This is a very silly question but I can't seem to figure it out.
Let $G$ be the graph with 6 nodes and 3 edges where 1 is connected to 4, 2 to 5 and 3 to 6. Le $ U = \{1,2,3\}$ and $V=\{4,5,6\}$. Then each edge connects one node in $U$ with one node in $V$. It's adjacency matrix is:
$$ \begin{pmatrix}
0 & 0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 0 & 1\\
1 & 0 & 0 & 0 & 0 & 0\\
0 & 1 & 0 & 0 & 0 & 0\\
0 & 0 & 1 & 0 & 0 & 0\\
\end{pmatrix} $$
Which to me looks like a bipartite adjacency matrix but of course $A^2 = Id$ so $A^k = Id$ for all $k \geq 2$ which contradicts that $A^{2n+1}$ has 0s in the diagonal for all n if A is the adjacency matrix of a bipartite graph.
Any help is greatly appreciated, I seem to be missing a very basic thing.
Best Answer
By $A^2=I,$ you won't have $A^k=I,$ instead you have $A^{2n}=I,$ $A^{2n+1}=A.$ This won't contradict the proposition about bipartite graph.