| Topic: |
Science > Physics |
| User: |
"Edward Green" |
| Date: |
05 Aug 2005 11:01:13 PM |
| Object: |
operator trace |
The trace of a matrix is the sum of its diagonal elements.
I've encountered the idea that operators may have traces. Linear
operators can often be represented by matrices, so, presumably, in case
the operator is represented by a matrix, the operator trace reduces to
the matrix trace.
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
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| User: "" |
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| Title: Re: operator trace |
05 Aug 2005 11:12:38 PM |
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2 extensions am aware of: there is the trace of a tensor (very
similar to that of a matrix) and the trace-class operator
http://mathworld.wolfram.com/Trace-ClassOperator.html
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
06 Aug 2005 12:41:14 PM |
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Edward Green wrote:
The trace of a matrix is the sum of its diagonal elements.
I've encountered the idea that operators may have traces. Linear
operators can often be represented by matrices, so, presumably, in case
the operator is represented by a matrix, the operator trace reduces to
the matrix trace.
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
There is also a trace defined on the space of so called trace-class
operators on a Hilbert space, which is a subset of the set of compact
operators. It basically involves computing the trace on the operators
with finite dimensional range, and then extending the definition to
operators that are close to finite rank operators (e.g. define a
trace-class norm as the sum of the singular values, and then take the
completion with respect to this norm).
To be honest, I have always seen trace defined with respect to some
coordinate system, and then one proves that it is independent of choice
of coordinates.
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| User: "Lee Rudolph" |
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| Title: Re: operator trace |
06 Aug 2005 01:06:17 PM |
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Stephen Montgomery-Smith <stephen@math.missouri.edu> writes:
To be honest, I have always seen trace defined with respect to some
coordinate system, and then one proves that it is independent of choice
of coordinates.
It's quite easy to dispense with the coordinate system, in
favor of an inner product (I'm speaking only of real coefficients,
though it's pretty clear that the complex case follows), when
the space V is finite-dimensional: as usual, define the "Rayleigh
quotient" of a linear operator A on V to be <Av,v>/<v,v>, where
<.,.> is the inner product and v is non-zero; then the average
value of the Rayleigh quotient on the unit sphere of V (defined
by that inner product) is tr(A)/dim(V). (Actually, tr(A)/dim(V)
is probably what should be called the "trace" of A, anyhow, from
the point of view of an algebraist--including a "von Neumann
algebra"ist--since one wants, when possible, to have the trace
of the identity be 1; doesn't one?) However, I've never
found a proof that this is independent of choice of inner
product which is any more elegant than the (inelegant, to my
mind) proof that the trace defined in terms of coordinates
is independent of the choice of coordinates.
The two characterizations of tr(A) are actually closely related.
A coordinate system (that is, an ordered basis of V) defines
an inner product in the usual way (i.e., it lets you pull back
the standard inner product on R^n) such that the original basis
is orthonormal for that inner product. Then the coordinate-bound
definition of tr(A) (as the "integral", i.e., sum of the Rayleigh
quotient of A over that basis), the integral of *that* integral
over the Grassmann manifold of all orthonormal bases, and the
average-value-over-the-unit-sphere definition, have some nice
obvious relation that I'm temporarily forgetting (derived from
the fibration of the Grassmannian over the sphere).
I've never thought about how, if at all, any such definition
could extend to "trace-class operators" on an infinite dimensional
Hilbert space. An extension to von Neumann algebras (of the
right type) might make more sense.
Experimentation suggests that *something* can be said over finite
fields, as well, but with lots of problems (starting with a need
to have the dimension prime to the characteristic, as far as I
could see the one time I spent a while actually thinking about it
instead of just flailing about wildly).
Lee Rudolph
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
06 Aug 2005 04:00:38 PM |
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Lee Rudolph wrote:
(Actually, tr(A)/dim(V)
is probably what should be called the "trace" of A, anyhow, from
the point of view of an algebraist--including a "von Neumann
algebra"ist--since one wants, when possible, to have the trace
of the identity be 1; doesn't one?)
I don't think that you always want the trace of the identity to be 1.
This would correspond to insisting in measure theory that the total
measure of the space is 1. But then you are not studying all of measure
theory, but only a subclass called probability theory. Similarly, if
you insist that the trace of the identity is 1 then you restrict your
study to type II_1 von-Neuman algebras. I guess if you are doing one of
the latest fads which is "non-commutative probability theory" then this
is appropriate. But it excludes the most natural von-Neuman algebra,
that is, B(H).
