How to find bases for subspaces

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Homework Statement



I'm given a subspace in F^5 (not sure how to note that online) and asked to find a basis and dimension for it. I know it should be really easy, but ...

Homework Equations



We're given subspace W1 = {a1,a2,a3,a4,a5) in F^5: a1-a3-a4=0} .

We also know from linear algebra that dim (W) </= dim (V), and that we know the dimension from the number of vectors in the basis.

The Attempt at a Solution



From the given "constraint", if I can call it that, I can put a1 in terms of a3 and a4 such that a1 = a3 + a4. It seems that from this point, there are 3 basic variables (a1, a2, and a5) and 2 free variables (a3, a4). From this point I made a sort of matrix configuration that

(a1,a2,a3,a4,a5) = t1 (1,0,1,0,1) + t2 (0,1,0,0,0) + t3 (0,0,1,0,0) + t4 (0,0,0,1,0) where t1,t2,t3,t4 are just arbitrary parameters -- I guess coefficients of the linear combination made by these vectors.

I guess I'm confused about this sort of problem compared to the theoretical part of this math -- obviously I don't quite understand what's going on here. :uhh:

I also know that the dimension of W1 must be less than or equal to 5 (it would be 4 for the basis I made above), since it is in the vector space F^5 and dim (W1) </= dim (V). The dimension part I can figure out, it's just finding the basis that I'm lost. I guess my answer could be right, but it doesn't match what any of my classmates have, so I'm assuming this is wrong. Any help/explanation of finding bases is much needed/appreciated!
 

Answers and Replies

  • #2
radou
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A vector in W is of the form (a3+a4, a2, a3, a4, a5). So, this implies (a1, a2, a3, a4, a5) = a3(1, 0, 1, 0, 0) + a4(1, 0, 0, 1, 0) + a2(0, 1, 0, 0, 0) + a5(0, 0, 0, 0, 1). So the set of vectors {(1, 0, 1, 0, 0),(1, 0, 0, 1, 0),(0, 1, 0, 0, 0),(0, 0, 0, 0, 1)} is a candidate for a basis. Of course, one has to show that the set is linearly independent (which is quite obvious here, but anyway).
 
  • #3
HallsofIvy
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You have the equation a1- a2- a3- a4= 0 so a1= a2+ a3+ a4. That means you can pick a2, a3, a4 arbitrarily and calculate a1. Your subspace is 3 dimensional.

An easy way to find a basis is to let each of a2, a3, and a4 equal 1 in turn while the others are 0.
First a2= 1, a3= a4= 0 so a1= 1. A basis vector is (1, 1, 0, 0).
Then a3=1, a2= a4= 0 so a1= 1. A basis vector is (1, 0, 1, 0).
Now a4= 1, a2= a3= 0 so a1= 1. A basis vector is (1, 0, 0, 1).
 
  • #4
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this makes sooo much more sense. So for a similar problem, also in F4, where the subspace W is given such that a2=a3=a4 and a1=-a5, every vector in W can be written as (a1,a2,a3,a4,a5)=(-a5,a2,a2,a2,a5)=a5(-1,0,0,0,1) + a2 (0,1,1,1,0). Then a basis is given by {(-1,0,0,0,1),(0,1,1,1,0)} and the dim (W)=2. At least I think so...
 
  • #5
radou
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this makes sooo much more sense. So for a similar problem, also in F4, where the subspace W is given such that a2=a3=a4 and a1=-a5, every vector in W can be written as (a1,a2,a3,a4,a5)=(-a5,a2,a2,a2,a5)=a5(-1,0,0,0,1) + a2 (0,1,1,1,0). Then a basis is given by {(-1,0,0,0,1),(0,1,1,1,0)} and the dim (W)=2. At least I think so...
That's correct.
 
  • #6
HallsofIvy
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this makes sooo much more sense. So for a similar problem, also in F4, where the subspace W is given such that a2=a3=a4 and a1=-a5, every vector in W can be written as (a1,a2,a3,a4,a5)=(-a5,a2,a2,a2,a5)=a5(-1,0,0,0,1) + a2 (0,1,1,1,0). Then a basis is given by {(-1,0,0,0,1),(0,1,1,1,0)} and the dim (W)=2. At least I think so...
Except that it is not in F4, it is in F5, or you wouldn't have that "a5"! Notice that you have 3 equations, a1=-a5, a2= a3, and a2= a4, (of course a3= a4 but that is not an independent equation) which reduces the dimension from 5 to 5-3= 2.
 

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