Matrix Groups & Textbook-RSA-Style Attacks over Matrices

Updated 19 Sep 2026

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  What it is

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    Invertible $n\times n$ matrices over a finite field form a group under matrix multiplication (the general linear group $GL_n$), and any single invertible matrix $M$ generates a cyclic subgroup with a well-defined multiplicative order — just like an ordinary field element, just computed differently (via the matrix's eigenvalues/characteristic polynomial rather than simple exponent-counting). This means "RSA" or "DH"-style constructions built out of matrix exponentiation ($C = M^E$) inherit the exact same vulnerabilities as textbook integer RSA whenever that order is computable: if you can compute $\text{ord}(M)$, you can invert the public exponent $E$ modulo it exactly like inverting $e$ modulo $\phi(N)$, recovering $M$ from $C$ directly with no "hard problem" involved at all.

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  When to apply

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    A challenge builds an "encryption" or key-exchange scheme using matrix exponentiation (over $GF(2)$ or $GF(p)$) instead of ordinary integer exponentiation — check first whether the matrix's multiplicative order is directly computable (Sage exposes this via .multiplicative_order(), itself computed from the matrix's characteristic polynomial/eigenvalue structure) before assuming any matrix-analogue of the discrete log problem is actually hard here.

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  Math

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    For invertible $M$ over a finite field, $\operatorname{ord}(M)$ is the smallest $k$ with $M^k=I$. If $C=M^E$ and $\gcd(E,\operatorname{ord}(M))=1$, then $C^{E^{-1}\bmod\operatorname{ord}(M)}=M$. There is no general “totient of the entries” shortcut; compute or derive the matrix order from its rational or characteristic structure.

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  Worked example

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    A small invertible $4\times4$ matrix over $GF(2)$ had multiplicative order 6 (found by direct repeated multiplication); inverting $E=7$ modulo 6 and raising the "ciphertext" $M^7$ to that inverse recovered the exact original matrix.

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  Python

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    def mat_order(M, mat_mul, identity):
            cur, I, k = M, identity, 1
            while cur != I:
                cur = mat_mul(cur, M)
                k += 1
            return k
     
    d = pow(E, -1, order)
    M_recovered = mat_pow(C, d)
    

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  SageMath

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    Computing a matrix's order over a large field/dimension by brute-force repeated multiplication (as above) is impractical — Sage computes it properly via the characteristic polynomial's factorization:

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    order = M.multiplicative_order()
    d = inverse_mod(E, order)
    M_original = C^d
    

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  Related

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    Groups (Group Theory Basics)

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    RSA with Non-Coprime Exponent (e Shares a Factor with φ(N))

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    Finite Fields (Galois Fields: GF(p) and GF(2^n))

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  Cards

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    What structural property do invertible matrices over a finite field share with ordinary field elements?

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      Both form groups under their respective multiplications, so every invertible matrix has a well-defined multiplicative order, just like every field element does.

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    Why does a directly-computable matrix order break "RSA over matrices" the same way a known φ(N) breaks textbook RSA?

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      Because inverting the public exponent modulo that order recovers a valid "private exponent" that undoes the "encryption" directly — the exact same mechanism as RSA decryption, just in a different group.

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