16 at 15:27

























      asked Nov 15 at 23:03









      SmileyCraft

      72318




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          1 Answer
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          The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.



          EDIT



          Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.






          share|cite|improve this answer























          • You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
            – SmileyCraft
            Nov 16 at 14:53










          • No problem. I edited my response, let me know if anything is unclear.
            – Aweygan
            Nov 16 at 16:19










          • First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
            – SmileyCraft
            Nov 16 at 17:10










          • I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
            – Aweygan
            Nov 16 at 17:24












          • Ah of course. Thank you!
            – SmileyCraft
            Nov 16 at 17:25











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          1 Answer
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          1 Answer
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          up vote
          1
          down vote



          accepted










          The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.



          EDIT



          Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.






          share|cite|improve this answer























          • You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
            – SmileyCraft
            Nov 16 at 14:53










          • No problem. I edited my response, let me know if anything is unclear.
            – Aweygan
            Nov 16 at 16:19










          • First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
            – SmileyCraft
            Nov 16 at 17:10










          • I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
            – Aweygan
            Nov 16 at 17:24












          • Ah of course. Thank you!
            – SmileyCraft
            Nov 16 at 17:25















          up vote
          1
          down vote



          accepted










          The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.



          EDIT



          Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.






          share|cite|improve this answer























          • You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
            – SmileyCraft
            Nov 16 at 14:53










          • No problem. I edited my response, let me know if anything is unclear.
            – Aweygan
            Nov 16 at 16:19










          • First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
            – SmileyCraft
            Nov 16 at 17:10










          • I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
            – Aweygan
            Nov 16 at 17:24












          • Ah of course. Thank you!
            – SmileyCraft
            Nov 16 at 17:25













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.



          EDIT



          Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.






          share|cite|improve this answer














          The claim is false, as is the proof you give for the commutative case (the fallacy lies in $|b|-h(a)<0$). Consider $mathcal A=mathbb C$, $a=4$, and $b=7$.



          EDIT



          Note that now the claim is equivalent to "If $0leq aleq b$, then $|a|leq|b|$." This can be proven as follows: By functional calculus, we have $bleq|b|$, and thus $aleq|b|$. Applying functional calculus again gives $|a|leq|b|$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 16 at 17:19

























          answered Nov 16 at 5:59









          Aweygan

          13.1k21441




          13.1k21441












          • You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
            – SmileyCraft
            Nov 16 at 14:53










          • No problem. I edited my response, let me know if anything is unclear.
            – Aweygan
            Nov 16 at 16:19










          • First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
            – SmileyCraft
            Nov 16 at 17:10










          • I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
            – Aweygan
            Nov 16 at 17:24












          • Ah of course. Thank you!
            – SmileyCraft
            Nov 16 at 17:25


















          • You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
            – SmileyCraft
            Nov 16 at 14:53










          • No problem. I edited my response, let me know if anything is unclear.
            – Aweygan
            Nov 16 at 16:19










          • First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
            – SmileyCraft
            Nov 16 at 17:10










          • I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
            – Aweygan
            Nov 16 at 17:24












          • Ah of course. Thank you!
            – SmileyCraft
            Nov 16 at 17:25
















          You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
          – SmileyCraft
          Nov 16 at 14:53




          You are totally correct. I stupidly made a mistake in my writing, I confused $a$ and $b$, since I first used different names and then changed them :/ I fixed it now. The question should make sense now.
          – SmileyCraft
          Nov 16 at 14:53












          No problem. I edited my response, let me know if anything is unclear.
          – Aweygan
          Nov 16 at 16:19




          No problem. I edited my response, let me know if anything is unclear.
          – Aweygan
          Nov 16 at 16:19












          First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
          – SmileyCraft
          Nov 16 at 17:10




          First, I think you swapped that $a$ and $b$ now. If that is the case, that is very understandable :P Second, I don't see how the statement is equivalent. Last, how does $bleq|a|$ give $|b|leq|a|$?
          – SmileyCraft
          Nov 16 at 17:10












          I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
          – Aweygan
          Nov 16 at 17:24






          I may have swapped them. I swapped them again, because this looks more natural. The statement I have is the contrapositive to the statement you made. The negation of the statement $a-bnotinmathcal A_+$ is $0leq a-b$ or $bleq a$, and the negation of $|a|<|b|$ is $|b|leq|a|$. Lastly, if $lambda$ is a positive real number, then $0leq bleqlambda$ means that $sigma(b)subset[0,lambda]$, and since the norm of $b$ is its spectral radius, we have $|b|leqlambda$. Now put $lambda=|a|$.
          – Aweygan
          Nov 16 at 17:24














          Ah of course. Thank you!
          – SmileyCraft
          Nov 16 at 17:25




          Ah of course. Thank you!
          – SmileyCraft
          Nov 16 at 17:25


















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