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How was the constant matrix in AES MixColumns constructed?

\begin{bmatrix} 02 & 03 & 01 & 01 \\ 01 & 02 & 03 & 01 \\ 01 & 01 & 02 & 03 \\ 03 & 01 & 01 & 02 \\ \end{bmatrix}

How can I construct in $GF(2^4)$ an invertible matrix for the MixColumns step?

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  • $\begingroup$ how was it constructed? with a circulant matrix containing the smallest possible values to get the job done $\endgroup$ Jun 16, 2016 at 20:33
  • $\begingroup$ There's an entire book (230 pages) about the design of AES, it's called: "The Design of Rijndael: AES – The Advanced Encryption Standard.", chances are the answer is in there somewhere... $\endgroup$
    – SEJPM
    Jun 16, 2016 at 21:51
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    $\begingroup$ Read at least the Rinjdael submission, section 7.3 $\endgroup$
    – fgrieu
    Jun 16, 2016 at 22:02

1 Answer 1

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Any full rank $n \times n$ circulant matrix $A$ with all entries nonzero generates an MDS code. The entries here were picked for efficient fast multiplication, as far as I can recall.

Do refer to the book suggested by @SEJPM for more.

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