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?

Alright then: You can view every Matrix as a linear funktion which transforms you the $R^p$ in a sub-Vector-Space with the dimension of its rank. The union (I mean the Union of Vector Spaces which is equal to the Linar combination not to the union of the Sets) of those planes( as I will call those sub-Spaces even if they do not have to be planes in the regular way) has to be the full space because you need to have the Identity on this space which is equal to $I_p$ but because the sum of all the ranks has to be the p the sum of all dimensions has to be the dimension of the hole space. Therefor the pairwise Intersection of two different Planes has to be $\{0\}$.
Therefor if you transform your space in one plane and then in another the result will be the trivial sub-space which contains just 0. That means for every point x the result of the linear function will be 0. This is the constant function which can be represented by the 0-Matrix. The komposition of Transformations can be represented by a matrix-multiplication. therefor the pairwise multiplication of those Matrices has to be 0
That was a lot of text.... in a proof you'd have to do it a little more in a formal way. but I think this can be better understood than some weird bijections
