$A,B$ Matrices $5{\times}5$, $p(B)<p(A)$
Solution space of $Bx=0$: $Sp\{(1,-2,5,0,1),(4,0,-3,4,1)\}$
Prove: $BA \neq 0$.
*$p(A) = \operatorname{rank}(A)$
My attempt: (stuck in the middle)
We know that:
$\#n-p(B)=\dim P$
Where:
#n - num of varaibles
$p(B) - \operatorname{rank} B$
$\dim P$ - the dimension of the solution set for $Bx = 0$
Therefore, we conclude: $p(B) = 3$
$\Rightarrow 3<p(A) \leq 5 = 4 \ or \ 5$
I though maybe i could conclude somehow that $A$ is invertible ,therefore $BA \neq 0$, but could'nt make it. (The rank(A) may be 4).
hint?
*Edit:
As suggested by the answer i will write the way i understood it for people that will look at this question in the future.
If $p(A) = 5$, therefore A invertible, therefore for every B square matrix $5x5$: $BA \neq 0$ (invertible matrix proteprties).
If $p(A) = 4$, We will assume $BA = 0$ and get to a contradiction.
If $p(A) = 4$, there are 4 independent colums in A.
Choose 4 different vectors$x_1,x_2,x_3,x_4$, such that $Ax_1$ represent the first independent column, $Ax_2$ the second independent and so on.
We assume $BA = 0$, therefore, multiply from the right by $x_1$ get: $BAx_1 = 0$ its a unique, independent solution for $Bx = 0$.
Again, take $BA = 0$, multiply from the right by $x_2$ we get: $BAx_2 = 0$, its another independent solution for $B_x = 0$
And so on.
We get therefore 4 independent solutions for $Bx = 0$.
But we know that the dimension of the solution set of $Bx = 0$ is 2.
Contradiction.
$BA \neq 0$.
Thanks for the help.
If rank of $A$ is $5$ and $BA=0$, it implies that $A$ is invertible and hence $B=0A^{-1}=0$, which is a contradiction, because nullity of $B$ is not $5$.
If rank of $A$ is $4$, then
$$B(Ax)=0\forall x\in\Bbb R^5$$ Hence, nullity of $B$ is atleast the rank of $A$, which is a contradiction (because kernel of $B$ contains atleast as many linearly independent vectors, as in the column space of $A$).