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if $\hspace{0.2cm}$$Z\in$ R$^{n\times n} $ $\hspace{0.2cm}$is the down-shift matrix with ones on the first subdiagonal and zeros elsewhere, and $L\in$ R$^{n\times n} $ $\hspace{0.2cm}$ is the lower triangular matrix with $1s$ at the non-zero entries, then the matrix $$A=\left( \begin{array}{ccccc} w_1 & w_1 &\cdots & w_1 \\ w_1 & w_2 & \cdots& w_2 \\ \vdots & \vdots & \ddots& \vdots \\ w_1 & w_2 & \cdots & w_n\\ \end{array} \right)$$ can be written as $\hspace{0.2cm}$ $A=L(D_w-ZD_wZ^T)L^T$$\hspace{0.2cm}$ where $\hspace{0.2cm}$$D_w=Diag[w_1,\cdots,w_n]$

If I have the matrix $$B=\left( \begin{array}{ccccc} w_1 & w_2 &\cdots & w_n \\ w_2 & w_2 & \cdots& w_n \\ \vdots & \vdots & \ddots& \vdots \\ w_n & w_n & \cdots & w_n\\ \end{array} \right)$$ How can I to factorize it in a similar way to the previous matrix in terms of $Z$ and $L$ ?

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Let $K$ denote the matrix with $1$s on the "antidiagonal", i.e. $$ K = \pmatrix{&&1\\&\cdots\\1} $$ We note that "flipping" the matrix $B$ gives us $$ KBK = \pmatrix{ w_n&w_n&\cdots&w_n\\ w_n&w_{n-1}&\cdots&w_{n-1}\\ \vdots & \vdots & \ddots & \vdots\\ w_n & w_{n-1} & \cdots & w_1} $$ Let $D_\hat w$ denote the matrix $\operatorname{Diag}[w_n,w_{n-1},\dots,w_1]$ (if you prefer, you could write $D_{\hat w} = KD_wK$). Using your decomposition, we have $$ KBK = L(D_{\hat w}-ZD_{\hat w}Z^T)L^T \implies\\ B = KL(D_{\hat w}-ZD_{\hat w}Z^T)L^TK $$ Or if you prefer, $$ B = (KL)(D_{\hat w}-ZD_{\hat w}Z^T)(KL)^T $$

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