Define $$A = \frac{1}{2h} \begin{pmatrix} 0 & 1 & & & & -1\\ -1 & 0 & 1 & & \\ &\ddots & \ddots & \ddots \\ & & & -1 & 0 & 1 \\ 1 & & & & -1 & 0 \end{pmatrix} \in \mathbb{R}^{d \times d},$$ where $d = \frac{1}{h}$. The task is to show that $\rho(A) = \sqrt{\lambda_{\text{max}}(A^*A)} \ge d$ when $d$ is divisible by $4$. I have a proof, but it is long and tedious, since I am following the definitions. I would like to know if there is a simpler way to approach this problem and if this proof is correct.
My proof: We first compute $A^*A$, which is given by $$ A^*A = \frac{1}{4h^2}\begin{pmatrix} 2 & 0 & -1 & & & -1 & 0\\ 0 & 2 & 0 & -1 & & & -1 \\ &\ddots & \ddots & \ddots & \ddots & \ddots \\ -1 & & & -1 & 0 & 2 & 0 \\ 0 & -1 & & & -1 & 0 & 2 \end{pmatrix}$$ where the rows of $A^*A$ are cycling over $(-1,0,2,0,-1)$. Now $A^*A v = \lambda v$ if and only if $-v_{j-2}+2v_j-v_{j+2} = \lambda v_j$, $j=1,\cdots,d$, $v_{-1} = v_{d+1}$, $v_{0} = v_d$, $v_{d+1} = v_{1}$, $v_{d+2} = v_{2}$. Now, I was stuck here for sometime and I guessed that the eigenvectors might be of the form $v_{j} = \sin(jk\theta)$ where $k \in \{ 1,\cdots,d\}$ and $j=1,\cdots,d$. Since $$\begin{align*} -v_{j-2}+2v_j-v_{j+2} &= -\sin((j-2)k\theta)+2\sin(jk\theta)-\sin((j+2)k\theta) \\ &= -2\sin(jk\theta) \cos{(2k\theta)}+2\sin(jk\theta)= 4\sin^2(k\theta)v_{j},\end{align*}$$ and $\lambda = 4\sin^2(k\theta)$. Now to satisfy the boundary conditions, I simply took $\theta = 2\pi h$. Thus, $$\rho(A) = \frac{1}{h}\max_{k=1,\dots,d} |\sin{(2\pi h k)}|,$$ and if $d$ is divisible by $4$, then $d = 4d'$ for some integer $d'$ and
$$\rho(A) \ge \frac{1}{h}|\sin{\left(\frac{2\pi}{4d'} d'\right)}| = \frac{1}{h} = d$$
NOTE: The matrix $A$ was obtained through the discretization of the eigenvalue problem $\frac{d}{dx} v = \lambda v$ through the central difference scheme with periodic boundary conditions $v(0) = v(1)$.