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 $a-bnotinmathcal{A}_+$.
spectral-theory c-star-algebras
asked Nov 15 at 23:03
SmileyCraft
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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 $a-bnotinmathcal{A}_+$.
spectral-theory c-star-algebras
asked Nov 15 at 23:03
SmileyCraft
72819
add a comment |
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 $a-bnotinmathcal{A}_+$.
spectral-theory c-star-algebras
asked Nov 15 at 23:03
SmileyCraft
72819
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
spectral-theory c-star-algebras
asked Nov 15 at 23:03
SmileyCraft
72819
asked Nov 15 at 23:03
SmileyCraft
72819
edited Nov 16 at 15:27
asked Nov 15 at 23:03
SmileyCraft
72819
asked Nov 15 at 23:03
SmileyCraft
72819
asked Nov 15 at 23:03
SmileyCraft
72819
72819
add a comment |
add a comment |
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 $|a|leq|b|$.
answered Nov 16 at 5:59
Aweygan
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
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $|a|leq|b|$.
answered Nov 16 at 5:59
Aweygan
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
|
show 1 more comment
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 $|a|leq|b|$.
answered Nov 16 at 5:59
Aweygan
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
|
show 1 more comment
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 $|a|leq|b|$.
answered Nov 16 at 5:59
Aweygan
13.1k21441
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 $|a|leq|b|$.
answered Nov 16 at 5:59
Aweygan
13.1k21441
edited Nov 16 at 17:19
answered Nov 16 at 5:59
Aweygan
13.1k21441
answered Nov 16 at 5:59
Aweygan
13.1k21441
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
|
show 1 more comment
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
|
show 1 more comment
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