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| User: "Timothy Murphy" |
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| Title: Re: operator trace |
06 Aug 2005 12:55:24 PM |
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Stephen Montgomery-Smith wrote:
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
The trace is equal to the sum of the eigenvalues.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
06 Aug 2005 12:55:47 PM |
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Timothy Murphy wrote:
Stephen Montgomery-Smith wrote:
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
The trace is equal to the sum of the eigenvalues.
Yes, of course, silly me. And for trace-class operators, this "sum" is
also well defined.
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| User: "Edward Green" |
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| Title: Re: operator trace |
06 Aug 2005 08:01:01 PM |
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Stephen Montgomery-Smith wrote:
Timothy Murphy wrote:
Stephen Montgomery-Smith wrote:
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
The trace is equal to the sum of the eigenvalues.
Yes, of course, silly me. And for trace-class operators, this "sum" is
also well defined.
Ah, thank you! That must be a best one-line Usenet answer. For
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Now I'm thinking and wondering what this means, and why it should be
useful: for many common operators, even those with totally discrete
spectrums, the sum will fail to converge -- for example, the
hamiltonian of a simple harmonic oscillator. What _is_ a simple
example of a "trace class operator"? An angular momentum projection
operator for given fixed total angular momentum?
[Thanks for all the responses and references, and particularly also to
Lee Rudolph, for his longer printable exposition which will no doubt
repay study.]
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| User: "Puppet_Sock" |
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| Title: Re: operator trace |
09 Aug 2005 09:26:59 AM |
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Edward Green wrote:
[snip]
Now I'm thinking and wondering what this means, and why it should be
useful: for many common operators, even those with totally discrete
spectrums, the sum will fail to converge -- for example, the
hamiltonian of a simple harmonic oscillator. What _is_ a simple
example of a "trace class operator"? An angular momentum projection
operator for given fixed total angular momentum?
Sure. You get into trouble with traces fairly quickly.
In many common cases, you can't represent the operators
as matrices, at least not finite dimensional ones.
Recall that, for A and B both square matrices of the
same dimension, the trace of AB is equal to the trace
of BA. So, think about cases where operators A and B
do not commute.
AB - BA != 0
So, what of trace(AB - BA), which must be zero?
Another way of saying this is, you can't get both of
these operators diagonal at the same time in a finite
basis.
Socks
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| User: "Gregory L. Hansen" |
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| Title: Re: operator trace |
07 Aug 2005 10:00:24 AM |
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In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Stephen Montgomery-Smith wrote:
Timothy Murphy wrote:
Stephen Montgomery-Smith wrote:
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
The trace is equal to the sum of the eigenvalues.
Yes, of course, silly me. And for trace-class operators, this "sum" is
also well defined.
Ah, thank you! That must be a best one-line Usenet answer. For
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
--
"Out of the way, you slime, a physicist is coming!"
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
07 Aug 2005 06:10:28 PM |
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Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Stephen Montgomery-Smith wrote:
Timothy Murphy wrote:
Stephen Montgomery-Smith wrote:
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
The trace is equal to the sum of the eigenvalues.
Yes, of course, silly me. And for trace-class operators, this "sum" is
also well defined.
Ah, thank you! That must be a best one-line Usenet answer. For
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
You can do it like this for compact operators. A non-zero eigenfactor
is any scalar lambda such that A-lambda I is not invertible. The
multiplicity is going to be the limit as n->infinity of the dimension of
the kernel of (A-lambda I)^n - the limit will exist because the sequence
will be eventually constant. The trace is going to be the sum of these
eigenvalues taken with multiplicity.
Well, something like this.
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| User: "Edward Green" |
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| Title: Re: operator trace |
07 Aug 2005 05:15:33 PM |
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Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
<...>
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
Are you sure about that?
Let A be an operator, X a vector, b a number, and let AX = bX
Now X is an eigenvector, b an eigenvalue, and that statement is
independent of basis.
The eigenvectors and eigenvalues are what they are, though they do
happen to define a representation in which the eigenvalues of the
operator appear on the diagonal. I'm not enough of a linear algebraist
to immediately say what happens to the trace in other representations.
Maybe it's invariant under orthogonal transformations -- that sounds
like a good thing to say. ;-)
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| User: "Gregory L. Hansen" |
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| Title: Re: operator trace |
07 Aug 2005 10:32:41 PM |
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In article <1123452933.165181.136740@f14g2000cwb.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
<...>
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
Are you sure about that?
Let A be an operator, X a vector, b a number, and let AX = bX
Now X is an eigenvector, b an eigenvalue, and that statement is
independent of basis.
