# Tag Info

## Hot answers tagged matrix-multiplication

5

One observation is that if we modify the problem so that $M, A, B$ are random invertible matrices, then it is easy to prove the security of the system. In fact, we can prove that the system is informationally secure; that is, for any observed $C_1, C_2$ pair, for any possible value of $K$, there is a unique set of values of $A, B, M$ that yield that $K$ ...

3

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 ...

3

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. ...

3

Multiplication of bits matrices works just like multiplication of number matrices, except the rule of addition is modified to: $1+1\mapsto 0$. Let $U$ (resp. $V$) be a square matrix of $n\times n$ elements noted $u_{l,c}$ (resp. $v_{l,c}$) with $1\le l\le n$ and $1\le c\le n$. The product $U\cdot V$ is a square matrix $W$ of $n\times n$ elements noted ...

2

You misunderstand $(02) \cdot 10000100$; it is not integer multiplication (resulting in a 9 bit integer); instead, it is multiplication in $GF(2^8)$ (which results in an element in $GF(2^8)$, which can be represented in 8 bits). AES uses a polynomial representation of $GF(2^8)$, using the polynomial $x^8 + x^4 + x^3 + x + 1$; what this means is that ...

2

You would need (at least) 3 pairs of vectors in order to determine the 3*3 matrix.

2

Let $k=\left(\begin{array}{cc}k_0 & k_1\\k_2 & k_3\end{array}\right)$ be the key. And I'll assume the transformation $a=0$, $b=1$, and so on. So you know $k\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}c_0\\c_1\end{array}\right)$ where you know $c_0$ and $c_1$. Thus $c_0=0k_0+1k_1$ and $c_1=0k_2+1k_3$. So that is two equations, ...

1

In literature there are 2 ways to show the affine transform for a given polynomial, and that depends on the location of the MSB in the input as a polynomial. The polynomial representation of the full transformation, in the format of the original Rijndael paper, is: $b(x) = a(x)(x^7 + x^6 + x^5 + x^4 + 1) + (x^7 + x^6 + x^2 + x) ~~mod~~ x^8 + 1$ Where ...

1

Both representations are essentially equivalent. If $$\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} k_{11} & k_{12} & k_{13} \\ k_{21} & k_{22} & k_{23} \\ k_{31} & k_{32} & k_{33} \end{bmatrix} \cdot \begin{bmatrix} p_1 \\ p_2 \\ p_3 \end{bmatrix},$$ then, equivalently  \begin{bmatrix} p_1 & p_2 & ...

1

Well, each party would know that the other party uses the same key, because they would probably have the same public key. Now, if some person I have never met before would use the same key as me, I would go into paranoid panic mode. That person could decrypt all messages that were only meant to be decrypted by me (and of course vice versa). However, if I ...

Only top voted, non community-wiki answers of a minimum length are eligible