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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?

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$A=B^TB$ 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 $n\times n$.)

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  • $\begingroup$ I see the key point of identifying the rotational matrix with the complex number $w$. Therefore I can also see that the "physical meaning" of $A$ and $C$ is the same. However, I cannot see straightforwardly why $PA = (C \oplus \bar C) P$ $\endgroup$
    – user51196
    Jul 29 '16 at 5:44
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    $\begingroup$ @noether All 2D rotations commute. Therefore, by a similarity transform via a block diagonal unitary matrix, you may convert $W$ and $W^T$ simultaneously into $\pmatrix{e^{i\theta}\\ &e^{-i\theta}}$ and $\pmatrix{e^{-i\theta}\\ &e^{i\theta}}$ and obtain a new tridiagonal matrix $A$, whose only nonzero diagonals are the main one and the second sub/super-diagonals. Now, if move the odd-indexed rows/columns to the top/left and move the even-indexed rows/columns to the bottom/right, we obtain $C\oplus \bar{C}$. $\endgroup$
    – user1551
    Jul 29 '16 at 7:07
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    $\begingroup$ @noether E.g. when $n=2$, the new tridiagonal $A$ is of the form (where $w=e^{i\theta}$ and $2c=w+w'$) $$A=\pmatrix{2c& 0& -w&0\\ 0&2c& 0&-w'\\ -w'& 0&2c&0\\ 0& -w& 0&2c}.$$ Now the $\{1,3\}$-submatrix is $C$ and the $\{2,4\}$-submatrix is $\bar{C}$. $\endgroup$
    – user1551
    Jul 29 '16 at 7:08
  • $\begingroup$ thank you for your time and effort in the explanation. I have checked and understood what you said. Everything is clear now. Your answer indeed is true ;). $\endgroup$
    – user51196
    Jul 29 '16 at 13:25

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