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Function Repository Resource:

 IntersectionBasis

Source Notebook

Find a basis for the intersection of subspaces of ℝ^n

Contributed by: Dennis M Schneider

ResourceFunction["IntersectionBasis"][list1,list2,…]

finds a basis for the intersection of the subspaces of spanned by list1,list2, ….

Details and Options

The method is based on expressing each subspace as the null space of a matrix and then finding the null space of the matrix obtained by stacking these matrices vertically.
The empty list spans the trivial subspace.

Examples

Basic Examples (4) 

The intersection of two three-dimensional subspaces of :

In[1]:=
ResourceFunction[ "IntersectionBasis"][{{1, 1, 1, 1}, {1, 2, 3, 4}, {0, 1, 2, 2}}, {{2, 3, 4, 7}, {1, 0, 1, 0}, {0, 1, 0, 3}}]
Out[1]=

Two subspaces are called disjoint if their intersection is the trivial space, and a basis for the trivial subspace is the empty list. These two subspaces are disjoint:

In[2]:=
ResourceFunction[ "IntersectionBasis"][{{1, 1, 1, 1}, {1, 2, 3, 4}}, {{2, 3, 4, 7}}]
Out[2]=

All vectors in the lists must have the same size:

In[3]:=
ResourceFunction[ "IntersectionBasis"][{{1, 1, 1, 1}, {1, 2, 3, 4}}, {{2, 3, 4, 7, 8}}]
Out[3]=

Again, the vectors in the list must have the same size; the sizes of the vectors in the second list are 4 and 0:

In[4]:=
ResourceFunction[ "IntersectionBasis"][{{1, 1, 1, 1}, {1, 2, 3, 4}}, {{2, 3, 4, 5}, {}}]
Out[4]=

Compare with the following. The first list has two elements, the second has one, and the third also has one:

In[5]:=
ResourceFunction[ "IntersectionBasis"][{{1, 1, 1, 1}, {1, 2, 3, 4}}, {{2, 3, 4, 5}}, {{}}]
Out[5]=

Publisher

Dennis M Schneider

Version History

  • 1.0.0 – 24 September 2019

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License Information