Linear algebra Subspaces

  • Thread starter Neoon
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  • #26
Office_Shredder
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when are saying (x*(1,-1,-1) is a solution for any x.) what do mean by that ?
We decided that if (a,b,c) is a solution, then b=-a, c=a. So the solution really looks like (a,-a,a) for any a. We can factor out the a to get a*(1,-1,1) which is just the set of linear combinations of the basis of a straight line where the basis is (1,-1,1)

True moo, it needs to have a vector in it also

Neonn, you're on the right track in your first post.

Like you said, x+y=w solves the case of s=0. (note z can be whatever you want

So our solution looks like (x,y,z,x+y)

Or we can make that basis by separating it

x*(1,0,0,1) + y*(0,1,0,1) + z*(0,0,1,0) (you should see or be able to prove that's a basis) and you can see summming those vectors gives you a solution to the original equation for any x,y,z.

Do the same for s=2
 
  • #27
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So the basis of the subspace when s=0 is:
x*(1,0,0,1), y*(0,1,0,1), z*(0,0,1,0)

And similarly the basis when s=2 is:
x*(1,0,0,1), y*(0,1,0,1), z*(0,0,1,2)

Both with dimension= 3

As per your comment before: I became an expert in subspaces, now, am I an expert in dimensions and bases or not yet?:smile:
 
  • #28
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That looks good to me

You can see how this method is much preferred to the one in your post (at least for simple problems like this).
 
  • #29
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Thanks Office Shredder
Thanks very much. I really appreciate your valuable help
 

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