Traceless hermitian matrices form groups?

[TheIsingGuy]

1. is the set of nxn traceless hermitian matrices under addition a group?
2. is the set of nxn traceless hermitian matrices under multiplication a group?
3. is the set of nxn traceless non-hermitian matrices under addition a group?

question 1-I thought that traceless means trace=0 is this right? so what would the identity element be? it can't be the null matrix because it doesn't have an inverse, can anyone help? I haven't got around to the other questions but help is probably needed coz i don't like matrices

 

[TheIsingGuy]

I just realized in the first quesiton, the composition law is actually addition, so that makes the inverse of the identiy just putting a minus sign on all of its elements, which doesn't change the diagonal, which mean its still traceless, so it must be a group.

for the second question closure isn't satisfied , the third one I am not sure what to do...

 

[weejee]

non-hermitian matrices don't include the identity.

 

[TheIsingGuy]

non-hermitian matrices don't include the identity.

yes of course, thanks a lot

 

[Avodyne]

Well, first of all, the identity element for addition is the matrix of all zeroes, not the identity matrix. Of course, this is also hermitian. But "non-hermitian" is often supposed to mean "not necessarily hermitian" rather than "definitely not hermitian". The answer depends on which meaning is implied.

 

[weejee]

Well, first of all, the identity element for addition is the matrix of all zeroes, not the identity matrix. Of course, this is also hermitian.

That was what I meant.

But "non-hermitian" is often supposed to mean "not necessarily hermitian" rather than "definitely not hermitian". The answer depends on which meaning is implied.

"Not necessarily hermitian" just means all matrices. Then, there is no point in using such term.

To me it seems safe to consider "non-hermitian" as "definitely not hermitian".