1. ## rank problem.....

Q:If A and B are 3 by 3 real matrices such that rank(AB)=1,then rank(BA) can't be
a)0
b)1
c)2
d)3

2. Hint: Think about determinants (det(BA) = det(AB)).

3. Originally Posted by Opalg
Hint: Think about determinants (det(BA) = det(AB)).
plz clarify it more...because when i use ur hint i got two answers (3,2)
but i have to select one.

4. Another big hint : if $\displaystyle rank (AB)=1$, $\displaystyle \det (AB)$ cannot be ...?

5. Originally Posted by arbolis
Another big hint : if $\displaystyle rank (AB)=1$, $\displaystyle \det (AB)$ cannot be ...?
it can't be zero...is it right??

6. Originally Posted by Mathventure
it can't be zero...is it right??
I thought so, but now I realize I'm confused and I think it's false. Disregard my post.
EDIT: If rank (AB) is one, there are 2 null rows in the equivalent reduced matrix of AB, right? I have a conclusion but I don't know if it's true. I'll wait someone to help us.
EDIT 2 : we were on the right track. det (AB)=0 since the its rank is 1. Hence det (BA) =0 and so it has at least a null row. What does it say about the rank?

7. Originally Posted by arbolis
I thought so, but now I realize I'm confused and I think it's false. Disregard my post.
its okay...