Finding Basis for S in LA: Solution

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They are equivalent.In summary, a basis for the given subspace S is [1/3 8/3 1 0] and [1/3 5/3 0 1], which are linearly independent and can be written as a linear combination of the variables x, y, z, and w. This can be found by solving the system of equations using substitution or row-reduction.
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Haystack
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Homework Statement



Let S = [x y z w] [itex]\in[/itex] [tex]R^4[/tex] , 2x-y+2z+w=0 and 3x-z-w=0

Find a basis for S.


Homework Equations




The Attempt at a Solution



I started by putting the system into reduced row form:

[2 -1 2 1]
[3 0 -1 -1]

[2 -1 2 1]
[0 3 -8 -5]

[6 0 -2 -2]
[0 3 -8 -5]

[1 0 -1/3 -1/3]
[0 1 -8/3 -5/3]

Now have:

x - 1/3z - 1/3w = 0
y - 8/3z - 5/3w = 0

Letting z = s, and w = t, we get:

x = 1/3s + 1/3t
y = 8/3s + 5/3t
z = s
w = t

And this gives:

s[1/3 8/3 1 0] and t[1/3 5/3 0 1]

Where the basis vectors are:

[1/3 8/3 1 0] and [1/3 5/3 0 1]

and are linearly independent.


Did I do this correctly? I'm really struggling with these concepts and I feel like I'm missing something. Thanks in advance.
 
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  • #2
I didn't check all your arithmetic, but assuming your numbers are OK, yes. That is exactly how to work the problem.
 
  • #3
Personally, I would prefer this:
2x-y+2z+w=0 and 3x-z-w=0

From the second equation, w= z- 3x
Putting that into the first equation, 2x- y+ 2z+ z- 3x= -x- y+ 3z= 0 so y= 3z- x.
Having solved for y and w in terms of x and z,
[x, y, z, w]= [x, 3z- x, z, z- 3x]= [x, -x. 0, -3x]+ [0, 3z, z, z]= x[1, -1, 0, -3]+ z[0, 3, 1, 1]
so a basis is {[1, -1, 0, -3], [0, 3, 1, 1]} which is essentially what you have, multiplied by 3.
 
  • #4
Thanks for the replies. Although, I'm not sure I understand your basis HallfofIvy. I can't seem to find how it's related to the one I ended up with. I understand how you got there with the substitution, but is it just another basis for S?
 
  • #5
It is just your method using basic algebra to solve the equations rather than "row-reduction".

Every member of this subspace is of the form <x, y, z, w> and we must have
2x-y+2z+w=0 and 3x-z-w=0

I just solved for y and w in terms of x and z and then used x and z as multipliers where you solved for x and y and used z and w as multipliers.
 

FAQ: Finding Basis for S in LA: Solution

1. What is the purpose of finding a basis for a subspace in linear algebra?

Finding a basis for a subspace in linear algebra is important because it allows us to represent the subspace in a simpler and more manageable way. By finding a basis, we can reduce the number of vectors needed to span the subspace and make it easier to perform calculations and solve problems involving the subspace.

2. How do you find a basis for a subspace in linear algebra?

To find a basis for a subspace in linear algebra, you can use the row reduction method to solve a system of equations representing the subspace. The pivot columns in the reduced row echelon form of the matrix will correspond to the basis vectors for the subspace.

3. Can there be more than one basis for a subspace in linear algebra?

Yes, there can be more than one basis for a subspace in linear algebra. This is because a subspace can be represented by an infinite number of basis sets. However, all of these basis sets will have the same number of vectors and will span the same subspace.

4. How does finding a basis for a subspace relate to linear independence?

Finding a basis for a subspace is closely related to linear independence. A basis for a subspace must consist of linearly independent vectors, meaning that none of the vectors can be written as a linear combination of the other vectors. Therefore, finding a basis is essentially finding a set of linearly independent vectors that span the subspace.

5. Can a basis for a subspace contain more vectors than the dimension of the subspace?

No, a basis for a subspace cannot contain more vectors than the dimension of the subspace. The dimension of a subspace is the maximum number of linearly independent vectors that can span the subspace, so a basis for the subspace must have the same number of vectors as its dimension.

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