def. of unitary

**turin** · Apr12-04, 03:18 PM

I have read two definitions of unitary.
A is unitary if:
#1: det(A) = 1
or
#2: A^adjointA = I

Are these definitions equivalent?

**matt grime** · Apr12-04, 03:24 PM

Absolutely not. And you want to be careful with adjoint, better to specify that you mean A* the conjugate trasnpose.

1 is sufficient for A to be SL

2 says, with adjoint understood as above it is unitary.

there are of course non-unitary matrices of det 1. exercise find a simple upper triangular real (integer) matrix to show this.

**turin** · Apr12-04, 05:20 PM

Originally Posted by matt grime

... you want to be careful with adjoint, better to specify that you mean A* the conjugate trasnpose.

What is the difference between "adjoint" and "conjugate transpose?"

Originally Posted by matt grime

1 is sufficient for A to be SL

What is "SL?"

Originally Posted by matt grime

there are of course non-unitary matrices of det 1. exercise find a simple upper triangular real (integer) matrix to show this.

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** · Apr12-04, 05:26 PM

Adjoint is context dependent. if the matrices are real then adjoint is just transpose - part of the definition for unitary is that it is complex, and preserves an Hermitian form (which is often how you'll see it defined). Just writing adjoint without specifying we mean complex at some point is omitting some information. One could argue that A^{adjoint}A =I defines the orthogonal (real) matrices.

SL(F) is the special linear group - the group of matrices with determinant 1 (with entries in the field F)

upper triangular means all entries below the main diagonal are zero.

You can do the exerices with a diagonal matrix too, though obviously not one with integer coefficients.

**turin** · Apr12-04, 06:10 PM

matt grime,
I did not understand the point you were trying to make about "adjoint." If the matrices are real, then the adjoint and transpose are the same thing, I get that. I don't see how this makes "adjoint" and "transpose conjugate" different (since the conjugate of a real number is itself).

**matt grime** · Apr12-04, 06:17 PM

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.

**turin** · Apr12-04, 06:26 PM

Originally Posted by matt grime

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.

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?

Please don't think I'm trying to be difficult. I was just taught that "adjoint" always means "transpose" and "complex conjugate" and I would like to be aware of any variation in usage for the sake of future conversations with the math people.

**matt grime** · Apr13-04, 05:27 AM

That is waht it means if the matirx is complex. Don't take this the wrong way, but as you asked if unitary meant det is 1 I assume you've not done much maths, thus I want to see that you understand that unitarity is a property of compelx matrices, and just saying adjoint doesn't do that.

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)

it occurs in many places, in the case of a complex vector space it means transpose conjugate and the bilinear map is the natural Hermitian inner product.

**turin** · Apr13-04, 01:24 PM

Originally Posted by matt grime

Don't take this the wrong way, but ... I assume you've not done much maths, ...

No problem. And, you are correct.

Originally Posted by matt grime

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)

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?

I don't understand the following notation

T:C-->C

**matt grime** · Apr13-04, 01:45 PM

I was trying not to have to specify what any of the things is, but C is some "space" X and Y are some elements of the space (it needn't be a vector space with vectors just a couple of sets with elements) and, T is a map from X to Y - that is what T:C--->C means.

(?,?) is a just some way of paring up elements of C to get something in some other set, such as the real or complex numbers. (I didn't mean to say bilinear, just binary ie two inputs).

*The* model for this is, yes, vector spaces (over R or C, this C isn't the other C in the post unfortunately) and vectors and linear maps and the inner product (or Hermitian product), but there are other places where adjoint gets used.

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. One is a subgroup of the other. It is nitpicking but if you don't make precise the terms you use then maths becomes hard.

**lethe** · Apr13-04, 01:54 PM

Originally Posted by matt grime

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.

Or if the matrices are quaternionic, then we have the symplectic group

**lethe** · Apr13-04, 01:57 PM

Originally Posted by turin

I have read two definitions of unitary.
A is unitary if:
#1: det(A) = 1
or
#2: A^adjointA = I

Are these definitions equivalent?

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)

so a matrix in SU(n) satisfies both of your conditions. a matrix that satisfies only condition #1 (but not necessarily condition #2) is in SL(n,C), SL means special linear.

if it satisfies condition #2 (but not necessarily condition #1) then the matrix is in U(n), for unitary.

then the intersection of SL(n,C) and U(n) is SU(n). matrices that satisfy both condition #1 and #2

**matt grime** · Apr13-04, 01:58 PM

That's just evil, but as the quaternions aren't a field I don't feel too bad for not addressing them.

**turin** · Apr13-04, 02:40 PM

Originally Posted by matt grime

I was trying not to have to specify what any of the things is, ...

Sorry. I'm probably the wrong guy for you to be helping, then.

Originally Posted by matt grime

T is a map from X to Y - that is what T:X--->Y means.

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.

Originally Posted by matt grime

It is nitpicking but if you don't make precise the terms you use then maths becomes hard.

I definitely don't mind the nitpicking. Thanks for the help.

**turin** · Apr13-04, 02:41 PM

lethe,
What does the C mean in SL(n,C)?

**Janitor** · Apr13-04, 03:03 PM

Here is something that Zee says in his book on Q.F.T.

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.