Suppose I have a Graph G(V,E) and A is the adjacency matrix of G. The graph thus formed by creating a GP series of adjacency matrix
$P=A+A^2+A^3+....+A^{n-1}$
Is this graph Eulerian?
A is such that diagonal elements are 0. I was thinking of the fact that P has multiple edges between a pair of vertices. So, P(i,j) = total no. of paths from i to j. So if I can somehow prove degree of all vertices = even, then it will be proved. But how to proceed. Please help.
If $G$ is $k$-regular $k$ odd then for odd integers $n$ the resulting graph $P$ will have odd degree (but of course may not be simple and $P$ may also have self-loops) for $n$ odd. So in general, no.
Even if you eliminate duplicate edges and self-loops in $P$, let $G$ be a $k$-regular graph w no small cycles--they exist and there are even explicit constructions in the literature. Then if $k$ is odd, then for each $n$ sufficiently small (less than half the length of the shortest cycle) there is an odd number of vertices $k + \sum_{i=1}^{n} k(k-1)^i$ within distance $n$ of each vertex in $P$.
And if you do decide to allow one self-loop then you could take $G$ as in the above paragraph fix a vertex $v$ and remove an edge of say distance $n$ from $v$ so that $v$ has odd degree in the resulting $P$.