# Linear Algebra - Ranks and dimension

1. Jan 6, 2012

### BitterX

1. The problem statement, all variables and given/known data
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
$AB\neq 0$

2. Relevant equations
$\rho(A)=n- \rho(P_a)$

3. 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 $AB\neq 0$

2. Jan 6, 2012

### micromass

Staff Emeritus
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?

3. Jan 7, 2012

### BitterX

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!

4. Jan 7, 2012

### micromass

Staff Emeritus
No, it's not.

No, it's not.

5. Jan 7, 2012

### 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?