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We have that $C_n\in\mathbb{C}^{n\times n}$ is a circulant matrix of order n. So we have $$ C_n = \begin{bmatrix} \alpha_0 & \alpha_{n-1} & \dots & \alpha_{2} & \alpha_{1} \\ \alpha_{1} & \alpha_0 & \alpha_{n-1} & & \alpha_{2} \\ \vdots & \alpha_{1}& \alpha_0 & \ddots & \vdots \\ \alpha_{n-2} & & \ddots & \ddots & \alpha_{n-1} \\ \alpha_{n-1} & \alpha_{n-2} & \dots & \alpha_{1} & \alpha_0 \\ \end{bmatrix}$$ The associated polynomial is $f(x) = \alpha_0 + \alpha_1 x + \ldots + \alpha_{n-1}x^{n-1}$

How do I show that the set of $n\times n$ circulant matrices, $C_n$, is a subspace of that vector space with dimension $n$?

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  • $\begingroup$ Note that (in the syntax of your earlier question) $f(Z_n)=C_n$. $\endgroup$ – Omnomnomnom Jan 15 '16 at 22:07
  • $\begingroup$ Nice, thank you for that $\endgroup$ – Wolfy Jan 15 '16 at 22:17
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Check the three subspace axioms:

  1. Contains the zero vector:

It is simple to see that $ \begin{bmatrix} 0 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & 0 \\ \end{bmatrix} $ is circulant.

  1. Closure under addition:

$ \begin{bmatrix} \alpha_0 & \alpha_{n-1} & \dots & \alpha_{2} & \alpha_{1} \\ \alpha_{1} & \alpha_0 & \alpha_{n-1} & & \alpha_{2} \\ \vdots & \alpha_{1}& \alpha_0 & \ddots & \vdots \\ \alpha_{n-2} & & \ddots & \ddots & \alpha_{n-1} \\ \alpha_{n-1} & \alpha_{n-2} & \dots & \alpha_{1} & \alpha_0 \\ \end{bmatrix} + \begin{bmatrix} \beta_0 & \beta_{n-1} & \dots & \beta_{2} & \beta_{1} \\ \beta_{1} & \beta_0 & \beta_{n-1} & & \beta_{2} \\ \vdots & \beta_{1}& \beta_0 & \ddots & \vdots \\ \beta_{n-2} & & \ddots & \ddots & \beta_{n-1} \\ \beta_{n-1} & \beta_{n-2} & \dots & \beta_{1} & \beta_0 \\ \end{bmatrix} = \begin{bmatrix} \alpha_0 + \beta_0 & \alpha_{n-1} + \beta_{n-1} & \dots & \alpha_{2} + \beta_{2} & \alpha_{1} + \beta_{1} \\ \alpha_{1} + \beta_{1} & \alpha_0 + \beta_0 & \alpha_{n-1} + \beta_{n-1} & & \alpha_{2} + \beta_{2} \\ \vdots & \alpha_{1} + \beta_{1}& \alpha_0 + \beta_0 & \ddots & \vdots \\ \alpha_{n-2} + \beta_{n-2} & & \ddots & \ddots & \alpha_{n1} + \beta_{n-1} \\ \alpha_{n-1} + \beta_{n-1} & \alpha_{n-2} + \beta_{n-2} & \dots & \alpha_{1} + \beta_{1} & \alpha_0 + \beta_0 \\ \end{bmatrix} $

  1. Closure under scalar multiplication:

$ c\cdot \begin{bmatrix} \alpha_0 & \alpha_{n-1} & \dots & \alpha_{2} & \alpha_{1} \\ \alpha_{1} & \alpha_0 & \alpha_{n-1} & & \alpha_{2} \\ \vdots & \alpha_{1}& \alpha_0 & \ddots & \vdots \\ \alpha_{n-2} & & \ddots & \ddots & \alpha_{n-1} \\ \alpha_{n-1} & \alpha_{n-2} & \dots & \alpha_{1} & \alpha_0 \\ \end{bmatrix} = \begin{bmatrix} c\cdot \alpha_0 & c\cdot \alpha_{n-1} & \dots & c\cdot \alpha_{2} & c\cdot \alpha_{1} \\ c\cdot \alpha_{1} & c\cdot \alpha_0 & c\cdot \alpha_{n-1} & & c\cdot \alpha_{2} \\ \vdots & c\cdot \alpha_{1}& c\cdot \alpha_0 & \ddots & \vdots \\ c\cdot \alpha_{n-2} & & \ddots & \ddots & c\cdot \alpha_{n-1} \\ c\cdot \alpha_{n-1} & c\cdot \alpha_{n-2} & \dots & c\cdot \alpha_{1} & c\cdot \alpha_0 \\ \end{bmatrix} $

So the set of circulant matrices (subset of the vector space $\mathbb{C}^{n\times n}$) is indeed a subspace.

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