Matrix multiplication in AES' MixColumns step.

Question

In Advanced Encryption Std, say after a ShiftRow operation, I want to perform MixColumns.

      State             MixColumn Matrix             Mixed
┏             ┓       ┏             ┓       ┏             ┓
┃ d4 e0 b8 1e ┃       ┃ 02 03 01 01 ┃       ┃ 04 e0 48 28 ┃ 
┃ bf b4 41 27 ┃       ┃ 01 02 03 01 ┃       ┃ 66 cb f8 06 ┃ 
┃ 5d 52 11 98 ┃       ┃ 01 01 02 03 ┃       ┃ 81 19 d3 26 ┃ 
┃ 30 ae f1 e5 ┃       ┃ 03 01 01 02 ┃       ┃ e5 9a 7a 4c ┃ 
┗             ┛       ┗             ┛       ┗             ┛

Example taken from this flash file. I'm puzzled by the description at Wikipedia.

The flash told that after MixColumns, the 1^st column of State is turned into the 1^st column of the Mixed matrix. How is this so?

First of all, how to multiply a column by a matrix? The shapes don't match, or is there anything special in cryptography? Well, if we transpose the column, I still don't understand: ($\cdot$ means dot product, and here are my Attempts)

# 1: [d4 bf 5d 30]$\cdot$[02 01 01 03] = d4+d4+bf+5d+30+30+30 = 54 != 04

# 2: [d4 bf 5d 30]$\cdot$[02 01 01 03] = d4+d4+bf+5d+((30<<1) ^ 30) = 14 != 04

# 3: [d4 bf 5d 30]$\cdot$[02 03 01 01] = d4+d4+bf+bf+bf+5d+30 = 72 != 04

# 4: [d4 bf 5d 30]$\cdot$[02 03 01 01] = d4+d4+((bf<<1)^30)^11b+5d+30 = 0F != 04

What's happening?

Answer 1

All the matrix entries are elements of the Galois field $GF(2^8)$, using field poynomial $x^8 + x^4 + x^3 + x + 1$. There is an excellent Wikipedia article Rijndael mix columns which explains this very clearly.