Lemma 1: Consider $A$ be an $e\times e$ circulant matrix, denoted with $A=circ(a_0,a_1,\cdots , a_{e-1})$, over the $GF(2^q)$. Let $V=[\gamma^{-ij}]$, $0\leq i,j<e$, be an $e\times e$ Vandermonde matrix where $\gamma$ is an element in $GF(2^q)$ of order $e$. Then, $VAV^{-1}$ is an $e\times e$ diagonal matrix, denoted by $A^{F}$, whose $t$h diagonal element $0\leq t <e$, equals:
$$ A^{F}=dig\,(\, \sum_{i=0}^{e-1}\,a_i \, , \, \sum_{i=0}^{e-1}\,\gamma^{i}a_i \, , \, \cdots \, , \, \sum_{i=0}^{e-1}\,\gamma^{(e-1)i}a_i\, ) $$
In particular, the diagonal vector of $A^{F}$ is the Fourier transform of $(a_0,a_1,\cdots , a_{e-1})$. [1]
In Lemma $1$, we suppose that $e\mid 2^q-1$ which implies that $e$ should be an odd number. Therefor, every circulant matrix of odd size, over $GF(2^q)$, can be diagonalized.
My question: How to diagonalize a circulant matrix over $GF(2^q)$ of even size?
My try: Although we have no elements of even order in $GF(2^q)$, I try as follows: Let $A$ be an $2n\times 2n$ circulant matrix over $GF(2^q)$. Consider $V=van(0,1,\gamma , \gamma^2,\cdots , \gamma^{2n-2})$ be a Vandermonde matrix of $2n$ size where $\gamma$ is an elemnet of order $2n-1$. I construct $VAV^{-1}$, but the result was not a diagonal matrix. I have no idea to answer this question.
Thanks for any suggestions.
In general you cannot. Consider a size $2$ circulant $$A=\pmatrix{a&b\\b&a}$$ over a field of characteristic $2$. Its characteristic polynomial is $X^2+a^2+b^2$, which has repeated zeros. It is only diagonalisable if it is already diagonal, that is if $b=0$.