Basis for the intersection of two spans

1. Aug 18, 2004

Try hard

Let S and T be two spans of vectors, what's the general method to find a basis for the intersection of S and T (SnT)? Thanks

2. Aug 19, 2004

Muzza

My thoughts:

Suppose, without loss of generality, that the vectors that span S and T are linearly independant (otherwise, we may remove the dependant vectors without changing the span). (I'm also assuming that the vectors come from the same vector space). Let S = Span{s_1, ..., s_n} and T = Span{t_1, ..., t_m}.

The intersection of S and T = {x; x = s_1a_1 + ... + s_na_n and x = t_1b_1 + ... t_mb_m, for some scalars a_1, ..., a_n, etc} = {the set of all solutions in a_1, etc to s_1a_1 + ... + s_na_n = t_1b_1 + ... t_mb_m}. Thus, finding a basis for the intersection is equivalent to finding a basis for the solution space of that particular equation. While I don't know how to proceed in the general case, it should be pretty easy for R^n, say. An example in R^3:

S = Span{ [1,1,1], [3,0,2] }
T = Span{ [0,2,1], [5,1,0] }

We're looking for a basis for the solution space to this equation:

a[1,1,1] + b[3,0,2] - c[0,2,1] - d[5,1,0] = [0,0,0]

<=>

[a + 3b + 5d, a - 2c + d, a + 2b - c] = [0,0,0]

<=>

{ a + 3b + 5d = 0
{ a - 2c - d = 0
{ a + 2b - c = 0

<=>

{ a = -17t
{ b = 4t
{ c = -9t
{ d = t, t is any scalar.

I.e. any vector in the intersection can be written on the form a[1,1,1] + b[3,0,2] = -17t[1,1,1] + 4t[3,0,2] = t[5,-17,-9], so { [5, -17, -9] } is a basis for the intersection.

Last edited: Aug 19, 2004
3. Sep 23, 2004

mathwonk

i.e. write equations for both spaces then find a basis of the simultaneous solutions of all the equations.

e.g. if V1,...,Vn is a basis for one space, and if W1,...,Wm is a basis for the second space, first find a basis X1,...,Xr for the space of vectors perpendicular to all the V's.

Then find a basis Y1,...Ys for the vectors perpendicular to all the W's.

Then finally find a basis for all the vectors perpendicular to all the X's and Y's.