I'm trying to show the following:
Given the following $n\times n$ symmetric circulant matrices $$A^*=\begin{pmatrix} 1 & -\mu_a & 0 & ...&0&-\mu_a \\ -\mu_a & 1 & -\mu_a & 0 & ...& 0\\ ...&...&...&...&...&...\\ 0&0&0&...&1&-\mu_a\\ -\mu_a&0&0&...&-\mu_a&1 \end{pmatrix} \quad \quad \bar{A}= \begin{pmatrix} -\mu_b&0&...&0\\ 0&-\mu_b&...&0\\ ...&...&...&...\\ 0&0&...&-\mu_b \end{pmatrix}$$ Consider the $nn\times nn$ block matrices $$A_1 = \begin{pmatrix} A^*&0&...&0\\ 0&A^*&...&0\\ ...&...&...&...\\ 0&0&...&A^* \end{pmatrix}\quad \quad A_2=\begin{pmatrix} \mathbb{Id}&\bar{A}&0&...&0&\bar{A}\\ \bar{A}&\mathbb{Id}&\bar{A}&...&0&0\\ 0&\bar{A}&\mathbb{Id}&...&0&0\\ ...&...&...&...&...\\ \bar{A}&0&0&...&\bar{A}&\mathbb{Id} \end{pmatrix} $$ Now, let be $A=A_1+A_2$, that is $A$ is a block circulants with circulant blocks. I'd like to show that $\sqrt{A}$ has the same property, that is also $\sqrt{A}$ is a block circulant with circulant blocks. I think this is true because I did some numerical examples. But how to show it? I tried in the following way but I can't get through.
Some properties from "Circulant Matrices" by Philip J.Davis
- Let $$ \pi = \begin{pmatrix}0&1&0&...&0\\ 0&0&1&...&0\\ ...\\ 1&0&0&...&0 \end{pmatrix}$$ Then $\pi$ is diagonalize by the matrix $F$, that is $\pi = F^* \Omega F $, with $\Omega$ a diagonal matrix and $F$ such that $F^*F = \mathbf{Id}$;
- If C is a circulant, then it is diagonalized by $F$, that is $C=F^* \Lambda F $, with $\Lambda$ diagonal;
- Let $\Lambda$ a diagonal matrix. Then $C=F^* \Lambda F $ is a circulant;
What I had tried is:
Rembering the property that given $A,B$ block matrices consider the product $AB=C$. Then a bloc of C, say $C_{ij}$ is given by $C_{ij}= \sum_{k=1}^{n} A_{ik}B_{kj}$, that is $C_{ij}$ is the sum of the products between the blocks of $A$ and $B$.
Now I want to show that there exists a matrix $X$ such that $X X = A$ and that $X$ has the same property of $A$ to be block circulant with circulant blocks. I take a block of $A$, say $A_{ij}$. $A_{ij}$ is circulant by definition so I can write ( using the properties above ) $$A_{ij}= F^* \mathbf{diag} (\lambda_1, ..., \lambda_n) F.$$ On the other hand, I have to find $X_{ik} , X_{kj}$ such that $A_{ij}= \sum_{k=1}^{n} X_{ik}X_{kj}$. I choose $$X_{ik}= F^* \mathbf{diag}\Bigg(\bigg(\dfrac{\lambda_1}{n}\bigg)^{\frac{1}{2}},...,\bigg(\dfrac{\lambda_n}{n}\bigg)^{\frac{1}{2}}\Bigg) F$$ and this expression is independent of $k$, so $X_{ik}=X_{jk}$ for every $k=1,...,n$. In this way, I have that $\sum_{k=1}^{n} X_{ik}X_{kj}=A_{ij}$ and $X_{ik}$ is a circulant. This is work for one single $A_{ij}$ but the argument falls when I consider all other blocks of $A$. In fact I should choose the $X_{ij}$'s such that the operation works for all blocks of $A$. Why this? How can I show my thesis? Some counter example?
