Norm closure of diagonalizable operator on Hilbert space
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Problem: Prove the norm closure of diagonalizable operator in $mathcal{B(H)}$ (bounded operators on Hilbert space which is not necessarily finite dimensional) is the set of normal operators.
A bounded operator $mathbb{T}$ is said to be Diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $mathbb{T}$.
What I am thinking: Since I already known that a normal operator can be approximated by a linear combination of pairwise commuting projections, that is, for $mathbb{T}$ is bounded normal operator, then there exists a sequence of projections ${P_n}$ summing to identity operator ${I}$ and ${lambda_n}$ such that $|| mathbb{T} - sum_{n}lambda_nP_n||<epsilon$.
And we know that for diagonalizable operator $mathbb{T}$ in general Hilbert space $mathcal{H}$, we have another equivalent definition: There exists an orthonormal basis ${e_i}_{iin J}$ of $mathcal{H}$ and values ${lambda_i}_{i in J}$ such that $mathbb{T}(x) = sum_{i in J}lambda_i <x,e_i>e_i$. So my idea is to express a diagonalizable operator as a linear combination of projections then combined with the fact that a normal operator can be approximate by a linear combination of projections to finish the proof.
I am not sure whether expressing a diagonalizable operator as linear combination of projections is correct in general Hilbert space. In finite dimension, which should be true. Any idea will be appreciated.
functional-analysis operator-theory operator-algebras
asked Dec 12 '18 at 1:35
apple apple
467
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add a comment |
$begingroup$
Problem: Prove the norm closure of diagonalizable operator in $mathcal{B(H)}$ (bounded operators on Hilbert space which is not necessarily finite dimensional) is the set of normal operators.
A bounded operator $mathbb{T}$ is said to be Diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $mathbb{T}$.
What I am thinking: Since I already known that a normal operator can be approximated by a linear combination of pairwise commuting projections, that is, for $mathbb{T}$ is bounded normal operator, then there exists a sequence of projections ${P_n}$ summing to identity operator ${I}$ and ${lambda_n}$ such that $|| mathbb{T} - sum_{n}lambda_nP_n||<epsilon$.
And we know that for diagonalizable operator $mathbb{T}$ in general Hilbert space $mathcal{H}$, we have another equivalent definition: There exists an orthonormal basis ${e_i}_{iin J}$ of $mathcal{H}$ and values ${lambda_i}_{i in J}$ such that $mathbb{T}(x) = sum_{i in J}lambda_i <x,e_i>e_i$. So my idea is to express a diagonalizable operator as a linear combination of projections then combined with the fact that a normal operator can be approximate by a linear combination of projections to finish the proof.
I am not sure whether expressing a diagonalizable operator as linear combination of projections is correct in general Hilbert space. In finite dimension, which should be true. Any idea will be appreciated.
functional-analysis operator-theory operator-algebras
asked Dec 12 '18 at 1:35
apple apple
467
$endgroup$
add a comment |
$begingroup$
Problem: Prove the norm closure of diagonalizable operator in $mathcal{B(H)}$ (bounded operators on Hilbert space which is not necessarily finite dimensional) is the set of normal operators.
A bounded operator $mathbb{T}$ is said to be Diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $mathbb{T}$.
What I am thinking: Since I already known that a normal operator can be approximated by a linear combination of pairwise commuting projections, that is, for $mathbb{T}$ is bounded normal operator, then there exists a sequence of projections ${P_n}$ summing to identity operator ${I}$ and ${lambda_n}$ such that $|| mathbb{T} - sum_{n}lambda_nP_n||<epsilon$.
And we know that for diagonalizable operator $mathbb{T}$ in general Hilbert space $mathcal{H}$, we have another equivalent definition: There exists an orthonormal basis ${e_i}_{iin J}$ of $mathcal{H}$ and values ${lambda_i}_{i in J}$ such that $mathbb{T}(x) = sum_{i in J}lambda_i <x,e_i>e_i$. So my idea is to express a diagonalizable operator as a linear combination of projections then combined with the fact that a normal operator can be approximate by a linear combination of projections to finish the proof.
