Polynomial evaluated on a Normal Bounded Linear Operator Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounded linear operator on a Hilbert spaceSpectrum of a bounded linear operatorNon-empty closed subset of the complex plane is the spectrum of a normal operatorThe spectral radius of normal operatorShow that in a complex Hilbert space, T normal bounded linear operator, $| T^2 | =| T | ^2$$sigma_p(T)$ is at most countable for a normal operatorBounded linear operator property and its spectral radiusResolvents and spectral/numerical radius.Normal operator with real spectrum is hermitian.Bounded function of compact normal operator on Hilbert space is normal
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Polynomial evaluated on a Normal Bounded Linear Operator
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Bounded linear operator on a Hilbert spaceSpectrum of a bounded linear operatorNon-empty closed subset of the complex plane is the spectrum of a normal operatorThe spectral radius of normal operatorShow that in a complex Hilbert space, T normal bounded linear operator, $| T^2 | =| T | ^2$$sigma_p(T)$ is at most countable for a normal operatorBounded linear operator property and its spectral radiusResolvents and spectral/numerical radius.Normal operator with real spectrum is hermitian.Bounded function of compact normal operator on Hilbert space is normal
$begingroup$
Let X be a complex Hilbert Space and A be a Normal Bounded Linear Operator.
Show that the radius of the spectrum of A is equal to the norm of A.
Deduce that if P is a polynomial, then the norm of P(A) is equal to the supremum, over complex numbers z in the spectrum of A, of |P(z)|.
I can do the first, but not the second part of this question !! Any help would be much appreciated.
functional-analysis hilbert-spaces spectral-theory
asked Mar 27 at 17:51
Maths ManMaths Man
111
$endgroup$
add a comment |
$begingroup$
Let X be a complex Hilbert Space and A be a Normal Bounded Linear Operator.
Show that the radius of the spectrum of A is equal to the norm of A.
Deduce that if P is a polynomial, then the norm of P(A) is equal to the supremum, over complex numbers z in the spectrum of A, of |P(z)|.
I can do the first, but not the second part of this question !! Any help would be much appreciated.
functional-analysis hilbert-spaces spectral-theory
asked Mar 27 at 17:51
Maths ManMaths Man
111
$endgroup$
add a comment |
$begingroup$
Let X be a complex Hilbert Space and A be a Normal Bounded Linear Operator.
Show that the radius of the spectrum of A is equal to the norm of A.
Deduce that if P is a polynomial, then the norm of P(A) is equal to the supremum, over complex numbers z in the spectrum of A, of |P(z)|.
I can do the first, but not the second part of this question !! Any help would be much appreciated.
functional-analysis hilbert-spaces spectral-theory
asked Mar 27 at 17:51
Maths ManMaths Man
111
$endgroup$
Let X be a complex Hilbert Space and A be a Normal Bounded Linear Operator.
Show that the radius of the spectrum of A is equal to the norm of A.
Deduce that if P is a polynomial, then the norm of P(A) is equal to the supremum, over complex numbers z in the spectrum of A, of |P(z)|.
I can do the first, but not the second part of this question !! Any help would be much appreciated.
functional-analysis hilbert-spaces spectral-theory
functional-analysis hilbert-spaces spectral-theory
asked Mar 27 at 17:51
Maths ManMaths Man
111
asked Mar 27 at 17:51
Maths ManMaths Man
111
asked Mar 27 at 17:51
Maths ManMaths Man
111
asked Mar 27 at 17:51
Maths ManMaths Man
111
asked Mar 27 at 17:51
Maths ManMaths Man
111
111
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
If $P$ is a polynomial, then $P(A)$ is normal, assuming $A$ is normal. Therefore, $|P(A)| = r_sigma(P(A))$ is the spectral radius of $P(A)$. But the spectrum of $P(A)$ is $P(sigma(A))$ by the spectral mapping theorem. Therefore,
$$
|P(A)|=sup_lambdainsigma(A)|P(lambda)|.
$$
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
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$begingroup$
If $P$ is a polynomial, then $P(A)$ is normal, assuming $A$ is normal. Therefore, $|P(A)| = r_sigma(P(A))$ is the spectral radius of $P(A)$. But the spectrum of $P(A)$ is $P(sigma(A))$ by the spectral mapping theorem. Therefore,
$$
|P(A)|=sup_lambdainsigma(A)|P(lambda)|.
$$
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
$begingroup$
If $P$ is a polynomial, then $P(A)$ is normal, assuming $A$ is normal. Therefore, $|P(A)| = r_sigma(P(A))$ is the spectral radius of $P(A)$. But the spectrum of $P(A)$ is $P(sigma(A))$ by the spectral mapping theorem. Therefore,
$$
|P(A)|=sup_lambdainsigma(A)|P(lambda)|.
$$
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
add a comment |
$begingroup$
If $P$ is a polynomial, then $P(A)$ is normal, assuming $A$ is normal. Therefore, $|P(A)| = r_sigma(P(A))$ is the spectral radius of $P(A)$. But the spectrum of $P(A)$ is $P(sigma(A))$ by the spectral mapping theorem. Therefore,
$$
|P(A)|=sup_lambdainsigma(A)|P(lambda)|.
$$
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
$endgroup$
If $P$ is a polynomial, then $P(A)$ is normal, assuming $A$ is normal. Therefore, $|P(A)| = r_sigma(P(A))$ is the spectral radius of $P(A)$. But the spectrum of $P(A)$ is $P(sigma(A))$ by the spectral mapping theorem. Therefore,
$$
|P(A)|=sup_lambdainsigma(A)|P(lambda)|.
$$
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
answered Mar 27 at 21:21
DisintegratingByPartsDisintegratingByParts
60.5k42681
60.5k42681
add a comment |
add a comment |
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