How Do You Find the Basis and Equation for Specific Subspaces in R^3?

  • Thread starter Jimmy84
  • Start date
  • Tags
    Subspaces
In summary, the first problem cannot be solved, the second problem asks to find a set B in R^3 such that each vector in S is a linear combination of vectors in B, and the third problem asks for how to write three equations as one vector equation.
  • #1
Jimmy84
191
0

Homework Statement



-Problem number 1

Given the set {u ,v} , where u=(1,2,1) and v=(0,-1,3) in R^3 find an equation for the space generated by this set.

-Problem number 2

The subspace S is defined as S= {(x,y,z) : x + 2y - z =0}
find a set B={u,v} in R^3 such that each vector in S is a linear combination of vectors in B.

Homework Equations





The Attempt at a Solution



I have no idea how to solve problem number 2.

I don't know how to find an equation for the space in problem one

I started the problem like this (a,b,c) =(1,2,1)x + (0,-1,3)y

and then I found x and y in terms of a and b. but I don't have an idea what is meant to find an equation for the space.

I would appreciate some help, thanks a lot.
 
Physics news on Phys.org
  • #2
  1. How about su+tv for real s and t?
  2. Find two vectors in the plane. To do this, try z=0, et cetera...
 
  • #3
algebrat said:
  1. How about su+tv for real s and t?
  2. Find two vectors in the plane. To do this, try z=0, et cetera...

the answer for the first problem is 7x -3y -z = 0 but I don't know how to solve for that

what do you mean s and t?
 
  • #4
Jimmy84 said:
the answer for the first problem is 7x -3y -z = 0 but I don't know how to solve for that

what do you mean s and t?

  1. s and t would give the parametrized version of the plane. To get the implicit version they give, try the cross product, which gives the normal.
 
  • #5
Jimmy84 said:
-Problem number 2

The subspace S is defined as S= {(x,y,z) : x + 2y - z =0}
find a set B={u,v} in R^3 such that each vector in S is a linear combination of vectors in B.
If you have three variables and only one equation, you can solve for one variable in terms of the others. For example, if you had x-2y-3z=0, you could solve for x and get x=2y+3z. Now let y=s and z=t, where s and t are your free parameters, so you have
\begin{align*}
x &= 2s + 3t \\
y &= s \\
z &= t
\end{align*} Can you see how to write those three equations as one vector equation?
 

1. What is a subspace?

A subspace is a subset of a vector space that satisfies the properties of a vector space. This means that it is closed under addition and scalar multiplication, and contains the zero vector.

2. How do you determine if a set is a subspace?

To determine if a set is a subspace, you must check if it satisfies the three properties of a vector space: closure under addition and scalar multiplication, and contains the zero vector. If all three properties are satisfied, then the set is a subspace.

3. Can a subspace be empty?

Yes, a subspace can be empty. This happens when the set does not contain the zero vector. However, it is important to note that the empty set itself is not a subspace.

4. What is the dimension of a subspace?

The dimension of a subspace is the number of linearly independent vectors in the subspace. This represents the minimum number of vectors needed to span the subspace.

5. How do you find a basis for a subspace?

To find a basis for a subspace, you can use the row reduction method to find the linearly independent vectors that span the subspace. These vectors will form the basis for the subspace.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
0
Views
450
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
15
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
5K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
463
Back
Top