Bound on eigenvalues of hollow, tridiagonal symmetric matrix

I have considered the following matrix

\begin{bmatrix} 0 & \frac{1}{2} & 0 & \dots & 0 \\ \frac{1}{2} & 0 & \frac{1}{2} & \dots & 0 \\ \vdots & \ddots & \ddots & \ddots & 0 \\ \vdots & \ddots & \ddots & \ddots & \frac{1}{2}\\ 0 & 0 & 0 & \frac{1}{2} & 0 \\ \end{bmatrix}

It is a hollow, symmetric, tridiagonal matrix with entries $\frac{1}{2}$ alongside the diagonal of zeros. Call this matrix $B$. (excuse the possibly nonstandard formatting)

I have convinced myself that this matrix has eigenvalues of absolute value strictly smaller than 1 for all sizes $n \times n$.

The Gershgorin circle theorem doesn't seem to help directly in this case, it doesn't produce the strict inequality I was hoping for.

I have also tried observing the row sums of $B^{m}$, which I am fairly certain fall strictly below 1 for all sizes $n \times n$ given a large enough power $m$. I could then bound any eigenvalue by the maximal row sum and translate my findings back to the original matrix.

Any help would be greatly appreciated.

Solution 1:

You can go through the proof of Gershgorin circle theorem to prove strict inequality.

Let $M_n$ denote the original matrix. Suppose there exists an eigenvalue $\lambda$ of $M_n$ with $|\lambda| \geq 1$. Let $v = (v_1, \dots, v_n)^T$ be an eigenvector for $\lambda$.

Since at least one of $v_i$ is nonzero, we may divide $v$ by the maximum of $|v_i|$ and assume that the maximum of $|v_i|$ is equal to $1$. Let $k \in \{1, \dots, n\}$ be the smallest index such that $|v_k| = 1$.

If $k = 1$, then from $M_nv = \lambda v$ we get $\frac 12 v_2 = \lambda v_1$. Taking absolute values, we have $|v_2| \geq 2$ which contradicts our assumption. The same argument shows that $k = n$ leads to contradiction.

Now assume that $1 < k < n$. Again from $M_nv = \lambda v$ we get $\frac 1 2(v_{k - 1} + v_{k + 1}) = \lambda v_k$. Taking absolute values, we have $|v_{k - 1} + v_{k + 1}| \geq 2$.

However both $v_{k - 1}$ and $v_{k + 1}$ have absolute value at most $1$, hence the above inequality implies that $|v_{k - 1}| = 1$. This contradicts the minimality of $k$.

The same technique can be applied to more general matrices.

By the way, if we denote by $P_n(t)$ the characteristic polynomial of the matrix $M_n$, then the polynomials $P_n(t)$ satisfy the recurrence relation $P_{n + 2} - tP_{n + 1} + \frac14P_n = 0$, together with initial conditions $P_0 = 1$ and $P_1 = t$.