An algebraic method for a graph decomposition problem

Let $K_{n}$ denote an $n$-clique and $K_{n,m}$ denote a complete bipartite graph. We are interested in the least number $m$, denoted as $m(n)$, such that $K_{n}$ can be edge-decomposed into $m$ complete bipartite graphs. Observe that $m(1)=0$, $m(2)=1$, $m(3)=2$, $m(4)=3$, etc. It is not difficult to deduce $m(n)\le n-1$ after some case studies, since we can obtain a decomposition of $K_{n+1}$ by adding $K_{1,n}$ to a decomposition of $K_{n}$, which implies $m(n+1)\le m(n)+1$. The nontrivial part lies in the reverse direction: how can we prove a strict lowerbound for $m(n)$? If we want to establish the lowerbound by analyzing the graphs directly, we have to show for all $K_n$ that there is no better decomposition than the one we offer, and this is far from an easy job.

R.$~$Graham proposed a very clever method to establish $m(n)\ge n-1$ based on elementary knowledge from linear algebra. He used $n$ variables $x_{1},...,x_{n}$ to represent the $n$ vertices $v_{1},...,v_{n}$ of $K_{n}$. An edge $\overline{v_{i}v_{j}}$ is selected iff $x_{i},x_{j}$ are both nonzero. The key observation is: if $(A_{1},B_{1}),...,(A_{m},B_{m})$ is a decomposition of $K_{n}$ into complete bipartite graphs, then the equation \[ \sum_{1\le i < j\le n}x_{i}x_{j}=\sum_{k=1}^{m}\sum_{x\in A_{k}}\sum_{y\in B_{k}}xy=\sum_{k=1}^{m}\left(\sum_{x\in A_{k}}x\right)\left(\sum_{y\in B_{k}}y\right) \] must holds for any assignment of $x_{1},...,x_{m}$. (An equation like this is called a double counting in the buzzwords of Combinatorics.)

With the observation in mind, he considered the following homogeneous system of linear equations: \[ \begin{cases} \sum_{i=1}^{n}x_{i}=0,\\ \sum_{x\in A_{k}}x=0, & k=1,...,m \end{cases} \] Now suppose that $m < n-1$. The system then has $n$ unknowns and $m+1 < n$ equations. This implies that the system has an infinitely number of solutions. In particular, there exists a nontrivial assignment for $x_{1},...,x_{n}$ such that \[ 2\cdot\sum_{1\le i < j\le n}x_{i}x_{j}=\left(\sum_{i=1}^{n}x_{i}\right)^{2}- \sum_{i=1}^{n}x_{i}^{2}=-\sum_{i=1}^{n}x_{i}^{2}=0, \] which is impossible since the last equality enforces $x_{1}=...=x_{n}=0$. It follows that $m(n)\ge n-1$ by contradiction. Hence, we can conclude $m(n)=n-1$.