Setup: Let $A$ be the $n \times n$ adjacency matrix of a path with $n$ nodes, so the $ij^\text{th}$ element of $A$ is $a_{ij} = 1(|i-j|=1)$. I.e. the elements just off the main diagonal are $1$, and everything else is $0$.
Now define $B$ as equal to $A$, but with row sums normalized to equal $1$. I.e., $b_{ij} = a_{ij} / a_{i+}$, where $a_{i+}$ is the $i^\text{th}$ row sum of $A$. Here are the matrices:
$$
A = \begin{bmatrix}
0&1 \\
1&0&1&& \mathbf 0 \\
&1&0& \ddots \\
&&\ddots&\ddots &1& \\
& \mathbf 0&&1&0&1 \\
&&&&1&0 \end{bmatrix},\,\,\,\,
B = \begin{bmatrix}
0&1 \\
1/2&0&1/2&& \mathbf 0 \\
&1/2&0& \ddots \\
&&\ddots&\ddots &1/2& \\
& \mathbf 0&&1/2&0&1/2 \\
&&&&1&0 \end{bmatrix}.
$$
Note that the 1st and last rows have 1 non-zero element (since the nodes at the ends of the path have only 1 neighbor), while the middle rows have two non-zero elements (since the middle nodes have a neighbor on each side).
My question: Can you prove that the max and min eigenvalues of $B$ are $1$ and $-1$, respectively, for any $n \ge 2$? I've checked that this is true (with a computer program) for all $n$ from 2 to 500. So I'm pretty confident that it's true in general, but I'm wondering if there's a proof.
Context: The reason I care about this is that $B$ appears in a statistical model I'm using (a CAR spatial model), and the max and min eigenvalues of $B$ determine the parameter space for a parameter in the model.
This is clearly closely related to the following asked-and-answered question: Computing eigenvalue of the adjacency matrix of a path. There they wanted all eigenvalues of $A$, whereas I want the max and min eigenvalues of $B$.
Also, although this might not be helpful, here's a plot of all $n$ eigenvalues of $B$, for $n = 2,\ldots,16$.
Best Answer
The row sums of your matrix are 1 and by Gersgorin's theorem this will be the largest eigenvalue. There is a diagonal matrix $D$ such that $DBD^{-1}$ is symmetric. (Exercise.) The matrix $B$ is a weighted adjacency matrix of a bipartite graph, so its spectrum is symmetric about zero, whence $-1$ is an eigenvalue.