Prove or disprove: if AB=0 or BA=0 with rank(A)=n-1, then rank(B)=1

[daniel_i_l]

Homework Statement

A and B are nxn matricies. B =/= 0 and rank(A)=n-1
Prove or disprove that:
1) if AB=0 then rank(B)=1
2) if BA=0 then rank(B)=1

Homework Equations

The Attempt at a Solution

If n=2 and we define A as:
[1 1]
[1 1]
and B as:
[1 -1
[-1 1]
then rank(A) = 1 = 2-1 and B=/=0 but AB=BA=0 and rank(B)=2 whick disproves both (1) and (2). Is that right? It seems odd that they'd give to similar questions with identical answers.

 

[radou]

What makes you think that r(B) = 2?

 

[daniel_i_l]

Oh ... rank(B) = 1 also... I guess I have to think about it somemore.
Thanks.

 

[daniel_i_l]

Ok, now I think that both of the statements are right:
1) Since the rank of A is n-1 then the dimension to the space of all the solutions to Ax=0 is
n - rank(A) = n-n+1=1. Since AB=0 then all the coloums of B are solution to Ax=0. But the dimension of that space is 1 so there cannot be more that 1 linearly independent vectors in the coloums of B so rank(B)<=1 but we know that rank(B)>0 so rank(B)=1

2)If we look at the equation BA=0 then the coloums of A are solutions to Bx=0. So the dimension of the space of the solutions to Bx=0 (P) has to be bigger or equal than rank of A:
dim(P) >= rank(A) --> n-rank(B) >= n-1 --> rank(B) <= 1
But rank(B)>0 so rank(B)=1

Are those answers right?
Thanks.

 

[pizzasky]

Looks good to me!