TITLE: Problem on bipartite graphs.
QUESTION [13 upvotes]: Let's $G$ is bipartite graph with sides $X=\{v_1,v_2,\ldots,v_n\}$ and $Y=\{u_1,u_2,\ldots,u_n\}.$ Let $|N_G(v_i)|=k$ and $|N_G(v_i)\cap N_G(v_j)|=d\lt k~,(i\ne j)$. Prove that $|N_G(u_i)|=k$ and $|N_G(u_i)\cap N_G(u_j)|=d,~(i\ne j)$. $N_G(v)$ is neighborhood of vertex $v$ in graph $G$.
I have no idea how I can start the proof.

REPLY [2 votes]: As I promised here is the solution.
Let's $A=(a_{ij})_{i,j=1}^n$ be the following matrix: $a_{ij}=1,$ if $\{v_i,u_j\}$ is edge of graph $G$ and $a_{ij}=0,$ otherwise. From the conditions it follows that $AA^T=dB+(k-d)I,$ where $B$ is a matrix with all entries equal $1$. It is easy to check (by finding determinant or by proving that columns independent) that $det(AA^T)\ne0,$ therefore $det(A)\ne0$. So we have $$A^TA=A^{-1}(AA^T)A=dA^{-1}BA+(k-d)I.$$ Now to complete the proof we need to prove that $A^{-1}BA=B,$ that is, $AB=BA$. Let $e^T=(1,1,\ldots,1)$, then $e$ is an eignvector for $A$ (therefore for $A^{-1}$ too) and also for $AA^T$. So $e$ is an eignvector for $A^{-1}(AA^T)=A^T$, $A^Te=k'e.$ As $k'n=e^TAe=kn$, $k'=k$. Therefore $eA=ke~\Rightarrow~BA=kB=AB.$