Test Tomorrow need help Linear independence, spanning, basis

  • Thread starter blakpete91
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  • #1

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Hello I'm taking linear algebra and have a couple of questions about linear independence, spanning, and basis

Let me start of by sharing what I think I understand.

-If I have a matrix with several vectors and I reduce it to row echelon form and I get a pivot in every column then I can assume that each vector is linearly independent of one another.

-if I have a matrix with several vectors and I reduce it to row echelon form and I get a pivot in every row then I can assume that each vector spans the other vectors.

-If I have a matrix with several vectors and I reduce it to row echelon form and I get a pivot in every row AND column then it is considered a linearly independent basis.
- If a vector is a scalar multiple of another vector then it is linearly Dependent

Am I understanding this correctly?

Thanks, Blake
 

Answers and Replies

  • #2
RUber
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Sounds about right.
A linearly independent set of vectors have the property that the only solution to ##\sum_{i=1}^n a_i \vec v_i =0## is for all the ##a_i## coefficients to be zero. If you can reduce a set of vectors to get a row of zeros, then they are not linearly independent. As you said, a pivot in every column, would imply there is no way to make a row all zeros.
A spanning set of vectors mean that there is nothing in the vector space that can't be reached by a linear combination of the vectors in the set. As you said, if you have a pivot in each row, you should easily be able to find a set of constants ##a_i## such that ##\sum_{i=1}^n a_i \vec v_i =\vec x## for any ##\vec x## in the space.
A basis set is by definition, a linearly independent spanning set. You cannot have more vectors than the dimension of the space you plan to span, and cannot have fewer.
Linear dependence is simply not linearly independent. Of course if one vector is a scalar multiple of another then the two vectors are collinear, and dependent. However, a set of vectors is linearly dependent if you can solve the problem ##\sum_{i=1}^n a_i \vec v_i =0## with anything other than all zeros. Or, in matrix form, if you can reduce one row to all zeros.
 
  • #3
Thank you for the quick detailed response.

If what you say is true how come the following example is linearly independent if the third column does not contain a pivot?

v1 = [3,1,2]
v2= [-1 8 4]

putting into matrix form
3 ,1 ,2
-1, 8 ,4

row echelon form

1 , 0.333, 0.666
0 , 1 , 0 .56

Notice 3rd column does not have a pivot in above row echelon form.

Thank you
 
  • #4
RUber
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That's correct. Since you only have two vectors, it is sufficient to only have two pivots. Since, if you have reduce one column to a one on top and all zeros below, then you know that no linear combination of the vectors below can eliminate it.
This is an example of a linearly independent but not spanning set.
 
  • #5
Ah ok I see why it is linear independent now but I still don't understand why it wouldn't span since there is a pivot in every row.
 
  • #6
RUber
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There is no way for 2 vectors to span a 3D space.
Dimension is more important than rules about where pivots are.
You can't span a space unless you have at least as many vectors as you have dimensions.
Furthermore, if you have more vectors than the dimension of your space, it is impossible for the set to be linearly independent.
 
  • #7
Light bulb just went off. Thank you for all your help. That last part made a lot of sense.
 
  • #8
RUber
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Great. Good luck on the test.
 

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