Consider a set of distinct vectors {$a,b,c,d,...$} in a subspace $C$ of $R^n$.


$C = \pi (C)$ where $\pi : R^n \rightarrow R^n$ defined by $\pi (x\_1,x\_2,...,x\_n)=(x\_2,...,x\_n,x\_1)$


How can I prove the following?



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> For any two vectors $a$ and $b$ belong to $C$, $a \circledast b \in C$ ($\circledast$ : circular convolution); if $C$ is invariant under $\pi$.
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circular convolution over Invariant subspace
If we write the standard basis vectors as $e\_1,\dots,e\_n$, with the cyclic convention that $e\_{n+1}=e\_1$, note that $\pi(e\_i)=e\_{i-1}$ and that $\pi$ is linear, so $\pi(\sum b\_i e\_i)=\sum b\_{i+1}e\_i$. By induction, $\pi^{\circ j}(\sum b\_i e\_i)=\sum b\_{i+j} e\_i$.
$c=a\*b=\sum\_{ij} a\_j b\_{i-j} e\_i=\sum\_j a\_j (\sum\_i b\_{i-j} e\_i)= \sum\_j a\_j \pi^{-\circ j}b$.