I am not sure whether expressing a diagonalizable operator as linear combination of projections is correct in general Hilbert space. In finite dimension, which should be true. Any idea will be appreciated.
functional-analysis operator-theory operator-algebras
asked Dec 12 '18 at 1:35
apple apple
467
$endgroup$
Problem: Prove the norm closure of diagonalizable operator in $mathcal{B(H)}$ (bounded operators on Hilbert space which is not necessarily finite dimensional) is the set of normal operators.
A bounded operator $mathbb{T}$ is said to be Diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $mathbb{T}$.
What I am thinking: Since I already known that a normal operator can be approximated by a linear combination of pairwise commuting projections, that is, for $mathbb{T}$ is bounded normal operator, then there exists a sequence of projections ${P_n}$ summing to identity operator ${I}$ and ${lambda_n}$ such that $|| mathbb{T} - sum_{n}lambda_nP_n||<epsilon$.
And we know that for diagonalizable operator $mathbb{T}$ in general Hilbert space $mathcal{H}$, we have another equivalent definition: There exists an orthonormal basis ${e_i}_{iin J}$ of $mathcal{H}$ and values ${lambda_i}_{i in J}$ such that $mathbb{T}(x) = sum_{i in J}lambda_i <x,e_i>e_i$. So my idea is to express a diagonalizable operator as a linear combination of projections then combined with the fact that a normal operator can be approximate by a linear combination of projections to finish the proof.
I am not sure whether expressing a diagonalizable operator as linear combination of projections is correct in general Hilbert space. In finite dimension, which should be true. Any idea will be appreciated.
functional-analysis operator-theory operator-algebras
functional-analysis operator-theory operator-algebras
asked Dec 12 '18 at 1:35
apple apple
467
asked Dec 12 '18 at 1:35
apple apple
467
asked Dec 12 '18 at 1:35
apple apple
467
asked Dec 12 '18 at 1:35
apple apple
467
asked Dec 12 '18 at 1:35
apple apple
467
467
add a comment |
add a comment |
1 Answer
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Yes, that's how it works. You need to check that a diagonalizable operator $T=sum_jlambda_jP_j$ is normal, which is easy to do.
And, as you say, normal operators are limits of diagonalizable operators.
The two aforementioned facts, together, tell you that the normal operators are precisely the closure of the set of diagonalizable operators.
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
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1 Answer
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1 Answer
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oldest
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active
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$begingroup$
Yes, that's how it works. You need to check that a diagonalizable operator $T=sum_jlambda_jP_j$ is normal, which is easy to do.
And, as you say, normal operators are limits of diagonalizable operators.
The two aforementioned facts, together, tell you that the normal operators are precisely the closure of the set of diagonalizable operators.
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
Yes, that's how it works. You need to check that a diagonalizable operator $T=sum_jlambda_jP_j$ is normal, which is easy to do.
And, as you say, normal operators are limits of diagonalizable operators.
The two aforementioned facts, together, tell you that the normal operators are precisely the closure of the set of diagonalizable operators.
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
Yes, that's how it works. You need to check that a diagonalizable operator $T=sum_jlambda_jP_j$ is normal, which is easy to do.
And, as you say, normal operators are limits of diagonalizable operators.
The two aforementioned facts, together, tell you that the normal operators are precisely the closure of the set of diagonalizable operators.
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
Yes, that's how it works. You need to check that a diagonalizable operator $T=sum_jlambda_jP_j$ is normal, which is easy to do.
And, as you say, normal operators are limits of diagonalizable operators.
The two aforementioned facts, together, tell you that the normal operators are precisely the closure of the set of diagonalizable operators.
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 12 '18 at 18:49
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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