Is there a command that explicitly gives me the vectors for the image given a matrix M? If not, lets suppose I have the matrix in row echelon, N. So I create an empty list. Basically, I want to say while i< # of rows, if row i of row echelon form matrix N has a 1, put M[i] in list. But I don't know how to say if a row has a 1( a pivot).
P.s., yes I need the rows. My matrix is like transposed.
In Sage, matrices have an image method.
The image is given as a vector space with basis.
Using random coefficients in a square matrix we are likely to get an invertible matrix whose image is the whole space.
sage: a = matrix(QQ, 4, [randint(-5, 5) for _ in range(16)])
sage: a
[-5 -1 -1 0]
[ 4 -1 1 5]
[-5 -4 3 3]
[ 1 1 -1 2]
sage: a.image()
Vector space of degree 4 and dimension 4 over Rational Field
Basis matrix:
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
Now change the last line to be a combination of the first three lines.
sage: b = copy(a)
sage: b
[-5 -1 -1 0]
[ 4 -1 1 5]
[-5 -4 3 3]
[ 1 1 -1 2]
sage: b[3] = 3*b[2] + b[1] - b[0]
sage: b
[ -5 -1 -1 0]
[ 4 -1 1 5]
[ -5 -4 3 3]
[ -6 -12 11 14]
sage: b.image()
Vector space of degree 4 and dimension 3 over Rational Field
Basis matrix:
[ 1 0 0 29/33]
[ 0 1 0 -97/33]
[ 0 0 1 -16/11]
As explained by @slelievre, you can use the image method to get the row space of a matrix, that is, the vector space spanned by its rows. Then, the basis method allows to explicitly get a basis as a list of vectors. For example, consider the matrix
$$
A=\left(\begin{array}{rrrrrr}
3 & 6 & 3 & 3 & -6 & -3 \\
4 & 8 & 4 & 4 & -8 & -4 \\
3 & 6 & 0 & 0 & -6 & -9 \\
-3 & -6 & -3 & -4 & 7 & 3
\end{array}\right).
$$
We first define this matrix and get its RREF:
sage: A = matrix(QQ,[[3, 6, 3, 3, -6, -3], [4, 8, 4, 4, -8, -4],
....: [3, 6, 0, 0, -6, -9], [-3, -6, -3, -4, 7, 3]])
sage: A.rref()
[ 1 2 0 0 -2 -3]
[ 0 0 1 0 1 2]
[ 0 0 0 1 -1 0]
[ 0 0 0 0 0 0]
The first three rows form a basis of $\operatorname{row} A$. This is what the image method also points out:
sage: A.image()
Vector space of degree 6 and dimension 3 over Rational Field
Basis matrix:
[ 1 2 0 0 -2 -3]
[ 0 0 1 0 1 2]
[ 0 0 0 1 -1 0]
Let us extract the basis:
sage: A.image().basis()
[
(1, 2, 0, 0, -2, -3),
(0, 0, 1, 0, 1, 2),
(0, 0, 0, 1, -1, 0)
]
As an alternative, the pivots_row method provides the rows in the matrix that are linearly independent (please, note, in the original matrix, not in the rref):
sage: p = A.pivot_rows(); p
(0, 2, 3)
Now you can extract these rows and get a basis of $\operatorname{row} A$:
sage: A[p,:].rows()
[(3, 6, 3, 3, -6, -3), (3, 6, 0, 0, -6, -9), (-3, -6, -3, -4, 7, 3)]
For the sake of completeness, let us check that these rows really span $\operatorname{row} A$:
sage: A[p,:].image() == A.image()
True
Asked: 2020-06-01 00:52:22 +0100
Seen: 2,376 times
Last updated: Jun 02 '20
Copyright Sage, 2010. Some rights reserved under creative commons license. Content on this site is licensed under a Creative Commons Attribution Share Alike 3.0 license.
Please provide code for a matrix you care about, that can be copy-pasted in a fresh Sage session.