How to check if m n-sized vectors are linearly independent?
Solution 1
Construct a matrix of the vectors (one row per vector), and perform a Gaussian elimination on this matrix. If any of the matrix rows cancels out, they are not linearly independent.
The trivial case is when m > n, in this case, they cannot be linearly independent.
Solution 2
Construct a matrix M
whose rows are the vectors and determine the rank of M
. If the rank of M
is less than m
(the number of vectors) then there is a linear dependence. In the algorithm to determine the rank of M
you can stop the procedure as soon as you obtain one row of zeros, but running the algorithm to completion has the added bonanza of providing the dimension of the spanning set of the vectors. Oh, and the algorithm to determine the rank of M
is merely Gaussian elimination.
Take care for numerical instability. See the warning at the beginning of chapter two in Numerical Recipes.
Solution 3
If m<n
, you will have to do some operation on them (there are multiple possibilities: Gaussian elimination, orthogonalization, etc., almost any transformation which can be used for solving equations will do) and check the result (eg. Gaussian elimination => zero row or column, orthogonalization => zero vector, SVD => zero singular number)
However, note that this question is a bad question for a programmer to ask, and this problem is a bad problem for a program to solve. That's because every linearly dependent set of n<m
vectors has a different set of linearly independent vectors nearby (eg. the problem is numerically unstable)
Solution 4
I have been working on this problem these days.
Previously, I have found some algorithms regarding Gaussian or Gaussian-Jordan elimination, but most of those algorithms only apply to square matrix, not general matrix.
To apply for general matrix, one of the best answers might be this: http://rosettacode.org/wiki/Reduced_row_echelon_form#MATLAB
You can find both pseudo-code and source code in various languages. As for me, I transformed the Python source code to C++, causes the C++ code provided in the above link is somehow complex and inappropriate to implement in my simulation.
Hope this will help you, and good luck ^^
Solution 5
If computing power is not a problem, probably the best way is to find singular values of the matrix. Basically you need to find eigenvalues of M'*M
and look at the ratio of the largest to the smallest. If the ratio is not very big, the vectors are independent.
Comments
-
tunnuz almost 2 years
Disclaimer
This is not strictly a programming question, but most programmers soon or later have to deal with math (especially algebra), so I think that the answer could turn out to be useful to someone else in the future.Now the problem
I'm trying to check if m vectors of dimension n are linearly independent. If m == n you can just build a matrix using the vectors and check if the determinant is != 0. But what if m < n?Any hints?
See also this video lecture.
-
tunnuz over 15 yearsCould you please explain better your solution? I should execute a gaussian elimination on what exactly?
-
Pierre over 15 yearsLet's say you have the 2 vectors (2 3) (4 6). They map to the following set of equations:
2x + 3y = a
and4x + 6y = b
. If you try a gaussian elimination of x, you end up with0x + 0y = 2a - b
. Having the zeros indicates that the two vectors are not independant. Generalize forM
andN
. -
tunnuz over 15 yearsCan I use partial pivoting with this gaussian elimination?
-
BetweenTwoTests over 15 yearsTrue, but that just means every output of the algorithm on dependent set is somewhat bogus
-
Tyson Williams almost 13 yearsAre you sure it is the transpose and not the conjugate transpose?
-
Thomas Ahle about 10 yearsSo this has at least
O(n^3)
complexity (depending on how we measure). Do you know if there is a (perhaps randomized)O(n^2)
algorithm? -
Pranav over 9 yearsCan you give any reference(s) on this approach of randomly removing components of vector?
-
Karlo about 7 yearsHow do you define 'not very big'?