Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Adjoint multiplicity

  1. Jun 11, 2008 #1
    if A* is the adjoint of A in Complx, Then is A* x A* = (A^2)* or something else??
  2. jcsd
  3. Jun 12, 2008 #2


    User Avatar
    Science Advisor
    Homework Helper

    Have you given this any thought?
  4. Jun 22, 2008 #3
    A*a* = (aa)*

    Nice Question, mindOnmath.

    Let A:H->H be a bounded operator on a complex Hilbert space, H, with inner product (x,y) when x and y are vectors in H.

    Claim: A*A* = (AA)*.

    Proof: For any two vectors x and y in H, we have the following sequence of equalities:
    (A*A*x, y) = (A*x,Ay) = (x,AAy) = ( (AA)*x,y). Q.E.D.

    Note that all finite dimensional vector spaces over the complex numbers are Hilbert spaces. The claim probably holds for other vector spaces in addition to Hilbert spaces, but, Hilbert spaces are a nice example. For example, the claim probably holds in the "Lp(R,mu)" spaces where R is the real numbers, mu is Lebesque measure, 1<p<infinity, and the "p" in "Lp" should be a superscript. For another example, the claim probably holds for at least some class of unbounded operators. Examples of bounded operators, A, on the infinite dimensional Hilbert space L2(R,mu) are given by integral operators.

    For a more sophisticated, and more intuitive, proof, apply either the spectral theorem of the Gelfand transform. The A becomes a function in Linfinity(X,mu) for some measure mu on the maximal ideal space, X, of the C* or W* algebra generated by A, operating by multiplication on L2(X,mu). This reduces the question to the case when H = the complex numbers, C. If lambda is a complex number operating by multiplication on other complex numbers by multiplication, then lambda* = the complex conjugate of lambda. So, this reduces the question to showing that (conj(lambda))(conj(lambda)) = conj(lambdaxlambda), and that is an easy computation:
    Let lambda = a+bi where i is the square root of -1.
    Then (conj(lambda))(conj(lambda)) = (a-ib)(a-ib) = aa-bb - 2abi.
    And, conj(lambdaxlambda) = conj((a+ib)(a+ib)) = conj(aa-bb + 2abi) = aa-bb - 2abi. QED.

    This second proof extends to some class of unbounded operators.
    Note that neither proof assumes that the Hilbert space is separable.

  5. Jun 22, 2008 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Not only is there no need to invoke separability of Hilbert spaces, but there is no need to even invoke Hilbert spaces at all, or boundedness of the operators or even for the notion of boundedness to make sense. The result is true whenever it makes sense, if you don't consider that a tautology (eg, adjoint functors).
  6. Jun 22, 2008 #5

    Good point.

    By the way, my second proof has a flaw. It only applies to normal operators because that is where you have the spectral theorem or the Gelfand transform, and these things are the same for C* algebras.

    In addition, I forgot to mention that for finite dimensional vector spaces, the second proof just amounts to diagonalizing the matrix.

    You said that "there is no need to invoke separability" and I certainly agree. I believe I claimed that the two proofs that I gave work for NON-separable Hilbert spaces as well as for separable ones.

    And you are right that the result holds in more even general spaces than I had realized when I wrote it. I was thinking that AA and A*A* operate in different spaces (sometimes - like especially if there is no inner product hanging around). But that thought was off-base because it is (AA)* and A*A* that are being compared and they operate on the same space. So, there is undoubtedly a third proof lying around somewhere - with a category theory flavor - that covers much more general spaces.

    Thanks for taking the time to respond to my post.

    Deacon John
  7. Jun 23, 2008 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    When I said "no need to invoke separability" I meant it in the sense of "no need to mention it at all - it's just extra information that it isn't necessary to know about in order to get the proof"
  8. Jun 23, 2008 #7
    Well I don't agree Matt. I think that it is more important to learn about mathematics than it is to answer specifi questions. In other words, I think questionsw like "What?" and "Why?" are more important than "How?," as in "How do you do this homework problem." In other words, the inclusion of additional information as "asides" in the investigation of a mathematical question should always be encouraged. Well, almost always. Like Serge Lang is reputed to have written in the introduction to his book on Elliptic Curves,

    "It is possible to write endlessly on the subject of elliptic curves. That is not a threat."

    I hope you, Matt, will one day be a great teacher of mathematics. That you will one day be one of those who "pass on the tourch of enlightenment" about our beautiful subject. I only did that professionally for a few years after I got my PdD in 1972, but, I have always done it on a small scall - on the "hobby" scale. On the other hand, the desire to do that for its own sake is stronger in the last few years.

    If you do join the honored ranks (honored by me anyway) of math teachers, there are a few things I would like you to remember.

    1) In general, aim your lectures to the top of the class.

    2) Remember, your job is to enable others to learn.

    3) Don't ever forget to pass on the tordh of passion for whatever love you have for or for whatever joy you find in the study of mathematics.

    Deacon John

    Deacon John
  9. Jun 23, 2008 #8

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure how choosing to answer the question in the most general way counts as being too specific.
  10. Jun 23, 2008 #9
    Matt, I love your passion for mathematics! I don't exactly understand what your last question is trying to ask.

    But, I'll take a shot anyway.

    Of course, trying to answer in the most general way is an excellent thing. Your comment about duality in general category theory was most enlightening and helful to me. Even though the questioneer had specified duality in relation to the complex numbers.

    On the other hand, mentioning specific examples and related concepts is also extremely valuable.

    Like von Neumann is reputed to have said to one of the students in one of his classes,

    "Young man, you don't understand mathematics, you just get used to it."

    The sooner a budding mathemetician gets to hear the names of advanced concepts the better. Even if he doesn't know what they are. Repetition helps. He or she might even become curious enough to look it up on Wikipedia. Before you know it, your generation will bring in the Riemann hypothesis for us!

    By the way, the interest in the Riemann hypothesis is NOT because it is so "difficult" or "deep" but raqther because one feelis that we are so close, that one "should" be able to settle it.

    There are plenty of more difficult questions that we have no idea how to even attack. Like the Mersenne prime question.

    Deacon John
  11. Jun 23, 2008 #10
    Sorry, Matt, talking about extraneous material, I can not help but add that the Mersenne prime question was the central question in the recent fantastic movie called "proof."

    I don't think many people know that. I had to listen closely and then do some research. I had forgotten about the Mersenne prime question.

    The "proof" in the movie "Proof" was either a proof that there are a finite number of Mersenne primes or a proof that there was an infinite number of Mersenne primes. I couldn't figure out which. Does anybody know?

    Of course, it was science fiction. No such proof is known today. That doesn't mean that one doesn't exist. The movie claimed the proof used the theory of random matricies. Now, that seems like a stretch to me! But, one never knows.

    Deacon John
  12. Jun 23, 2008 #11

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It is a stretch. I say this with the greatest respect to my friends, colleagues, and mentors who work on Random Matrices. But. Random matrices have not proved (directly, so far as I am aware) anything in number theory. What they have done is provide a useful way to translate ideas from applied mathematics to number theory and vice versa.
  13. Jun 24, 2008 #12
    I don't know anybody personally who works in that field, but, that was my impression from what I've read about them. DJ
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Adjoint multiplicity
  1. Adjoint Functors (Replies: 3)

  2. Adjoint operator (Replies: 8)

  3. Adjoint of an operator (Replies: 9)

  4. Adjoint operators (Replies: 10)