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I have read two definitions of unitary.
A is unitary if:
#1: det(A) = 1
or
#2: AadjointA = I
Are these definitions equivalent?
A is unitary if:
#1: det(A) = 1
or
#2: AadjointA = I
Are these definitions equivalent?
What is the difference between "adjoint" and "conjugate transpose?"matt grime said:... you want to be careful with adjoint, better to specify that you mean A* the conjugate trasnpose.
What is "SL?"matt grime said:1 is sufficient for A to be SL
What is an "upper triangular" matrix? If it is what I think it is, then the determinant is just the product of the trace elements. This will not be a different value for the transpose of the matrix.matt grime said:there are of course non-unitary matrices of det 1. exercise find a simple upper triangular real (integer) matrix to show this.
This just seems a little strange to me. If the components of A are completely real-valued, then even though the condition were to specify that A is orthogonal, it would also be unitary. If the components were not all completely real-valued, are you saying that "adjoint" can strictly mean "transpose" in this case without complex conjugation?matt grime said:If you just say that the matrix A satisfies A^{adjoint}A=I, we do not know if you mean orthogonal or unitary, at least indicate that you mean complex matrices.
No problem. And, you are correct.matt grime said:Don't take this the wrong way, but ... I assume you've not done much maths, ...
I don't follow this. What is a "bilinear map?" I'm assuming that X and Y are like vectors and that T is like an operator?matt grime said:Let C be some space (what kind we won't say), with a bilinear map (?,?) on the elements of that space. THe (right) adjoint of a map T:C-->C is a map T* satisfying (TX,Y)=(X,T*Y)
Or if the matrices are quaternionic, then we have the symplectic groupmatt grime said:The main thing is if you ask:
What is the group of matrices described by the law A^{adjoint}A=I=AA^{adjoint}?
Then the answer is either the unitary group OR the orthogonal group depending on whether the matrices are real or complex.
You might be thinking of the group SU(n). the "S" in SU(n) means special linear, which means det(A) = 1. the "U" in SU(n) means unitary, which means A^adjoint*A=1 (where here, "adjoint" mean transpose conjugate)turin said:I have read two definitions of unitary.
A is unitary if:
#1: det(A) = 1
or
#2: AadjointA = I
Are these definitions equivalent?
Sorry. I'm probably the wrong guy for you to be helping, then.matt grime said:I was trying not to have to specify what any of the things is, ...
I think that it makes sense to me in this way. But I am still confused by T:C-->C. That is, it seems like T should be the identity in this case.matt grime said:T is a map from X to Y - that is what T:X--->Y means.
I definitely don't mind the nitpicking. Thanks for the help.matt grime said:It is nitpicking but if you don't make precise the terms you use then maths becomes hard.
The symbol * denotes complex conjugation, and {dagger} hermitean conjugation: the former applies to a number and the latter to an operator... When there is no risk of confusion I abuse the notation, using {dagger} when I should use *... For a matrix M, then of course M^{dagger} and M^* should be carefully distinguished from each other.
how about sesquilinear? i like that word.matt grime said:(I didn't mean to say bilinear, just binary ie two inputs).
you're too discriminating! you'll never cut it as a physicist!matt grime said:That's just evil, but as the quaternions aren't a field I don't feel too bad for not addressing them.
unitary : orthogonal : symplectic :: complex : real : quaternionic
my favorite notation convention (one that is, i think, shared by many physicists) is for [itex]z[/itex] and [itex]\overline{z}[/itex] to be your independent complex variables, and restrict yourself to the 2d plane where [itex]z^*=\overline{z}[/itex]matt grime said:No one will ever agree on notation it seems. In Hilbert space theory adjoint is often just written * because overline denotes complex conjugation (damn physicists mucking it up), and who on Earth wants to just take the conjugate of a matrix?
Well, I am interested, and I followed the link, but I did not get an explanation. I understood -17% (I now feel dumber). What is a quaternion? I've heard of it, and I've even seen people bring it up, but i have never understood it. Apparently, it is an extension of the complex field. But I thought that the complex field was as high as one needed to go in order to accomplish ... something, I forgot what it was called, something like "algebraic closure," maybe. So, anyway, can you give an algebraic excercise or something that would help us (me) understand quaternions, there purpose, etc.?lethe said:... in case anyone is interested see this page for an explanation ...
turin said:Well, I am interested, and I followed the link, but I did not get an explanation. I understood -17% (I now feel dumber). What is a quaternion? I've heard of it, and I've even seen people bring it up, but i have never understood it. Apparently, it is an extension of the complex field. But I thought that the complex field was as high as one needed to go in order to accomplish ... something, I forgot what it was called, something like "algebraic closure," maybe. So, anyway, can you give an algebraic excercise or something that would help us (me) understand quaternions, there purpose, etc.?
do you know the difference between a group and an algebra? an algebra has addition, subtraction, and multiplication (here, we also have division, so it's almost a field). a group has only multiplication and inversion.turin said:Could I say, with any sort of meaning, that the Pauli matrices (with the 2x2 identity) are a realization of the quaternions? I'm trying to learn this group theory stuff, and there are some suspicious similarities here.
How would one have a matrix with quaternion elements, which are themselves matrices apparently? Can you give an example? Would it be a matrix of matrices? Would the vectors on which it operates also need matrix components?
turin said:Excellent, lethe! I just wish I new that I was appreciating the isomorphism to the complexes. I checked commutability. What else should I check to convince myself? Oh ya, I just checked inversion.
i am not sure where, in your previous post, you would like to insert the word "representation"BTW, I think I meant representation in my previous post. I haven't figured out what is the difference.
In place of "realization."lethe said:i am not sure where, in your previous post, you would like to insert the word "representation"
A unitary matrix is a square matrix whose conjugate transpose is equal to its inverse. In other words, it satisfies the condition U*U^H = I, where U* is the conjugate transpose of U and I is the identity matrix.
For a unitary matrix U, the determinant of U (det(U)) is always equal to 1. Additionally, the adjoint of U (adj(U)) is equal to the inverse of U (U^-1).
Yes, since the determinant of a unitary matrix is always equal to 1 and the adjoint is equal to the inverse, the determinants of a unitary matrix and its adjoint are always equivalent.
If det(A) = 1, then adj(A) = I. This means that the adjoint of a unitary matrix A is equal to the identity matrix if and only if the determinant of A is equal to 1.
No, a non-unitary matrix cannot have both a determinant of 1 and an adjoint equal to the identity matrix. This is because for a non-unitary matrix, the determinant can be any non-zero complex number, while the adjoint is only equal to the inverse for a unitary matrix.