How can one show that the Euclidean norm of adjacency matrix $A$ of a Tree graph of order $n$ is given by $\sqrt{2} \sqrt{n-1}$?
I think the following hint can work: $\Vert A \Vert = \sqrt{ \displaystyle \sum_{ i \le n } \sum _{ i \le n } \vert (A)_{ij} \vert} =\sqrt{\text{trace} (AA^T)}$
If $A$ is adjacency matrix, then $A^T=A$. etc.
I am not very sure after this point.
Thanks before hand
If you mean by "of order $n$" that your tree has $n$ vertices, then observe the following. Clearly, your tree has $n-1$ edges. The adjacency matrix consists of $0$'s and $1$'s. More precisely, $A_{i,j} = 1$ if and only if node $i$ and node $j$ are connected by an edge. Further, observe that your matrix is symmetric, i.e. $A_{i,j} = A_{j,i}$. Each edge must appear twice in the adjacency matrix. $A$ must thus contain exactly $2(n-1)$ many $1$'s.
Hence: $$\lVert A \rVert = \sqrt{2(n-1)}$$