Basis for given subspace - check answers?

In summary: So this set is independent of the original set.basically, this is saying that the vectors in the original set (A) do not depend on the vectors in the new set (B).
  • #1
Carmen12
8
0

Homework Statement



Find the basis for the subspaces of R3 and R4 below.

Homework Equations



A) All vectors of the form (a,b,c), where a=0
B) All vectors of the form (a+c, a-b, b+c, -a+b)
C) All vectors of the form (a,b,c), where a-b+5c=0

The Attempt at a Solution



I honestly had no idea what I was doing. I just scratched what I could make of it down. I'm going to do more research online but I'm terrified of the time limit coming up so I thought I'd post my work here meanwhile. I hope you all don't mind!
Here's what I got (don't laugh please!):

A) {(0,1,1),(1,1,0)}
B) {(0,-1,0,1),(1,0,-1,0)}
C) {(5,0,1),(1,0,5)}

-Carmen :eek:
 
Physics news on Phys.org
  • #2
Carmen12 said:

Homework Statement



Find the basis for the subspaces of R3 and R4 below.

Homework Equations



A) All vectors of the form (a,b,c), where a=0

A) {(0,1,1),(1,1,0)}

-Carmen :eek:

start with A)
(a,b,c), where a=0 means

(0,b,c) for any b or c...

your first vector is in the set, but your 2nd has a = 1 so is not

it may help to picture it geometrically in which case it is the yz plane
 
  • #3
A) Basis spans your subspace. So your answer indicates, for example that,

[tex]
\left[ \begin{array}{cccc}0 & 1 \\ 1 & 1 \\ 1 & 0 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 2 \\ 5 \end{array} \right]
[/tex]

should have a solution, but under ur basis, this eqn has no solution. so ur basis cannot span this subspace

Now if we use

[tex]
\left[ \begin{array}{cccc}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ a \\ b \end{array} \right]
[/tex]

this guy has a solution for all a, b. Now you have to show this is independent. Do this by showing[tex]
\left[ \begin{array}{cccc}0 & 0 \\ 1 & 0 \\ 0 & 1 \end{array} \right] x = \left[ \begin{array}{cccc}0 \\ 0 \\ 0 \end{array} \right]
[/tex]

has only the trivial soln. which it does.
 
Last edited:

1. What is a subspace?

A subspace is a subset of a vector space that satisfies all the properties of a vector space. This means that it must contain the zero vector, be closed under vector addition and scalar multiplication, and contain all linear combinations of its vectors.

2. How do you determine if a given set of vectors form a basis for a given subspace?

To determine if a set of vectors form a basis for a given subspace, you can check if the vectors are linearly independent and if they span the entire subspace. This can be done by setting up a system of equations and solving for the coefficients of the linear combination.

3. What is the importance of a basis in linear algebra?

A basis is important in linear algebra because it allows us to represent any vector in a given subspace using a unique set of coordinates. This makes it easier to perform calculations and transformations on vectors in the subspace.

4. Can a subspace have more than one basis?

Yes, a subspace can have multiple bases. This is because a basis is not unique, and there can be many different sets of vectors that satisfy the properties of a basis for a given subspace.

5. How does the dimension of a subspace relate to the number of vectors in a basis?

The dimension of a subspace is equal to the number of vectors in a basis for that subspace. This means that the dimension of a subspace is a measure of the minimum number of vectors needed to span the entire subspace.

Similar threads

  • Calculus and Beyond Homework Help
Replies
0
Views
408
  • Calculus and Beyond Homework Help
Replies
4
Views
928
  • Calculus and Beyond Homework Help
Replies
18
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
513
  • Linear and Abstract Algebra
Replies
6
Views
710
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
Back
Top