In a regular rooted tree graph with $k$-fold branching, since there are no edges connecting vertices within each sphere, then $L_S = 0$ so that $$\label{eq3,5}
L = \Pi + \Pi^\ast.$$ To diagonalize the operator $L$ given by , invariant subspaces $M_n$, $n \geq 0$, for $\Pi$ are firstly constructed in [@af]. More precisely, we have the following lemma:

[[@af Lemma 1]]{}\[l3,1\] The Hilbert space $\ell^2(\mathcal{V})$ can be decomposed as an orthogonal direct sum $$\label{eq3,6}
\ell^2(\mathcal{V}) = \bigoplus_{n = 0}^\infty M_n = 
\bigoplus_{n = 0}^\infty \bigoplus_{r = n}^\infty Q_{n,r},$$ where the subspaces $M_n$ are $L$-invariant.

By construction in this Lemma \[l3,1\], we have $Q_{0,0} := \ell^2(S_0)$ and $\ell^2(S_r) = \bigoplus_{\ell = 0}^r Q_{\ell,r}$. A schematic interpretation yields a triangular diagram as in Figure 4.1 below.

\[fig 2\]

![Orthogonal subspace decomposition](tree_2.png)

According to Lemma \[l3,1\], for any $n \geq 0$, the subspace $M_n$ is invariant for the operator $L$. Thus, $L$ can be decomposed as $$\label{eq3,60}
L = \bigoplus_{n = 0}^{\infty} L_n,$$ the operators $L_n$, $n \geq 0$, being the restriction of $L$ to $M_n$. So, in order to diagonalize the operator $L$, it suffices to do it for each operator $L_n$ for $n \geq 0$. Consider a vector $\phi \in 