tracial state of a orthogonal projection
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Suppose $Ain M_n(mathbb{C})$,$A$ has eigenvalues$lambda_1,cdots,lambda_n$,$P$ is the orthogonal projection from $mathbb{C}^n$ onto the span of eigenvectors associated with $lambda_1,cdots,lambda_n$,how to computer the tracial state of $P$.I saw a reference book,it writes:$tr_n(P)=frac{1}{n}$(cardinality of $lambda_1,cdots,lambda_n)$.I feel a little confused.
operator-theory matrix-calculus operator-algebras c-star-algebras trace
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
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add a comment |
$begingroup$
Suppose $Ain M_n(mathbb{C})$,$A$ has eigenvalues$lambda_1,cdots,lambda_n$,$P$ is the orthogonal projection from $mathbb{C}^n$ onto the span of eigenvectors associated with $lambda_1,cdots,lambda_n$,how to computer the tracial state of $P$.I saw a reference book,it writes:$tr_n(P)=frac{1}{n}$(cardinality of $lambda_1,cdots,lambda_n)$.I feel a little confused.
operator-theory matrix-calculus operator-algebras c-star-algebras trace
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
$endgroup$
add a comment |
$begingroup$
Suppose $Ain M_n(mathbb{C})$,$A$ has eigenvalues$lambda_1,cdots,lambda_n$,$P$ is the orthogonal projection from $mathbb{C}^n$ onto the span of eigenvectors associated with $lambda_1,cdots,lambda_n$,how to computer the tracial state of $P$.I saw a reference book,it writes:$tr_n(P)=frac{1}{n}$(cardinality of $lambda_1,cdots,lambda_n)$.I feel a little confused.
operator-theory matrix-calculus operator-algebras c-star-algebras trace
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
$endgroup$
Suppose $Ain M_n(mathbb{C})$,$A$ has eigenvalues$lambda_1,cdots,lambda_n$,$P$ is the orthogonal projection from $mathbb{C}^n$ onto the span of eigenvectors associated with $lambda_1,cdots,lambda_n$,how to computer the tracial state of $P$.I saw a reference book,it writes:$tr_n(P)=frac{1}{n}$(cardinality of $lambda_1,cdots,lambda_n)$.I feel a little confused.
operator-theory matrix-calculus operator-algebras c-star-algebras trace
operator-theory matrix-calculus operator-algebras c-star-algebras trace
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
asked Dec 5 '18 at 15:19
mathrookiemathrookie
832512
832512
add a comment |
add a comment |
1 Answer
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There is no general answer to this. If $lambda_1,ldots,lambda_n$ are distinct, then the eigenvectors of $A$ span $mathbb C^n$ and so $P=I$. If $A=lambda P$, then $operatorname{tr}_n(P)=k/n$, where $k$ is the rank of $P$.
So all possible values $1/n,2/n,ldots,1$ are possible.
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
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1 Answer
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1 Answer
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$begingroup$
There is no general answer to this. If $lambda_1,ldots,lambda_n$ are distinct, then the eigenvectors of $A$ span $mathbb C^n$ and so $P=I$. If $A=lambda P$, then $operatorname{tr}_n(P)=k/n$, where $k$ is the rank of $P$.
So all possible values $1/n,2/n,ldots,1$ are possible.
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
add a comment |
$begingroup$
There is no general answer to this. If $lambda_1,ldots,lambda_n$ are distinct, then the eigenvectors of $A$ span $mathbb C^n$ and so $P=I$. If $A=lambda P$, then $operatorname{tr}_n(P)=k/n$, where $k$ is the rank of $P$.
So all possible values $1/n,2/n,ldots,1$ are possible.
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
add a comment |
$begingroup$
There is no general answer to this. If $lambda_1,ldots,lambda_n$ are distinct, then the eigenvectors of $A$ span $mathbb C^n$ and so $P=I$. If $A=lambda P$, then $operatorname{tr}_n(P)=k/n$, where $k$ is the rank of $P$.
So all possible values $1/n,2/n,ldots,1$ are possible.
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
$endgroup$
There is no general answer to this. If $lambda_1,ldots,lambda_n$ are distinct, then the eigenvectors of $A$ span $mathbb C^n$ and so $P=I$. If $A=lambda P$, then $operatorname{tr}_n(P)=k/n$, where $k$ is the rank of $P$.
So all possible values $1/n,2/n,ldots,1$ are possible.
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
answered Dec 5 '18 at 17:29
Martin ArgeramiMartin Argerami
124k1177176
124k1177176
add a comment |
add a comment |
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