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I choose at random an invertible square matrix A of size 128 in GF(2). I want to use this matrix as a substitution box. Is this a non linear transformation ?

I've seen that substitution boxes are the non linear parts of a block cipher algorithm, but the product between A and x is linear ? Not ? There is something that I don't understand.

Thank you.

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2 Answers

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We call an operation F linear if the following holds:

$F(X+Y) = F(X) + F(Y)$

for all $X, Y$ within the appropriate set, and for some group operator $+$.

Now, if we consider matrix multiplication by a fixed matrix $A$, we do have the identity:

$A \cdot (X+Y) = A \cdot X + A \cdot Y$

for arbitrary vectors $X$, $Y$, and where $+$ is vector addition. Hence, matrix multiplication by any fixed matrix $A$ is linear.

When you are designing a block cipher, it is critically important that, for any group operator $+$, there be some component that is nonlinear with respect to that operator. Hence, matrix multiplication is probably not an ideal selection.

You appear to have some understanding of this; so what are you confused about?

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Multipliying a 128-bit vector by a random square matrix of size 128 in $GF(2)$ yields a 128-bit vector. Output bit $j$ is the exclusive-OR of some of the input bits, as determined by the coefficients set in line $j$ of the matrix.

Each output bit is a linear combination of some of the input bits. The transformation qualifies as linear, for whatever definition of that. One definition often used in cryptanalysis is that a transformation $F$ is linear when $\forall X, \forall Y, F(X\oplus Y)=F(X)\oplus F(Y)\oplus F(0)$, where $0$ is the neutral element for $\oplus$, here the all-zero vector. Chaining linear transformations leads to a linear transformation, and this is bad for security in a block cipher. Further, here, we have $F(0)=0$, which might be an undesirable regularity.

Also, in the original statement, the transformation is was unlikely to be a bijection, which is undesirable for a block cipher algorithm of the permutation-substitution kind.

Update: in an AES round, the non-linearity comes from using an 8-bit substitution based on the multiplicative inverse in $GF(2^8)$, within the SubBytes step. Everything else is strictly linear.

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Thank you for your answer. In this case, in AES, it's not the P-box alone (an affine transformation) that produces the non linearity but its combination with an inverse modulo a primitive polynomial ? Fgrieu, In fact I was speaking about invertible square matrix, so this is bijective. Thank you again. –  user7060 Jun 3 '13 at 16:51
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