If $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$? 
Is it true that if $A$ and $B$ are $2\times 2$ matrices and $AB=0$ then $A=0$ or $B=0$. Prove it, or prove the contrary.  

I tried saying that if:
$$A= \begin{pmatrix}
        0 & 0  \\
        0 & 0  \\
        \end{pmatrix}\quad\text{and}\quad B= \begin{pmatrix}
        e & f  \\
        g & h  \\
        \end{pmatrix}.$$
The product will be 0. But, if $A$ is non zero, i tried doing something like this:
$$A= \begin{pmatrix}
        a & b  \\
        c & d  \\
        \end{pmatrix},\quad
    B = \begin{pmatrix}
        e & f  \\
        g & h  \\
        \end{pmatrix}
\implies
AB = \begin{pmatrix}
        ae+bg & af+bh  \\
        ce+dg & cf+dh  \\
        \end{pmatrix}$$
is equal to \begin{pmatrix}
        0 & 0  \\
        0 & 0  \\
        \end{pmatrix}
But then, how should i proceed? I don't know if these are the right steps to follow, i've tried searching, but i didn't find the right examples.
 A: For example you may try with
$$A=B=\begin{bmatrix}
        0 & 1  \\
        0 & 0  \\
        \end{bmatrix}.$$
A: Note that your own calculation shows that if $a=g=1$ but all other variables vanish, then the product $AB$ will vanish, too. (There are other possibilities. Robert's solution is $b=f=1$.)

You may be interested to know that the failure of $AB=0$ to imply $A=0$ or $B=0$ when $A$ and $B$ are matrices means that "matrix algebras" have "zero divisors." This is one reason matrix algebra is harder than ordinary algebra.
The great master Halmos in Finite-Dimensional Vector Spaces, for example:

[...] the two nasty properties that the multiplication of linear transformations has [are] non-commutativity and the existence of divisors of zero.

A: The answer to your question is "no". 
If the first matrix has rank $1$ it annihilates a subspace of dimension $1$. All you need then is for a second matrix of rank $1$ to annihilate the subspace that the first preserves, and preserve the subspace that it annihilates (because once the first matrix annihilates it, it is gone).
You get a simple set of matrices if you choose a sensible basis (as others have done).
You are working in two dimensions here. I'm hoping that my comments will help you to find examples in more dimensions, and to build some intuition and understanding rather than just doing calculations.
