# Rank(A)= Trace(AA*) ?

1. Jun 21, 2011

### arthurhenry

Rank(A)= Trace(AA*) ??

I have two questions and I hope it is acceptable...Seemingly unrelated, though I came to wonder about the first while thinking the second. Thanks

1)Is this statement true? or is there a statement that relates Rank(A) and Trace(??)

2) AB-BA=I (When does this identity hold if at all? Field can be closed or Z/Z2, or etc)

Last edited: Jun 21, 2011
2. Jun 21, 2011

### micromass

Re: Rank(A)= Trace(AA*) ??

Hi arthurhenry!

The first one is not true. It is true that Rank(A)=Rank(AA*). But it isn't in general true that Rank(AA*)=Trace(AA*). For example, take

$$A=\left(\begin{array}{cc} 2 & 0\\ 0 & 2\\ \end{array}\right)$$

Then AA* has rank 2, but the trace is 8.

The second one is not true too. Take A the zero matrix and B an arbitrary matrix.

3. Jun 21, 2011

### arthurhenry

Re: Rank(A)= Trace(AA*) ??

Perhaps I was not clear, I will phrase it correctly:

2)Does there exist matrices A and B such that AB-BA=I holds?

In particular, what is the answer to the question in the case the field is Complex NUmbers and in the case the field is Z/Z2 ?

1) Is rank(A) equal to Trace(A*A) ?

not "is the rank(A*A) equal to Trace (A*A)?" As you have pointed out this one is definittely incorrect.

4. Jun 21, 2011

### micromass

Re: Rank(A)= Trace(AA*) ??

Well, try it yourself. Take general 2x2-matrices and calculate AB-BA. Then solve the system to see whether they can equal I...

We have that rank(A)=rank(A*A), so the same example applies.