Subspace Intersection Problem: Proving W1 and W2 in R^n

  • Thread starter Thread starter mlarson9000
  • Start date Start date
  • Tags Tags
    Subspace
mlarson9000
Messages
46
Reaction score
0

Homework Statement



Let W1 and W2 be two subspaces of R^n. Prove that their intersection is also a subspace.

Homework Equations





The Attempt at a Solution



I know that in R^2 and R^3 the intersection would be the origin, which would be the zero vector, which would be a subspace, but I don't know how to make a general argument about this.
 
Physics news on Phys.org
Let W be the intersection of W1 and W2. Is the vector space W closed under addition and scalar multiplication?
 
VeeEight said:
Let W be the intersection of W1 and W2. Is the vector space W closed under addition and scalar multiplication?

Wouldn't it have to be? If W is in both W1 and W2, which are both subspaces and therefore closed under addition and scalar multiplication, wouldn't W be also? If so, I still don't know exactly how to say that in Math speak.
 
mlarson9000 said:
Wouldn't it have to be?
Well, if not, you're going to have a tough time proving it.
mlarson9000 said:
If W is in both W1 and W2, which are both subspaces and therefore closed under addition and scalar multiplication, wouldn't W be also?
Take a couple of arbitrary vectors u1 and u2 in W, and show that their sum is also in W. Then take an arbitrary scalar s, and show that su1 is in W. That's how you would do it it "math speak."
mlarson9000 said:
If so, I still don't know exactly how to say that in Math speak.
 
mlarson9000 said:
Wouldn't it have to be? If W is in both W1 and W2, which are both subspaces and therefore closed under addition and scalar multiplication, wouldn't W be also? If so, I still don't know exactly how to say that in Math speak.
Suppose u and v are in W. Then u and v are both is W1 and, since W1 is a subspace, u+ v is in W1. Also u and v are both in W2 ...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top