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

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Homework Help Overview

The discussion revolves around determining the dimension of the span of three vectors in R^3, specifically examining the vectors x₁, x₂, and x₃. The original poster presents a calculation involving the determinant of a matrix formed by these vectors, initially concluding that they are linearly independent and thus suggesting that the span should have a dimension of 3. However, there is a discrepancy with the answer provided in their textbook, which states the dimension is 2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of a zero determinant, questioning the linear independence of the vectors. Some suggest using row reduction as an alternative method to clarify the relationships between the vectors. Others explore the conditions under which the span could have dimensions of 1 or 2.

Discussion Status

There is an ongoing exploration of the relationships between the vectors, with some participants providing guidance on interpreting the determinant's result. Multiple interpretations of the span's dimension are being considered, and participants are actively questioning assumptions about linear combinations and independence.

Contextual Notes

Participants note that the original poster miscalculated the determinant, which is central to the discussion. There is also mention of the constraints imposed by homework rules and the need for careful consideration of vector relationships in determining the span's dimension.

amolv06
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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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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.
 
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
 
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.
 
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
 
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.
 
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
 

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