On a linearly independent set of matrices whose pairwise products are zero Is the following statement true/false?

Suppose that $A_1,A_2,\ldots ,A_m$ are $m$ linearly independent real $n\times n$ matrices such that $A_iA_j=0$ for all $i\neq j$. Then $m\le n$.

I tried to find counter-example for $n=2$. I tried using the matrices
$$
\begin{bmatrix} 1 & 0\\ 0& 0\end{bmatrix},
\ \begin{bmatrix} 0 & 0\\ 1& 0\end{bmatrix},
\ \begin{bmatrix} 0 & 0\\ 0& 1\end{bmatrix}
$$
but they are not working. All the examples I have done are not working.
Is the result true? Please help.
 A: We do not necessarily have $m \leq n$, although you (probably) need $n$ to be a little larger than $2$ to make it work.
Fix an $n$ and consider the set of $n\times n$ matrices which are zero everywhere except for the entries which are closer to the top right corner than to the diagonal (I will call these the "free entries"). So for $n = 2$ or $3$, that means just the top right corner, for $n = 4$ or $n = 5$ it means the top right corner and the two elements next to the top right corner, and so on. The product of any two such matrices is necessarily $0$, and the number of free entries in such a martix is
$$
\frac{\left\lfloor\dfrac n2\right\rfloor\left(\left\lfloor\dfrac n2\right\rfloor + 1\right)}{2}
$$We see that $8$ is the first $n$ for which there are more than $n$ free entries ($10$). This means that we can make $m = 10$ by, for instance, letting $A_i$ have all zeroes except one of the free entries, which is $1$. Then clearly they are all linearly independent.
I don't know whether this problem is impossible for $n = 7$, or anything lower, but at least it is possible for $n = 8$
A: The lower bound in Arthur's answer can be sharpened. Since
$$
\pmatrix{
0_{k\times k}&B_{k\times(n-k)}\\
0_{(n-k)\times k}&0_{(n-k)\times(n-k)}}^2=0,
$$
we can always find a linearly independent set of $m=\lfloor\frac n2\rfloor\times\lceil\frac n2\rceil$ square matrices of size $n$ whose pairwise products are zero. So, we can have $m>n$ for every $n\ge5$. E.g. for $n=5$, we can choose $A_j$ by setting all but one asterisks below to zero:
$$
\pmatrix{0&0&\ast&\ast&\ast\\
0&0&\ast&\ast&\ast\\
0&0&0&0&0\\
0&0&0&0&0\\
0&0&0&0&0}.
$$
A: The statement is true under the additional condition that the matrices be diagonalizable (note that the counterexamples in the existing answers involve matrices that are not diagonalizable). For it's well-known that matrices that are diagonalizable and commute are simultaneously diagonalizable. So, take $P$ such that $P^{-1}A_iP=D_i$ is diagonal for all $i$. Then for $i\ne j$, $$D_iD_j=P^{-1}A_iPP^{-1}A_jP=P^{-1}A_iA_jP=P^{-1}0P=0$$ so $D_i$ and $D_j$ can't have any common nonzero entries. So each diagonal location can have a nonzero entry in at most one of the $D_i$, so there are at most $n$ of the $D_i$, and we're done. 
This is best possible, as you can see by letting $A_i$ be the matrix with zero everywhere except in the $(i,i)$ position. 
