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Inverse function / mapping considering vector multiplication by matrix also touches symetric encryption

Consider, there's a simple matrix as a mapping from R3 ➝ R3

R3 ➝ R3: v(3) x M(3,3) = v(3)

⎡ 1 ⎤   ⎡ 4 0 0 ⎤   ⎡ 4 ⎤ 
⎢ 2 ⎥ x ⎢ 0 3 0 ⎥ = ⎢ 6 ⎥

⎣ 3 ⎦   ⎣ 0 0 2 ⎦   ⎣ 6 ⎦

Then it ∃ also an inverse function as a simple mapping matrix from R3 ← R3

v(3) x M(3,3) = v(3) R3 ← R3

⎡ 4 ⎤    ⎡ ¼ 0 0 ⎤     ⎡ 1 ⎤

⎢ 6 ⎥  x ⎢ 0 ½ 0 ⎥  =  ⎢ 2 ⎥ 
⎣ 6 ⎦    ⎣ 0 0 ⅓ ⎦     ⎣ 3 ⎦

Suppose, there is a trivial matrix(3x2) as a mapping from R2 ➝ R3

R2 ➝ R3: v(2) * M(3x2) = v(3)

⎡ 1 ⎤   ⎡ 4 0 ⎤    ⎡ 4  ⎤
⎣ 2 ⎦ x ⎢ 0 3 ⎥  = ⎢ 6  ⎥ 



⎢⎢         ⎣ a b ⎦    ⎣ ab ⎦

Then also there ∃ a matrix(2x3) an inverse function as reverse mapping from R2 ← R3

R3 ➝ R2: v(3) * M(2x3) = v(2)

⎡ 4  ⎤   ⎡ ¼ 0 0 ⎤   ⎡ 1 ⎤ 
⎢ 6  ⎥ x ⎣ 0 ½ 0 ⎦ = ⎣ 2 ⎦
⎣ ab ⎦

Question for all mathematicians:

For which non-trivial matrices does also ∃ an inverse function to / reverse mapping?

Simple Matrices are

a 0 0 0
0 b 0 0
0 0 c 0
0 0 0 d
or
0 0 a 0
0 b 0 0
0 0 0 c
d 0 0 0

Non simple matrices are if there is more then one variable per row or column.

Kind regards,

Heinrich Elsigan.

post scriptum: I know, that when mapping higher to lower dimension, that there's certainly not any inverse mapping, because of loss of information / complexity.

My question is, how big is the numnber of not simple matrices having an inverse mapping (needed for symetric encryption school contest)

Hyperbolic PDE friend
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Heinrich Elsigan
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  • Am I understanding correctly that we are talking about square matrices (i.e. $n \times n$). And are you familiar with the concept of the determinant? – Hyperbolic PDE friend May 13 '24 at 13:39
  • Also, for your case of "simple matrices": If $a, b, c, d$ are all $\neq 0$, the simple matrix is always invertible – Hyperbolic PDE friend May 13 '24 at 13:41
  • Thank you for your answer @HyperbolicPDEfriend I was talking about square matrices and when a,b,c,d are all ≠0. I wanted to now if there exist non simple square matrices who are invertible (maybe in some modular arithmetic cases, deutsch: Restklassenring, Restklassenkörper). Do you have an answer? – Heinrich Elsigan May 13 '24 at 16:15
  • I see. What exact Restklassenring are you talking about, i.e. $\mathbb{Z}/n \mathbb{Z}$ for what $n \in \mathbb{N}_0$? – Hyperbolic PDE friend May 13 '24 at 18:26
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    This might be of interest – Hyperbolic PDE friend May 13 '24 at 18:36
  • please use MathJax to typeset your math. here is a guide. – paulina May 13 '24 at 19:02
  • Yes, Restklassenring is Z/nZ fhttps://en.wikipedia.org/wiki/Modular_arithmetic#Integers_modulo_n – Heinrich Elsigan May 13 '24 at 22:54
  • Your $\Bbb{R}^2\to\Bbb{R}^3$ example is not a matrix multiplication. And it looks non-sensical due to the appearance of the product $ab$. I'm quite sure it isn't invertible. – Jyrki Lahtonen May 14 '24 at 03:29
  • If you want to invert matrices with entries in the residue class ring $\Bbb{Z}_n$, then the usual algorithm will work, if $n$ is a prime number, and the determinant is not divisible by $n$ (in that ring $n=0$, so the algorithm would call for division by zero at some point). See this old example. If $n$ is not a prime, the algoritm may or may not work, depending on luck (something more precise can be worked out). The problem is that you cannot divide by anything that has a common factor with $n$. – Jyrki Lahtonen May 14 '24 at 03:35
  • See Wikipedia on matrix multiplication. – Jyrki Lahtonen May 14 '24 at 04:01
  • Basically what I was complaining about is that the number of rows of the matrix $A$ on the left must be equal to the number of columns of the matrix $B$ on the right for the product $AB$ to make sense. – Jyrki Lahtonen May 17 '24 at 06:02

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