Let $G$ be a graph which has an edge decomposition into $r$ complete bipartite graphs. $A$ is the adjacency matrix of $G$, then prove $r\geq n$, where $n$ is the number of positive eigenvalues of $A$.
Here is the answer: Let $u_i$ and $v_i$ be the characteristic vectors of both sides of a bipartition of the $i$-th complete bipartite graph. Then that graph has adjacency matrix $D_i = u_iv_i^T+v_iu_i^T$ , and $A = \sum D_i$. Let $w$ be a vector orthogonal to all $ui$. Then $w^TAw = 0$ and it follows that $w$ cannot be chosen in the span of eigenvectors of $A$ with positive eigenvalue.
I don't know what does the characteristic vectors mean, but I guess if the vertex set of the $i$-th complete bipartite graph is $\{1,2\}\cup \{3,4\}$, then $u_i$ should be $(1,1,0,\cdots,0)^T$and $v_i=(0,0,1,1,0,\cdots,0)^T$.
I am having confusion in understanding how can we get $r\geq n$ from $w^TAw = 0$?
I have solve it. Let $V_1 $ be the orthocomplement space of $span<u_1,\cdots,u_r>$ and $V_2 $ be the the span of eigenvectors of A with positive eigenvalues. Thus we get $\dim V_1\geq\dim V-r$ and $\dim V_2=n$. Use the dimension formula :$\dim V_1+\dim V_2=\dim(V_1+V_2)+\dim(V_1\cap V_2)$, then we get: $\dim V-r+n\leq \dim V+0$.