For a set of symmetric matrices $A_i$ of order p, show that if the sum of their ranks is p, $A_iA_j=0$ Here's what I know.
Matrices $A_i$ for $i=1,...,k$ are all symmetric p by p matrices. 
$\sum\limits_{i=1}^k A_i = I_p$ where $I_p$ is the p by p identity matrix
$\sum\limits_{i=1}^k rank(A_i) = p$
With this, I have to find a way to show that for all $i \neq j$, $A_iA_j=0$. 
I assume this is solved by showing that $rank(A_iA_j)=0$ but every attempt that I've made to do that goes nowhere. Can anyone point me in the right direction?

Hint. Use the inequality
$$
\operatorname{rank}(A+B+C)
\le\operatorname{rank}(A+B)+\operatorname{rank}(C)
\le \operatorname{rank}(A)+\operatorname{rank}(B)+\operatorname{rank}(C)
$$
to show that $\operatorname{rank}(A_i+A_j)=\operatorname{rank}(A_i)+\operatorname{rank}(A_j)$ and in turn $\operatorname{col}(A_i)\cap\operatorname{col}(A_j)=0$ for every $i\ne j$. Now, if the matrices are real symmetric, their null spaces are orthogonal to their column spaces.
