Matrix Multiplication and Rank? I'm having trouble with this question: 

Is it possible for $A$ and $B$ to be $3\times3$ rank $2$ matrices with $AB = 0$? Prove your
      answer.

I searched the internet and found this: Rank (A) = Rank (AB)
So since Rank (A) = Rank (B) = 2 it follows that this is false since Rank (AB) has to be 2 as well? And a 0 matrix won't have a rank? Somehow I feel this is not the correct explanation, and the answer to this question talks about some column space and null space of the matrix but I can't figure out what it means.
Any help is greatly apreciated!
 A: Use the Rank-Nullity theorem. If you don't get an opening, use the following hint: 
Hint: Let's assume $A,B$ are $n\times n$ matrices whose product $AB=0$.
Considered as matrix $A$ acting on the columns of $B$, the nullspace of $A$ has to contain the column space of $B$, in order to get to a zero product.
That means the nullity of $A$ (dimension of its nullspace) has to be at least the (column) rank of $B$.
Specialize to the data in your problem, where $A$ and $B$ have rank $2$, and where by Rank-Nullity Theorem, the nullity of $A$ is ... ? Hope you get it after this.
A: If $A$ is of rank $2$, which means it is not full-rank, then, its columns are linearly dependent. For example, if 
$A=
 \begin{bmatrix}
    A_{1} & A_{2} & A_{2} \end{bmatrix}$
then you can find coefficients $\alpha_i$, such that
$\alpha_1A_1+\alpha_2A_2+\alpha_3A_3=0$
The set of all vectors $[\alpha_1, \alpha_2, \alpha_3]^T$ that gives us the above equality, is the null-space of $A$. As $A$ is of rank $2$, then the null-space is one-dimentional. So, all other such vectors are the scaled versions of the first one. 
Now, assume we want to make a matrix $B$ of rank $2$ that gives $AB=0$. All columns of $B$ should be of the form $\beta [\alpha_1, \alpha_2, \alpha_3]^T$, where $\beta$ is a constant. Then, you can see that $B$ is of rank $1$, as all the columns are a multiple of the first column.