The eigenvectors and eigenvalues are what they are, though they do
happen to define a representation in which the eigenvalues of the
operator appear on the diagonal. I'm not enough of a linear algebraist
to immediately say what happens to the trace in other representations.
Maybe it's invariant under orthogonal transformations -- that sounds
like a good thing to say. ;-)
Sure, you have an X and a b in the abstract. But if you wanted a trace,
you need actual numbers, not something in the abstract. Once you find the
trace in any basis it will be the same in any other basis, but you have to
find it in some basis. As far as I know. I mean, if X were a velocity
we'd usually say it has some magnitude and <X|Y> equals something, which
gives information about it without resorting to coordinates. But in QM
problems we usually don't start by defining a vector in the Hilbert space,
we start by defining the properties of an interesting operator.
--
"What are the possibilities of small but movable machines? They may or
may not be useful, but they surely would be fun to make."
-- Richard P. Feynman, 1959
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| User: "Edward Green" |
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| Title: Re: operator trace |
08 Aug 2005 11:32:51 PM |
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Gregory L. Hansen wrote:
In article <1123452933.165181.136740@f14g2000cwb.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
<...>
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
Are you sure about that?
Let A be an operator, X a vector, b a number, and let AX = bX
Now X is an eigenvector, b an eigenvalue, and that statement is
independent of basis.
<...>
Sure, you have an X and a b in the abstract. But if you wanted a trace,
you need actual numbers, not something in the abstract. Once you find the
trace in any basis it will be the same in any other basis, but you have to
find it in some basis. As far as I know.
Well, according to the definition as the sum of the eigenvalues if you
merely knew all the eigenvalues you would know the trace, and that is
independent of basis. One _might_ have taken from your lines
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
the implication that the eigenvalues (the "specific set") depended on
the representation. But they don't.
As to whether you have to resort to a specific representation to find
the eigenvalues -- I won't argue that, and I don't know. But aren't
there cases where by being very clever you can find the eigenvalues by
playing with the commutation relations?
As for being the same in any basis, that is certainly implied in the
eigenvalue definition which we may hope is equivalent to a basis
dependent definition (adding diagonal matrix elements). I suppose it
is also true then.
Now, just what does the damn thing mean physically...
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| User: "Gregory L. Hansen" |
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| Title: Re: operator trace |
09 Aug 2005 09:12:20 AM |
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In article <1123561971.044840.158650@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Gregory L. Hansen wrote:
In article <1123452933.165181.136740@f14g2000cwb.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
<...>
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
Are you sure about that?
Let A be an operator, X a vector, b a number, and let AX = bX
Now X is an eigenvector, b an eigenvalue, and that statement is
independent of basis.
<...>
Sure, you have an X and a b in the abstract. But if you wanted a trace,
you need actual numbers, not something in the abstract. Once you find the
trace in any basis it will be the same in any other basis, but you have to
find it in some basis. As far as I know.
Well, according to the definition as the sum of the eigenvalues if you
merely knew all the eigenvalues you would know the trace, and that is
independent of basis. One _might_ have taken from your lines
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
the implication that the eigenvalues (the "specific set") depended on
the representation. But they don't.
Yes, I think you're right about that. That is, you get the eigenvalues in
the basis that diagonalizes the operator, but finding the eigenvalues is
the easy part, and the first step to finding the eigenvectors. So we
could know in the abstract that there's a basis, but not have to find it.
As to whether you have to resort to a specific representation to find
the eigenvalues -- I won't argue that, and I don't know. But aren't
there cases where by being very clever you can find the eigenvalues by
playing with the commutation relations?
Again, I think you're right about that. But again I think there's the
caveat that there's a basis in there somewhere (e.g. angular momentum
eigenstates) even if you don't have to find it to find the trace.
As for being the same in any basis, that is certainly implied in the
eigenvalue definition which we may hope is equivalent to a basis
dependent definition (adding diagonal matrix elements). I suppose it
is also true then.
Now, just what does the damn thing mean physically...
In my small exposure to crystallography they kept trying to throw traces
at me as if there were some meaning to them. Or characters, or something
like that. It's a conserved quantity, but you knew that.
--
"What's another word for thesaurus?" -- Steven Wright
"Let me look in my synonymicon." -- Thaddeus Stout
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| User: "Edward Green" |
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| Title: Re: operator trace |
09 Aug 2005 11:17:39 PM |
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Gregory L. Hansen wrote:
Edward Green <spamspamspam3@netzero.com> wrote:
Now, just what does the damn thing mean physically...
In my small exposure to crystallography they kept trying to throw traces
at me as if there were some meaning to them.
The dastards.
Or characters, or something
like that. It's a conserved quantity, but you knew that.
