If $r(A)=omega(A)$, is $r(A)=|A|$? The 2019 Stack Overflow Developer Survey Results Are Inequality of norm and numerical radius for normal (or even self-adjoint) operators on Hilbert spaceShowing that $r(T)leq omega(T)$?Joint numerical radius of operators$|S|=sup leq 1,?$Why $|S|=sup_|Sx|?$why $w(B^*AB)leq |B|^2w(A)?$Example of normal operator on infinite-dimensional Hilbert spacesThe operator norm and the spectral radiusThe equality $r(A)=omega(A)=|A|$ is not true in general.invertibility of a bounded linear operator
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If $r(A)=omega(A)$, is $r(A)=|A|$?
The 2019 Stack Overflow Developer Survey Results Are Inequality of norm and numerical radius for normal (or even self-adjoint) operators on Hilbert spaceShowing that $r(T)leq omega(T)$?Joint numerical radius of operators$|S|=sup;;,?$Why $|S|=sup_x|Sx|?$why $w(B^*AB)leq |B|^2w(A)?$Example of normal operator on infinite-dimensional Hilbert spacesThe operator norm and the spectral radiusThe equality $r(A)=omega(A)=|A|$ is not true in general.invertibility of a bounded linear operator
$begingroup$
Let $mathcalB(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
It is well know that
$$r(A)leqomega(A)leq|A|,$$
for every $AinmathcalB(F)$, where $r(A)$, $omega(A)$ and $|A|$ denote respectively the spectral radius, the numerical radius and the norm of $A$.
So clearly if $r(A)=|A|$ then $r(A)=omega(A)$.
If $r(A)=omega(A)$, is $r(A)=|A|$?
functional-analysis operator-theory examples-counterexamples
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
$endgroup$
add a comment |
$begingroup$
Let $mathcalB(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
It is well know that
$$r(A)leqomega(A)leq|A|,$$
for every $AinmathcalB(F)$, where $r(A)$, $omega(A)$ and $|A|$ denote respectively the spectral radius, the numerical radius and the norm of $A$.
So clearly if $r(A)=|A|$ then $r(A)=omega(A)$.
If $r(A)=omega(A)$, is $r(A)=|A|$?
functional-analysis operator-theory examples-counterexamples
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
$endgroup$
add a comment |
$begingroup$
Let $mathcalB(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
It is well know that
$$r(A)leqomega(A)leq|A|,$$
for every $AinmathcalB(F)$, where $r(A)$, $omega(A)$ and $|A|$ denote respectively the spectral radius, the numerical radius and the norm of $A$.
So clearly if $r(A)=|A|$ then $r(A)=omega(A)$.
If $r(A)=omega(A)$, is $r(A)=|A|$?
functional-analysis operator-theory examples-counterexamples
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
$endgroup$
Let $mathcalB(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.
It is well know that
$$r(A)leqomega(A)leq|A|,$$
for every $AinmathcalB(F)$, where $r(A)$, $omega(A)$ and $|A|$ denote respectively the spectral radius, the numerical radius and the norm of $A$.
So clearly if $r(A)=|A|$ then $r(A)=omega(A)$.
If $r(A)=omega(A)$, is $r(A)=|A|$?
functional-analysis operator-theory examples-counterexamples
functional-analysis operator-theory examples-counterexamples
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
asked Mar 30 at 10:02
SchülerSchüler
1,5571421
1,5571421
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=Boplus Cin M_3(mathbb C)$, where $B=1$ and $C=2E_12$. That is,
$$
A=beginbmatrix 1&0&0\0&0&2\0&0&0endbmatrix.
$$
Then
$sigma(A)=0,1$ and thus $r(A)=1$
the numerical range of $A$ is the convex hull of the ranges of $B$ ($=1)$ and of $C$ (disk of radius $1$ centered at the origin). So $omega(A)=1$.
the norm of $A$ is the maximum of the two norms, so $|A|=2$.
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=Boplus Cin M_3(mathbb C)$, where $B=1$ and $C=2E_12$. That is,
$$
A=beginbmatrix 1&0&0\0&0&2\0&0&0endbmatrix.
$$
Then
$sigma(A)=0,1$ and thus $r(A)=1$
the numerical range of $A$ is the convex hull of the ranges of $B$ ($=1)$ and of $C$ (disk of radius $1$ centered at the origin). So $omega(A)=1$.
the norm of $A$ is the maximum of the two norms, so $|A|=2$.
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=Boplus Cin M_3(mathbb C)$, where $B=1$ and $C=2E_12$. That is,
$$
A=beginbmatrix 1&0&0\0&0&2\0&0&0endbmatrix.
$$
Then
$sigma(A)=0,1$ and thus $r(A)=1$
the numerical range of $A$ is the convex hull of the ranges of $B$ ($=1)$ and of $C$ (disk of radius $1$ centered at the origin). So $omega(A)=1$.
the norm of $A$ is the maximum of the two norms, so $|A|=2$.
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=Boplus Cin M_3(mathbb C)$, where $B=1$ and $C=2E_12$. That is,
$$
A=beginbmatrix 1&0&0\0&0&2\0&0&0endbmatrix.
$$
Then
$sigma(A)=0,1$ and thus $r(A)=1$
the numerical range of $A$ is the convex hull of the ranges of $B$ ($=1)$ and of $C$ (disk of radius $1$ centered at the origin). So $omega(A)=1$.
the norm of $A$ is the maximum of the two norms, so $|A|=2$.
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
No, because the three numbers are maxima, so you can "hide" things in blocks. For instance, let $A=Boplus Cin M_3(mathbb C)$, where $B=1$ and $C=2E_12$. That is,
$$
A=beginbmatrix 1&0&0\0&0&2\0&0&0endbmatrix.
$$
Then
$sigma(A)=0,1$ and thus $r(A)=1$
the numerical range of $A$ is the convex hull of the ranges of $B$ ($=1)$ and of $C$ (disk of radius $1$ centered at the origin). So $omega(A)=1$.
the norm of $A$ is the maximum of the two norms, so $|A|=2$.
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
answered Mar 30 at 11:16
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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