Spectral radius of sum of commuting operators
Prove that if two operators on a Hilbert space commute then the spectral radius of their sum is at most the sum of their spectral radii.
I tried to use Gelfand's formula along with the binomial formula to get somehow bound $|(A+B)^n|^{1/n}$ by $|A^k|^{1/k}+|A^l|^{1/l}$ where $l,k$ are functions of $n$ but have not been able to make progress yet.
operator-theory hilbert-spaces
asked Dec 2 '18 at 18:09
Manan
1,085611
add a comment |
Prove that if two operators on a Hilbert space commute then the spectral radius of their sum is at most the sum of their spectral radii.
I tried to use Gelfand's formula along with the binomial formula to get somehow bound $|(A+B)^n|^{1/n}$ by $|A^k|^{1/k}+|A^l|^{1/l}$ where $l,k$ are functions of $n$ but have not been able to make progress yet.
operator-theory hilbert-spaces
asked Dec 2 '18 at 18:09
Manan
1,085611
add a comment |
Prove that if two operators on a Hilbert space commute then the spectral radius of their sum is at most the sum of their spectral radii.
I tried to use Gelfand's formula along with the binomial formula to get somehow bound $|(A+B)^n|^{1/n}$ by $|A^k|^{1/k}+|A^l|^{1/l}$ where $l,k$ are functions of $n$ but have not been able to make progress yet.
operator-theory hilbert-spaces
asked Dec 2 '18 at 18:09
Manan
1,085611
Prove that if two operators on a Hilbert space commute then the spectral radius of their sum is at most the sum of their spectral radii.
I tried to use Gelfand's formula along with the binomial formula to get somehow bound $|(A+B)^n|^{1/n}$ by $|A^k|^{1/k}+|A^l|^{1/l}$ where $l,k$ are functions of $n$ but have not been able to make progress yet.
operator-theory hilbert-spaces
operator-theory hilbert-spaces
asked Dec 2 '18 at 18:09
Manan
1,085611
asked Dec 2 '18 at 18:09
Manan
1,085611
asked Dec 2 '18 at 18:09
Manan
1,085611
asked Dec 2 '18 at 18:09
Manan
1,085611
asked Dec 2 '18 at 18:09
Manan
1,085611
1,085611
add a comment |
add a comment |
1 Answer
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The spectral radius remains unchanged when calculating it in the closed sub-algebra.
So let's calculate the spectral radius in the unital abelian Banach algebra $mathscr B$ generated by $A$ and $B$.
begin{align*}sigma(A+B) & ={tau(A+B)|tau mbox{ is a character of } mathscr B}\ & = {tau(A)+tau(B)|tau mbox{ is a character of }mathscr B}\ & subset sigma(A)+sigma(B).end{align*}
It follows what you want.
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
The spectral radius remains unchanged when calculating it in the closed sub-algebra.
So let's calculate the spectral radius in the unital abelian Banach algebra $mathscr B$ generated by $A$ and $B$.
begin{align*}sigma(A+B) & ={tau(A+B)|tau mbox{ is a character of } mathscr B}\ & = {tau(A)+tau(B)|tau mbox{ is a character of }mathscr B}\ & subset sigma(A)+sigma(B).end{align*}
It follows what you want.
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
add a comment |
The spectral radius remains unchanged when calculating it in the closed sub-algebra.
So let's calculate the spectral radius in the unital abelian Banach algebra $mathscr B$ generated by $A$ and $B$.
begin{align*}sigma(A+B) & ={tau(A+B)|tau mbox{ is a character of } mathscr B}\ & = {tau(A)+tau(B)|tau mbox{ is a character of }mathscr B}\ & subset sigma(A)+sigma(B).end{align*}
It follows what you want.
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
add a comment |
The spectral radius remains unchanged when calculating it in the closed sub-algebra.
So let's calculate the spectral radius in the unital abelian Banach algebra $mathscr B$ generated by $A$ and $B$.
begin{align*}sigma(A+B) & ={tau(A+B)|tau mbox{ is a character of } mathscr B}\ & = {tau(A)+tau(B)|tau mbox{ is a character of }mathscr B}\ & subset sigma(A)+sigma(B).end{align*}
It follows what you want.
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
The spectral radius remains unchanged when calculating it in the closed sub-algebra.
So let's calculate the spectral radius in the unital abelian Banach algebra $mathscr B$ generated by $A$ and $B$.
begin{align*}sigma(A+B) & ={tau(A+B)|tau mbox{ is a character of } mathscr B}\ & = {tau(A)+tau(B)|tau mbox{ is a character of }mathscr B}\ & subset sigma(A)+sigma(B).end{align*}
It follows what you want.
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
answered Dec 3 '18 at 9:03
C.Ding
1,3931321
1,3931321
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
add a comment |
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
Actually we haven't covered Banach algebras, so I am unable to understand the proof. I was hoping for a proof using Gelfand's formula..
– Manan
Dec 3 '18 at 16:48
add a comment |
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