Help verifying the norm of the resolvent of a matrix The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Inverse of a certain differential operator (resolvent)Matrix norm in Banach spaceConvergence of the spectrum under norm resolvent convergenceFind the closest element in a span of matrices?Resolvent of a matrixNorm of the ResolventInfimum of spectrum differentiable if resolvent norm-differentiable?Can the level set of the resolvent norm be constant on a set of positive measure?Clarification: Is it true trace of $(A^TA)$ is induced matrix norm squared?It can happen that the norm 1 of a matrix and the infinite norm are different?
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Help verifying the norm of the resolvent of a matrix
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Inverse of a certain differential operator (resolvent)Matrix norm in Banach spaceConvergence of the spectrum under norm resolvent convergenceFind the closest element in a span of matrices?Resolvent of a matrixNorm of the ResolventInfimum of spectrum differentiable if resolvent norm-differentiable?Can the level set of the resolvent norm be constant on a set of positive measure?Clarification: Is it true trace of $(A^TA)$ is induced matrix norm squared?It can happen that the norm 1 of a matrix and the infinite norm are different?
$begingroup$
I'm reading a document where it is said that if
$$A=beginpmatrix0 & 1\0 & 0 endpmatrix$$
then the norm of the resolvent for $z neq 0$ is given by
$$|R(z,A)|= fracsqrt2sqrt^2-sqrt^2.$$
I think that if $zneq 0$ then
$$ R(z,A)=(A-zI)^-1=beginpmatrix-1/z & -1/z^2\0 & -1/z endpmatrix$$
and because of that
$$|R(z,A)|=fracsqrtz.$$
Am I wrong?.
linear-algebra functional-analysis spectral-theory
asked Apr 8 at 8:13
krenickkrenick
374
$endgroup$
add a comment |
$begingroup$
I'm reading a document where it is said that if
$$A=beginpmatrix0 & 1\0 & 0 endpmatrix$$
then the norm of the resolvent for $z neq 0$ is given by
$$|R(z,A)|= fracsqrt2sqrt^2-sqrt^2.$$
I think that if $zneq 0$ then
$$ R(z,A)=(A-zI)^-1=beginpmatrix-1/z & -1/z^2\0 & -1/z endpmatrix$$
and because of that
$$|R(z,A)|=fracsqrtz.$$
Am I wrong?.
linear-algebra functional-analysis spectral-theory
asked Apr 8 at 8:13
krenickkrenick
374
$endgroup$
add a comment |
$begingroup$
I'm reading a document where it is said that if
$$A=beginpmatrix0 & 1\0 & 0 endpmatrix$$
then the norm of the resolvent for $z neq 0$ is given by
$$|R(z,A)|= fracsqrt2sqrt^2-sqrt^2.$$
I think that if $zneq 0$ then
$$ R(z,A)=(A-zI)^-1=beginpmatrix-1/z & -1/z^2\0 & -1/z endpmatrix$$
and because of that
$$|R(z,A)|=fracsqrtz.$$
Am I wrong?.
linear-algebra functional-analysis spectral-theory
asked Apr 8 at 8:13
krenickkrenick
374
$endgroup$
I'm reading a document where it is said that if
$$A=beginpmatrix0 & 1\0 & 0 endpmatrix$$
then the norm of the resolvent for $z neq 0$ is given by
$$|R(z,A)|= fracsqrt2sqrt^2-sqrt^2.$$
I think that if $zneq 0$ then
$$ R(z,A)=(A-zI)^-1=beginpmatrix-1/z & -1/z^2\0 & -1/z endpmatrix$$
and because of that
$$|R(z,A)|=fracsqrtz.$$
Am I wrong?.
linear-algebra functional-analysis spectral-theory
linear-algebra functional-analysis spectral-theory
asked Apr 8 at 8:13
krenickkrenick
374
asked Apr 8 at 8:13
krenickkrenick
374
edited Apr 8 at 8:23
krenick
asked Apr 8 at 8:13
krenickkrenick
374
asked Apr 8 at 8:13
krenickkrenick
374
asked Apr 8 at 8:13
krenickkrenick
374
374
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.
answered Apr 8 at 9:32
gerwgerw
20.1k11334
$endgroup$
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.
answered Apr 8 at 9:32
gerwgerw
20.1k11334
$endgroup$
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
add a comment |
$begingroup$
You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.
answered Apr 8 at 9:32
gerwgerw
20.1k11334
$endgroup$
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
add a comment |
$begingroup$
You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.
answered Apr 8 at 9:32
gerwgerw
20.1k11334
$endgroup$
You have computed the Frobenius norm of the resolvent, whereas the first formula uses the spectral norm of the resolvent.
answered Apr 8 at 9:32
gerwgerw
20.1k11334
answered Apr 8 at 9:32
gerwgerw
20.1k11334
answered Apr 8 at 9:32
gerwgerw
20.1k11334
answered Apr 8 at 9:32
gerwgerw
20.1k11334
20.1k11334
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
add a comment |
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
1
1
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
Thank you. But, how can I find the spectral norm of the resolvent?.
$endgroup$
– krenick
Apr 8 at 11:35
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
I don't know how can I calculate the spectral norm. Can you tell me what I have to do, please?
$endgroup$
– krenick
Apr 8 at 14:52
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
$begingroup$
The spectral norm of $A$ coincides with the largest spectral value of $A$, i.e., the largest eigenvalue of $A^top A$.
$endgroup$
– gerw
Apr 8 at 19:50
add a comment |
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