The dimension of the span of three linearly independent R^3 vectors

In summary: I was unclear in my previous post. I meant to refer to the first two cases. If the dimension is 1, then all three vectors are multiples of one another. If the dimension is 0, then all three vectors are zero vectors.
  • #1
amolv06
46
0

Homework Statement



Given

[tex]x_{1}= \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, x_{2}= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, x_{3}= \begin{pmatrix} 2 \\ 6 \\ 4 \end{pmatrix}[/tex]

Find the dimension of [tex]Span(x_{1}, x_{2}, x_{3})[/tex]



Homework Equations



If V is a vector space of dimension n > 0 then any set of n linearly independent vectors spans V.

If the x vectors are combined into a matrix and the matrix is nonsingular then the vectors are linearly independent.

[tex]dim R^{3} = 3[/tex]



The Attempt at a Solution



[tex]det \begin{vmatrix} 2 & 3 & 2 \\ 1 & -1 & 6 \\ 3 & 4 & 4 \end{vmatrix} = -28.[/tex]

This shows that the three vectors are linearly independent. This should mean that any vector in [tex]R^{3}[/tex] should be able to be represented as a linear combination of the three given vectors. Therefore, shouldn't [tex]Span(x_{1}, x_{2}, x{3})[/tex] have a dimension of 3? The answers section in the back of my book says it has 2 dimensions. I can't quite figure out why.
 
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  • #2
Heh, I'm being silly. I miscalculated the determinant. It should be 0. Looks like we can represent anyone of the vectors with any of the other 2.
 
  • #3
amolv06 said:

Homework Statement



Given

[tex]x_{1}= \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}, x_{2}= \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix}, x_{3}= \begin{pmatrix} 2 \\ 6 \\ 4 \end{pmatrix}[/tex]

Find the dimension of [tex]Span(x_{1}, x_{2}, x_{3})[/tex]







Homework Equations



If V is a vector space of dimension n > 0 then any set of n linearly independent vectors spans V.

If the x vectors are combined into a matrix and the matrix is nonsingular then the vectors are linearly independent.

[tex]dim R^{3} = 3[/tex]



The Attempt at a Solution



[tex]det \begin{vmatrix} 2 & 3 & 2 \\ 1 & -1 & 6 \\ 3 & 4 & 4 \end{vmatrix} = -28.[/tex]

This shows that the three vectors are linearly independent. This should mean that any vector in [tex]R^{3}[/tex] should be able to be represented as a linear combination of the three given vectors. Therefore, shouldn't [tex]Span(x_{1}, x_{2}, x{3})[/tex] have a dimension of 3? The answers section in the back of my book says it has 2 dimensions. I can't quite figure out why.



what's wrong is that your calc is wrong----> det=0 in this case,

do it another time, pay attection to the signs.

regards
marco
 
  • #4
Be careful. Knowing that the determinant is 0 only tells you that the dimension of the span is less than 3. Either one of the vectors can be written as a combination of the other two (and so the span has dimension 2) or two can be written as multiples of the other one (in which case the span would have dimension 1). I would use row reduction rather than finding the determinant to distinguish those possibilities.
 
  • #5
HallsofIvy said:
Be careful. Knowing that the determinant is 0 only tells you that the dimension of the span is less than 3. Either one of the vectors can be written as a combination of the other two (and so the span has dimension 2) or two can be written as multiples of the other one (in which case the span would have dimension 1). I would use row reduction rather than finding the determinant to distinguish those possibilities.

Yes you are right, in general is like you say; but with three vector if the determinant is zero you can be sure that two are independent just looking at the numbers, i.e. choose two that are not multiple of each others. In this case is easy.

ciao
marco
 
  • #6
HallsofIvy said:
Either one of the vectors can be written as a combination of the other two (and so the span has dimension 2)

I'm actually working a problem similar to this, could you please explain this more? Does writing something as a combination involves multiplying the first vector by a scalar to get the second vector? So if I can multiply a vector by a scalar and get the other two then the dimension is 2?

I'm pretty lost.
 
  • #7
With a 3 x 3 matrix whose determinant is zero, the set of three vectors is linearly dependent. This means that one of the following must be true:
  1. One vector is a scalar multiple of another vector, such that there are two vectors that are linearly independent.
  2. One vector is a linear combination of the other two, such that there are two vectors that are linearly independent.
  3. Two vectors are (possibly different) scalar multiples of the third.
  4. All three vectors are zero vectors.
The dimensions of the span{v1, v2, v3} in these situations would be, respectively:
2
2
1
0
 

Related to The dimension of the span of three linearly independent R^3 vectors

What is the dimension of the span of three linearly independent R^3 vectors?

The dimension of the span of three linearly independent R^3 vectors is 3.

How do you determine if three vectors in R^3 are linearly independent?

To determine if three vectors in R^3 are linearly independent, you can use the determinant of the matrix formed by the three vectors. If the determinant is non-zero, then the vectors are linearly independent.

What does it mean for three vectors in R^3 to be linearly independent?

If three vectors in R^3 are linearly independent, it means that none of the vectors can be written as a linear combination of the other two. In other words, they are not collinear or coplanar.

Can three vectors in R^3 span a higher dimensional space?

No, three vectors in R^3 can only span a three-dimensional space. This is because the dimension of the span of three vectors is equal to the number of vectors.

Why is it important to have linearly independent vectors in a vector space?

Having linearly independent vectors in a vector space is important because it allows for a unique representation of any vector in that space. It also ensures that the vectors are not redundant and can form a basis for the vector space.

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