# Finding the matrix of a linear affine transformation and its inverse

How do i get the matrix representation of an affine transformation and it's inverse in sage?

I am more so interested in doing this for random affine transformations as i am using them in a multivariate cryptography scheme but for example the following affine transformation over GF(3)

[1 0 0 2 0 2]     [1]
[2 2 1 1 2 0]     [1]
[2 0 0 2 2 2]     [0]
x |-> [2 2 0 1 1 1] x + [1]
[1 0 2 0 2 0]     [0]
[0 0 0 2 0 2]     [2]
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Can you provide an example of an affine transformation you are interested in?

( 2018-03-19 11:07:53 -0500 )edit

I have edited my question with an example, any help would be appreciated.

( 2018-03-19 11:43:47 -0500 )edit

Do you mean like a projective coordinate representation? In that case if you have Mx+b you could make a block matrix out of M and b like block_matrix([[M,b],[0,1]]) or something, and then get that inverse. But I don't know if this is what you mean by the matrix representation of an affine transformation.

( 2018-03-19 12:22:25 -0500 )edit

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An affine transformation t is given by some square matrix a and some vector b, and maps x to a * x + b.

One can represent such a transformation t by an augmented matrix, whose first n columns are those of a and whose last column has the entries of b. We also denote this matrix by t.

Then the n first columns represent the linear part a of the transformation t, and its last column represents the translation part, the vector b.

Since the transformation t maps a vector x to a * x + b, provided a is invertible, with inverse A, then so is t, and its inverse T maps x to A * x - A * b.

Explanation: if y = t(x), then y = a * x + b, so y - b = a * x, and if a is invertible with inverse A, then A * y + A * (-b) = A * a * x, which means x =A * y + A * (-b), so the linear part isA and the translation part isA * (-b).

Therefore the matrix for t_inv, also denoted as T, has A as its first n columns and A * (-b) as its last column.

Here is an implementation of a function that returns the inverse of an affine transformation. The function includes a documentation string which summarizes the discussion above, and an example based on the one in your question.

def affine_inverse(t, as_block_matrix=False):
"""
Return the inverse of this affine transformation

The affine transformation is given by some n by (n + 1)
matrix t whose n first columns represent the linear
part a of the transformation, and whose last column represents
the vector b, so that the transformation maps a vector x
to a * x + b. Provided a is invertible, with inverse A,
then so is t, and its inverse T maps x to A * x - A * b.

Optionally, T` is returned as a block matrix.

EXAMPLE::

sage: k = GF(3)
sage: alist = [
....: [1, 0, 0, 2, 0, 2],
....: [2, 2, 1, 1, 2, 0],
....: [2, 0, 0, 2, 2, 2],
....: [2, 2, 0, 1, 1, 1],
....: [1, 0, 2, 0, 2, 0],
....: [0, 0, 0, 2, 0, 2],
....: ]
sage: blist = (1, 1, 0, 1, 0, 2)
sage: a = matrix(k, alist)
sage: n = a.nrows()
sage: b = matrix(k, n, 1, blist)
sage: t = a.augment(b)
sage: affine_inverse(t)
[1 0 0 0 0 2 1]
[1 0 2 2 0 2 2]
[2 0 1 0 2 0 1]
[2 1 0 2 1 0 1]
[2 0 2 0 0 2 0]
[1 2 0 1 2 2 1]
sage: affine_inverse(t, as_block_matrix=True)
[1 0 0 0 0 2|1]
[1 0 2 2 0 2|2]
[2 0 1 0 2 0|1]
[2 1 0 2 1 0|1]
[2 0 2 0 0 2|0]
[1 2 0 1 2 2|1]
"""
k = t.base_ring()
n = t.nrows()
a = t.delete_columns([n])
b = t.delete_columns(range(n))
A = ~a
if as_block_matrix:
return block_matrix(k, 1, 2, [A, A * -b])
return A.augment(A * -b)
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Okay, so it is what I thought. Nice example.

( 2018-03-27 20:58:43 -0500 )edit