Then there is the notorious signature of the metric.
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| User: "Edward Green" |
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| Title: Re: operator trace |
07 Aug 2005 05:21:47 PM |
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Gregory L. Hansen wrote:
In article <1123373697.951191.261630@g47g2000cwa.googlegroups.com>,
Edward Green <spamspamspam3@netzero.com> wrote:
<...>
familiar cases in quantum mechanics the representation of the operator
in an appropriate basis is a matrix with the eigenvalues on the
diagonal -- so this plausibly extends the concept and answers my
question.
Does it? I thought your question might have translated to something like
"How do we determine the trace without a specific set of eigenvalues?"
Can't have eigenvalues until you pick a representation.
Are you sure about that?
Let A be an operator, X a vector, b a number, and let AX = bX
Now X is an eigenvector, b an eigenvalue, and that statement is
independent of basis.
The eigenvectors and eigenvalues are what they are, though they do
happen to define a representation in which the eigenvalues of the
operator appear on the diagonal. I'm not enough of a linear algebraist
to immediately say what happens to the trace in other representations.
Maybe it's invariant under orthogonal transformations -- that sounds
like a good thing to say. ;-)
.
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| User: "Shmuel Seymour J. Metz" |
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| Title: Re: operator trace |
09 Aug 2005 07:24:55 AM |
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In <dd57m8$c7p$2@rainier.uits.indiana.edu>, on 08/07/2005
at 03:00 PM, (Gregory L. Hansen)
said:
Does it? I thought your question might have translated to something
like "How do we determine the trace without a specific set of
eigenvalues?"
No; "matrix representation " is not equivalent to "specific set of
eigenvalues".
Can't have eigenvalues until you pick a representation.
If it's a linear operator then you already have a representation.
There are other issues if the operator is not Hermitian, or if the
vector space it operates on is not finite dimensional.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to
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| User: "Gregory L. Hansen" |
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| Title: Re: operator trace |
09 Aug 2005 09:37:43 AM |
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In article <42f8aea7$4$fuzhry+tra$mr2ice@news.patriot.net>,
Shmuel (Seymour J.) Metz <spamtrap@library.lspace.org.invalid> wrote:
In <dd57m8$c7p$2@rainier.uits.indiana.edu>, on 08/07/2005
at 03:00 PM, (Gregory L. Hansen)
said:
Does it? I thought your question might have translated to something
like "How do we determine the trace without a specific set of
eigenvalues?"
No; "matrix representation " is not equivalent to "specific set of
eigenvalues".
Can't have eigenvalues until you pick a representation.
If it's a linear operator then you already have a representation.
There are other issues if the operator is not Hermitian, or if the
vector space it operates on is not finite dimensional.
The eigenvectors form the basis that diagonalizes the matrix, although
you can find eigenvalues without finding eigenvectors.
--
"He who only sees business in business is a fool."
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| User: "Shmuel Seymour J. Metz" |
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| Title: Re: operator trace |
10 Aug 2005 08:02:19 AM |
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In <ddaf3n$us5$4@rainier.uits.indiana.edu>, on 08/09/2005
at 02:37 PM, (Gregory L. Hansen)
said:
The eigenvectors form the basis that diagonalizes the matrix,
There might not be such a basis. Again, There are other issues if the
operator is not Hermitian, or if the vector space it operates on is
not finite dimensional. Consider a rotation in a real 2-dimensional
vector space or a shift operator in an infinite dimensional vector
space.
Also, there is no "the matrix" until you pick a basis.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to
.
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| User: "Shmuel Seymour J. Metz" |
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| Title: Re: operator trace |
10 Aug 2005 05:26:32 PM |
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In <ddaf3n$us5$4@rainier.uits.indiana.edu>, on 08/09/2005
at 02:37 PM, (Gregory L. Hansen)
said:
The eigenvectors form the basis that diagonalizes the matrix,
There might not be such a basis. Again, There are other issues if the
operator is not Hermitian, or if the vector space it operates on is
not finite dimensional. Consider a rotation in a real 2-dimensional
vector space or a shift operator in an infinite dimensional vector
space.
Also, there is no "the matrix" until you pick a basis.
--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>
Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to
.
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
05 Aug 2005 11:08:08 PM |
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Edward Green wrote:
The trace of a matrix is the sum of its diagonal elements.
I've encountered the idea that operators may have traces. Linear
operators can often be represented by matrices, so, presumably, in case
the operator is represented by a matrix, the operator trace reduces to
the matrix trace.
