About the eigenvalues of a block Toeplitz (tridiagonal) matrix I have found the following $n\times n$ squared matrix in one stability analysis problem (i.e. I have to identify the sign of its eigenvalues)
$$
A(\theta) = \begin{bmatrix}
W(\theta)+W(\theta)^T & -W(\theta) & 0 & \dots & 0 & 0 \\
-W(\theta)^T & W(\theta)+W(\theta)^T & -W(\theta) & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \vdots & \vdots \\
0 & 0 & 0 & \dots & -W(\theta)^T & W(\theta)+W(\theta)^T
\end{bmatrix},
$$
where $W(\theta) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ is a rotational matrix with $\theta\in(-\frac{\pi}{2}, \frac{\pi}{2})$. Therefore the block diagonal is $W + W^T = (2\cos\theta) I$, with $I$ being the identity matrix.
For the special case $\theta = 0$ we have that
$$
A(0) = 2\begin{bmatrix}
1 & -0.5 & 0 & \dots & 0 & 0 \\
-0.5 & 1 & -0.5 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \dots & \vdots & \vdots \\
0 & 0 & 0 & \dots & 0.5 & 1
\end{bmatrix} \otimes I,
$$
where $\otimes$ denotes the Kronecker product. The eigenvalues of $A(0)$ has analytical solution (https://en.wikipedia.org/wiki/Tridiagonal_matrix) and it can be checked that all of them are positive, in fact it can be decomposed as $B^TB$ (with $B$ being full row rank).
So I am wondering whether I can decompose $A(\theta) = B^TB$, or if I can show that the eigenvalues are still positive (or a counter-example) with my constraint in $\theta$, i.e., the main diagonal of $A$ is always positive. Any ideas or suggestions?

$A=B^TB$ for some $B$ if and only if $A$ is positive semidefinite. Let $c=\cos\theta$ and $w=e^{i\theta}$. Then $A$ is unitarily similar to $C\oplus \overline{C}$, where
$$
C=\pmatrix{
2c&-w\\
-\bar{w}&\ddots&\ddots\\
&\ddots&\ddots&-w\\
&&-\bar{w}&2c}.
$$
Now $C$ is a Hermitian tridiagonal Toeplitz matrix. Its eigenvalues are given by
$$
\lambda_k = 2c+2\cos\left(\frac{k\pi}{n+1}\right);\ k=1,2,\ldots,n.
$$
Therefore, $C$ and in turn $A$ are positive semidefinite iff $c\ge-\cos\left(\frac{n\pi}{n+1}\right)$, i.e. iff $\ |\theta|\le\frac{\pi}{n+1}$. (Here I suppose $A$ is $2n\times2n$ -- or $n$ blocks by $n$ blocks -- and $C$ is $n\times n$.)
