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if A* is the adjoint of A in Complx, Then is A* x A* = (A^2)* or something else??

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if A* is the adjoint of A in Complx, Then is A* x A* = (A^2)* or something else??

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morphism

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Have you given this any thought?

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Nice Question, mindOnmath.

Let A->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.

DeaconJohn

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matt grime

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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

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matt grime

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"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

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matt grime

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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

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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

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matt grime

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