How do I determine if a given set of vectors can generate a vector space?

[Shackleford]

How do I show the vectors, polynomials, and matrices generate the given sets?

A subset of a vector space generates the vector space if the span of the subset is the vector space. The span is the set of all linear combinations.

For 6, do I show the vector are linearly independent and can thus generate F³? I assume this is simply R³. What about for 6 and 8?

http://i111.photobucket.com/albums/n149/camarolt4z28/IMG_20110608_203711.jpg

 

[tylerc1991]

For #6: What is the span of {(1,1,0),(1,0,1),(0,1,1)}?

does there exist an a,b,c such that

a(1,1,0) + b(1,0,1) + c(0,1,1) = (x,y,z)

for all vectors (x,y,z) in F^3?

 

[Shackleford]

For #6: What is the span of {(1,1,0),(1,0,1),(0,1,1)} ?

I'm assuming they're linearly independent. If so, then it's R³.

 

[Shackleford]

For #6: What is the span of {(1,1,0),(1,0,1),(0,1,1)}?

does there exist an a,b,c such that

a(1,1,0) + b(1,0,1) + c(0,1,1) = (x,y,z)

for all vectors (x,y,z) in F^3?

I don't see why there couldn't be.

a + b = x
a + c = y
b + c = z

 

[tylerc1991]

I don't see why there couldn't be.

a + b = x
a + c = y
b + c = z

But what are the specific a,b, and c given an x,y, and z?

 

[Shackleford]

But what are the specific a,b, and c given an x,y, and z?

a = (1/2)(x + y - z)
b = (1/2)(x - y + z)
c = (1/2)(-x + y + z)

 

[tylerc1991]

a = (1/2)(a + b - c)
b = (1/2)(a - b + c)
c = (1/2)(- a + b + c)

I am getting the same thing, only with the a,b, and c replaced by x,y, and z respectively .

OK so you have shown that any vector in F^3 can be represented by a linear combination of (1,1,0),(1,0,1), and (0,1,1). Thus span{(1,1,0),(1,0,1),(0,1,1)} = F^3.

Also, F^3 does not equal R^3. F is a general representation of a field.

 

[Shackleford]

I am getting the same thing, only with the a,b, and c replaced by x,y, and z respectively .

OK so you have shown that any vector in F^3 can be represented by a linear combination of (1,1,0),(1,0,1), and (0,1,1). Thus span{(1,1,0),(1,0,1),(0,1,1)} = F^3.

Also, F^3 does not equal R^3. F is a general representation of a field.

Oops. I fixed it. This result seems trivial. That's why I was confused. What's a case in which you could not represent any arbitrary vector in F³ as a linear combination? If there were only two given vectors...

 

[tylerc1991]

Oops. I fixed it. This result seems trivial. That's why I was confused. What's a case in which you could not represent any arbitrary vector in F³ as a linear combination? If there were only two given vectors...

Right, if the only vectors given were (1,1,0) and (0,0,1), then we could not represent (3,2,1) for instance.

 

[Shackleford]

Right, if the only vectors given were (1,1,0) and (0,0,1), then we could not represent (3,2,1) for instance.

Or if we were given three vectors with one being relatively linearly dependent. How would that show up in determining the coefficients? Some impossible relationship? Or some common relationship?

 

[tylerc1991]

Or if we were given three vectors with one being relatively linearly dependent. How would that show up in determining the coefficients? Some impossible relationship? Or some common relationship?

You have to show that the given set of vectors spans the given vector space.

For instance, the set of vectors {(0,0,1)(0,1,0)(1,0,0)(1,1,1)} generate R^3, but they are not linearly independent.

The set of vectors {(0,1,1),(1,0,0)} are linearly independent but they do not span R^3.

 

[Shackleford]

You have to show that the given set of vectors spans the given vector space.

For instance, the set of vectors {(0,0,1)(0,1,0)(1,0,0)(1,1,1)} generate R^3, but they are not linearly independent.

The set of vectors {(0,1,1),(1,0,0)} are linearly independent but they do not span R^3.

I guess that's where I'm fuzzy. The only time a given set of vectors will not span a vector space is in a case where it's too small, i.e. not enough vectors.