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
There is the study of so called von-Neuman algebras, which are
C*-algebras of linear operators on a Hilbert space equipped with a
trace. There is a lot of literature on this subject, but it really is a
difficult and technical subject.
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| User: "Zdislav V. Kovarik" |
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| Title: Re: operator trace |
07 Aug 2005 03:25:22 PM |
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On Sat, 5 Aug 2005, Edward Green wrote:
The trace of a matrix is the sum of its diagonal elements.
I've encountered the idea that operators may have traces. Linear
operators can often be represented by matrices, so, presumably, in case
the operator is represented by a matrix, the operator trace reduces to
the matrix trace.
But is there a more general concept of operator trace which does not
explicitly depend on a matrix representation?
Most responses (maybe all, I may not received all of them yet) considered
traces of operators on Hilbert space.
All you really need in the infinite-dimensional case is a normed vector
space V (the norm does not have to come from an inner product, and the
space need not be complete, nor does it have to have a topological
basis!). For safety's sake, we stipulate that all operators in the
discussion below are bounded with respect to the norm on V.
Finite rank operators have a well-defined trace:
a rank-one operator A=xf (x from V, f from V*, the dual) has a
well-defined trace tr(xf) = f(x). It takes some tedious bookkeeping
algebra to prove that the trace extends linearly to finite-rank operators,
and is independent of the representation using rank-one summands.
Denote F1 the set of bounded finite-rank operators V->V of operator
norm <=1. Then for every linear operator T:V->V and every X from F1,
TX and XT are of finite rank, and tr(TX)=tr(XT) (routine proof).
To put the norm to work, we define the absolute trace of T (need not be
finite) by
||T||_1 = sup{abs(tr(XT)) : X from F1 }
If this absolute trace of T is finite, we call T a trace class operator on
V, and by further work on norms we can prove that the trace extends
linearly and continuously (with respect to the absolute trace, which is a
norm) from finite rank operators to all trace class operators.
Look Ma, no coordinates.
Remark: This definition, restricted to Hilbert spaces, defines the same
set of trace class operators. (An exercise I saw somewhere.)
Also, on Hilbert spaces, the trace class operators are the pre-dual of the
bounded operators. Has anyone studied the Banach space setup of this fact?
Cheers, ZVK(Slavek).
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
07 Aug 2005 06:12:38 PM |
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Zdislav V. Kovarik wrote:
Also, on Hilbert spaces, the trace class operators are the pre-dual of the
bounded operators. Has anyone studied the Banach space setup of this fact?
I have the impression that this has been studied to death, but its only
tangential to my area of study.
For example, I think that B(H) is an example of a Banach space that
fails the approximation property.
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| User: "Zdislav V. Kovarik" |
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| Title: Re: operator trace |
07 Aug 2005 06:21:46 PM |
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On Sun, 7 Aug 2005, Stephen Montgomery-Smith wrote:
Zdislav V. Kovarik wrote:
Also, on Hilbert spaces, the trace class operators are the pre-dual of the
bounded operators. Has anyone studied the Banach space setup of this fact?
I have the impression that this has been studied to death, but its only
tangential to my area of study.
For example, I think that B(H) is an example of a Banach space that
fails the approximation property.
B(H) does not enter the competition because it is not separable. The
approximation property, as studied by Enflo and others, assumes a
separable space.
Cheers, ZVK(Slavek).
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| User: "Stephen Montgomery-Smith" |
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| Title: Re: operator trace |
07 Aug 2005 10:06:18 PM |
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Zdislav V. Kovarik wrote:
B(H) does not enter the competition because it is not separable. The
approximation property, as studied by Enflo and others, assumes a
separable space.
Cheers, ZVK(Slavek).
MR0631090 (83a:46033)
Szankowski, Andrzej
$B({\cal H})$ does not have the approximation property.
Acta Math. 147 (1981), no. 1-2, 89--108.
46B20 (47D15)
Review in linked PDF Add citation to clipboard Document Delivery
Service Journal Original Article More links
References: 0 Reference Citations: 9 Review Citations: 1
Let $B(\scr H)$ denote the space of bounded linear operators in an
infinite-dimensional Hilbert space $\scr H$. The result is formulated in
the title; a Banach space $X$ is said to have the approximation property
(a.p.) if the identity operator on $X$ can be approximated uniformly on
compact subsets of $X$ by linear operators of finite rank. $B(\scr H)$
is the first example of a "natural" Banach space without the a.p. The
idea underlying the proof is basically the same as in earlier
constructions of "artificial" examples (the author has had important
contributions in this field), but technical problems arising in this
case are numerous and considerably more difficult.
Reviewed by T. Figiel
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