Maximum eigenvalue of an adjacency matrix Let $G = (V,E)$ be an undirected graph without multiple edges.
Additionally, let $|V| = n + 1$, where $n \geq 3$.
$G$ is created in such a way, that $n$-first vertices create a cycle and $n+1$ vertex is connected only to the previous $n$-vertices.
Prove that the maximum eigenvalue of adjacency matrix generated by this graph is equal:
$$
\lambda_{0} = \sqrt{n+1} + 1
$$

Example. Let $n+1 = 10$. The $G$ looks like this - [1]. The adjacency matrix $M$ for $G$ is the following:
$$
M = 
\left(\begin{array}{rrrrrrrrrr}
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\
1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \\
1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 0
\end{array}\right)
$$
The characteristics polynomial $W$ of $M$:
$$
W = \lambda^{10} - 18\lambda^8 - 18\lambda^7 + 72\lambda^6 + 90\lambda^5 - 93\lambda^4 - 126\lambda^3 + 27\lambda^2 + 34\lambda - 9
$$
Lastly, real roots of $W$ are the following:
$$
\begin{array}{lar}
\lambda_0 = 1 + \sqrt{10} \\
\lambda_1 = -1  \\
\lambda_2 = 1 - \sqrt{10} \\
\end{array}
$$
Of course, $\lambda_0$ satisfies the equality $\lambda_0 = \sqrt{n+1} + 1 = \sqrt{10} + 1 \approx 4,162277 $.

Hint: using the symmetries in the graph, try out vectors that equal 1 on the first $n$ components and equal a variable $x$ on the last component. The eigenvalue equation gives you two equations and two unknowns; without giving too much away, you’ll end up with a quadratic that you can solve to get the desired eigenvalue.
To prove it is maximal, try using Perron-Frobenius and Courant-Fischer; what do you notice about the entries of the corresponding eigenvector?
