Given a vector $v = (a_0, a_1, a_2, ... , a_n)$, we call a $k$-circulant matrix $C_k(v)$, where $k \geq n$, the following $k*k$ matrix : $$ C_k(v)= \begin{bmatrix} a_1 & a_2 & ... & a_n & 0 & ... & 0 \\ 0 & a_1 & a_2 & ... & a_n & 0 & ... \\ \cdot & & & & \cdot & & \\ & \cdot & & & & \cdot & \\ & & \cdot & & & & \cdot \\ a_3 & a_4 & ... & a_n & 0 & ... & 0 \\ a_2 & a_3 & a_4 & ... & a_n & 0 & ... \end{bmatrix} $$ (basically, you pad the vector with zeros and then you form a circulant matrix with it.)
For simplicity, given two natural numbers $n,k$, define the block-matrix $$I_{n,k} = \begin{bmatrix} I_n & I_n & ... & I_n \end{bmatrix} $$ where $I_n$ repeats k times (I_n is the identity matrix of order $n$.)
I found out that, given a vector $v = (a_1,a_2,...,a_n)$ of length $n$ and a natural number $k$, we have the following relation: $$ C_n(v) \cdot I_{n,k} = I_{n,k}\cdot C_{nk}(v)$$
This really isn't that hard to prove, since if for the vector $v$ we consider the following matrices: $$ B = \begin{bmatrix} 0 & 0 & 0 & 0 &... & 0 \\ a_n & 0 & 0 & 0 & ... & 0 \\ a_{n - 1}& a_n & 0 & 0 & ... & 0 \\ a_{n - 2} & a_{n-1} & a_n & 0 & ... & 0 \\ & & \cdot & & & \\ & & &\cdot & & \\ & & & & \cdot & \\ a_{2} & a_{3} & ... & a_{n-1} & a_n& 0 \\ \end{bmatrix} $$
$$A = \begin{bmatrix} a_1 & a_2 & a_3 & ... & a_n \\ 0 & a_1 & a_2 & ... & a_{n-1} \\ & & \cdot \\ & & & \cdot \\ & & & & \cdot \\ 0 & 0 & ... & 0 & a_1 \end{bmatrix}$$
we observe that $C_n(v) = A+B$, $$C_{n,k} = \begin{bmatrix} A & B & 0 & 0 & ... & 0\\ 0 & A & B & 0 & ... & 0\\ && \cdot \\ &&& \cdot \\ &&&& \cdot \\ B & 0 & 0 & 0 & ... & A \end{bmatrix}, $$
and so the relation to prove becomes $$ [ A + B ] \cdot \begin{bmatrix} I_n & I_n & ... & I_n \end{bmatrix} = \begin{bmatrix} I_n & I_n & ... & I_n \end{bmatrix} \cdot \begin{bmatrix} A & B & 0 & 0 & ... & 0\\ 0 & A & B & 0 & ... & 0\\ && \cdot \\ &&& \cdot \\ &&&& \cdot \\ B & 0 & 0 & 0 & ... & A \end{bmatrix}, $$
which is obvious.
However, here comes the question: I have tested that given two vectors $v = (a_1, a_2, a_3, ... a_n) $ and $w = (b_1, b_2, ... , b_n)$, the following is also true: $$ C_n(v) \cdot C_n(w) \cdot I_{n,k} = I_{n,k} \cdot C_{nk}(v) \cdot C_{nk}(w) $$
I haven't been able to prove this one; I know that the product of two circulant matrices is also a circulant matrix, but it doesn't seem like there is a vector $z = (c_1,c_2,...,c_n)$ such that $C_n(v) \cdot C_n(w) = C_n(z)$ and $ C_{nk}(v) \cdot C_{nk}(w) = C_{nk}(z)$.
Any ideas would be appreciated.
I figured it out. Basically we have $$C_n(w) \cdot \begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) \end{bmatrix} = \begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) \end{bmatrix} \cdot C_{nk} (w) $$
By splitting $C_{nk} (w) $ into matrices $X$ and $Y$, we want to show that $$\begin{bmatrix} C_{n}(w)C_n(v) & C_{n}(w) C_n(v) & ... & C_{n}(w)C_n(v) \end{bmatrix} = \begin{bmatrix} C_n(v) & C_n(v) & ... & C_n(v) \end{bmatrix} \cdot \begin{bmatrix} X & Y & 0 & 0 & ... & 0 \\ 0 & X & Y & 0 & ... & 0 \\ &&\cdot \\ &&&\cdot \\ &&&&\cdot \\ Y & 0 & 0 & 0 & ... & X \end{bmatrix} ,$$ which is obvious, since $ X+Y = C_n(w)$,