Linear Algebra - Ranks and dimension

[BitterX]

Homework Statement

given two matrices A, B of 5x5 order:
\rho(A)<\rho(B)

the nullspace for A is Sp{(0, 3, -1, 2, 1),(4, -2, 1, 4, 0)}
prove that
AB\neq 0

Homework Equations

\rho(A)=n- \rho(P_a)

The Attempt at a Solution

so I now that the nullity is 2 and so \rho(A)=3
and because of that 4\leq \rho(B)\leq 5

but I don't know how to prove that AB\neq 0

 

[micromass]

Can you find an x such that ABx\neq 0?? That is: can you find an x such that Bx is not contained in the nullspace of A?

 

[BitterX]

Thanks for the reply
the null space of A is Sp[(1,0,0,0,0),(0,1,0,0,0)]
the nullity of B is 1 but it can be either Sp(1,0,0,0,0) or Sp(0,0,0,0,1) etc. or am I wrong?
I guess I am because this way it can be contained in the null space of A or not, and there's no way to determine. let's say that the null space of B is Sp(0,0,1,0,0) so it's not contained in the null space of A,

and let's call it Bx=(0,0,s,0,0) |s \in \mathbb{F}
so I just say that

ABx = (0,0,a_{33}s,a_{43}s,a_{53}s)

and if AB was 0 for every x it would have been (0,0,0,0,0)?
if so how can I determine if Bx is contained in Ax or not?
Thanks again!

 

[micromass]

the null space of A is Sp[(1,0,0,0,0),(0,1,0,0,0)]

No, it's not.

the nullity of B is 1

No, it's not.

What made you think that?

 

[BitterX]

I guess I'm wrong, let me tell you what my train of thought was:

I took the original null space given, put it in a matrix and got to that, that's the basis of the
null space.
so the nullity is 2, and then rank(A) is 3 (because it's a 5x5 matrix)
because rank(B)>rank(A) , rank(B) is either 4 or 5. and it's nullity is 1 or 0.

I thought about a different way to prove it, but I need to prove that nullspace A, nullspace B,
are subspaces of nullspace AB.

if so: nullity(AB)=nullity(A)+nullity(B)-nullity(A\cap B)

which means:
nullity(AB)\leq nullity(A)+nullity(B)

because we know the nullity of both A and B (assuming I'm right)
nullity(AB)\leq 3
using the rank nullity theorem:

\rho(AB)=n-nullity(AB)

\rho(AB) \geq 2

what do you think?