An eigenvalue $\lambda_{1} $ is said to be a main eigenvalue if it has an associated eigenvector $x_{1} $ whose sum of entries is nonzero, in other words the projection of $e $, the all ones vector, onto the eigenspace $W_{\lambda_{1}}$ corresponding to $\lambda_{1} $ should be nonzero.
A graph $G $ is $k $-regular if each of its vertices is of degree $k $.
It is known that a graph $G$ on $n$ vertices is said to be $k$-regular graph if and only if its corresponding adjacency matrix has exactly one main eigenvalue.
Now I'm trying to prove the sufficient condition, i.e. a graph $G $ with one main eigenvalue is $k $-regular. I notice that the projection of $e $ onto the rest of the eigenspaces other than $W_{\lambda_{1}} $ is zero (since there exists only one maineigenvalue). Moreover, $e$ can be written as a linear combination of the basis of the eigenspaces corresponding to the $\lambda_{1},\lambda_{2},...,\lambda_{n} $ eigenvalues of $G $, hence, $e= proj_{W_{1}=span (x_{1})}(e)x_{1}= c x_{1}$ with c a nonzero constant. So $G $ is regular.
Is my proof true? Thank you
My explicit proof:
Let $\{\lambda_{1}^{(m_{1})}, \lambda_{2}^{(m_{2})},..., \lambda_{s}^{(m_{s})}\}$ be the spectrum of $G$ counting multiplicities, and for $j=1, 2,..., s$, let $\{w_{j1},w_{j2},...,w_{jm_{j}} \}$ be the eigenvectors corersponding to $\lambda_{j}$ hence the eigenspace corresponding to each eigenvalue $\lambda_{j}$ is $W_{j}= span(w_{j1},..., w_{jm_{j}})$ i.e. $(w_{j1},..., w_{jm_{j}})$ makes up the basis of $W_{j}$. We consider $\lambda_{1}$ to be the only main eigenvalue of $G$ (so $m_{1}=1$) and $w_{11}$ is its only corresponding eigenvector whose sum of entries is nonzero.
\Thus by Lemma 1.4.1, $proj_{W_{1}=span(w_{11})}(\textbf{e})= (w_{11}, \textbf{e})w_{11} \ne 0$. We deduce for $j=2,...,s,$ $ proj_{W_{j}}(\textbf{e})= 0$.\
Now, \textbf{e} can be written as a linear combination of basis of the eigenspaces $W_{j}$ for $j=1,...,s$, thus $\textbf{e}= \sum_{j=1}^{s}\sum_{k=1}^{m_{j}}(w_{jk}, \textbf{e})w_{jk}=\underbrace{(w_{11}, \textbf{e})}_{= c \ne 0} w_{11}+
\sum_{j=2}^{s}\underbrace{proj_{W_{j}}(\textbf{e})}_{0}= cw_{11}$
Therefore, $A(G)\textbf{e}= A(G)cw_{11}= cA(G)w_{11}= c\lambda_{1}w_{11}= \lambda_{1}cw_{11}= \lambda_{1} \textbf{e}$.