Linear algebra. completely lost.?

  • Thread starter ykaire
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  • #1
ykaire
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1. i am given a matrix
A= 1 0 2 1
1 1 3 1
2 3 8 -2
-3 3 -5 1
and then it asks why do I know that span {a1, a2, a3, a4} are a subset of R^4.



Homework Equations





The Attempt at a Solution


Is it as easy as saying "because there 4 vectors?"
 

Answers and Replies

  • #2
chiro
Science Advisor
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Hey ykaire and welcome to the forums.

Have you reduced your system down to row echelon form? What can you conclude from this?
 
  • #3
ykaire
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hi! thank you :)

ok, I just reduced the matrix and i got:

1 0 2 1
0 1 1 0
0 0 1 -4
0 0 0 0

This means that it is consistent, so a solution does exist. So, would that mean that the span would also exist in R4?
 
  • #4
chiro
Science Advisor
4,815
134
hi! thank you :)

ok, I just reduced the matrix and i got:

1 0 2 1
0 1 1 0
0 0 1 -4
0 0 0 0

This means that it is consistent, so a solution does exist. So, would that mean that the span would also exist in R4?

Yes this shows that you have three linearly independent vectors in R^4.

Now the next thing you have to answer is how many independent parameters you need to describe the span of this system.

Hint: You will have to start off with one free parameter. Can you show by looking at the results how many free parameters you need to describe the span?
 
  • #5
ykaire
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I can tell that there will be one free variable.
 
  • #6
ykaire
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but I'm not exactly sure what to do with it?
do i have to solve the other parameters in terms of the free parameter?
 
  • #7
ykaire
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Yes this shows that you have three linearly independent vectors in R^4.

Now the next thing you have to answer is how many independent parameters you need to describe the span of this system.

Hint: You will have to start off with one free parameter. Can you show by looking at the results how many free parameters you need to describe the span?

but I'm not exactly sure what to do with it?
do i have to solve the other parameters in terms of the free parameter?
 
  • #8
36,702
8,694
Assuming that a1, a2, a3 and a4 are the columns of A, all you need to say to answer the original question is to note that all four vectors are elements of R4, so of course they span a subset of R4.

To answer the question asked in post #1, it's not because there are four vectors, but it is because each vector belongs to R4, so the set of all linear combinations of these vectors (i.e., the span of these vectors) represents some subset (subspace in fact) of R4.

You don't need to row-reduce the matrix or do any of that other stuff.
 
  • #9
chiro
Science Advisor
4,815
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but I'm not exactly sure what to do with it?
do i have to solve the other parameters in terms of the free parameter?

The spanning set is a set of vectors where any linear combination of the those vectors in the spanning set gives a vector in your space.

You did a row reduction and you got three vectors: this means that there are three linearly independent vectors given the vectors you start off (the space you are spanning), and all of them are from R^4.

So with in this mind, what do you think the spanning set will be?
 
  • #10
ykaire
15
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Assuming that a1, a2, a3 and a4 are the columns of A, all you need to say to answer the original question is to note that all four vectors are elements of R4, so of course they span a subset of R4.

To answer the question asked in post #1, it's not because there are four vectors, but it is because each vector belongs to R4, so the set of all linear combinations of these vectors (i.e., the span of these vectors) represents some subset (subspace in fact) of R4.

You don't need to row-reduce the matrix or do any of that other stuff.

oh, ok thank you so much :)
 
  • #11
36,702
8,694
The spanning set is a set of vectors where any linear combination of the those vectors in the spanning set gives a vector in your space.

You did a row reduction and you got three vectors: this means that there are three linearly independent vectors given the vectors you start off (the space you are spanning), and all of them are from R^4.

So with in this mind, what do you think the spanning set will be?
Note that the problem didn't ask for a spanning set.
ykaire said:
why do I know that span {a1, a2, a3, a4} are a subset of R^4
 
  • #12
ykaire
15
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oh, ok thank you so much :)

The spanning set is a set of vectors where any linear combination of the those vectors in the spanning set gives a vector in your space.

You did a row reduction and you got three vectors: this means that there are three linearly independent vectors given the vectors you start off (the space you are spanning), and all of them are from R^4.

So with in this mind, what do you think the spanning set will be?

now, i am also asked if R^4 = span {a1, a2, a3, a4}
what is the difference between this question and the last one?
 
  • #13
ykaire
15
0
Note that the problem didn't ask for a spanning set.

yeah, i see.
now, i am also asked if R^4 = span {a1, a2, a3, a4}
what is the difference between this question and the last one?
 
  • #14
chiro
Science Advisor
4,815
134
now, i am also asked if R^4 = span {a1, a2, a3, a4}
what is the difference between this question and the last one?

Now you should look at your row reduced matrix and assess whether this set of spanning vectors allows you to descrive all possible vectors in R^4.

Hint: What's the minimum number of vectors in a basis to describe a four-dimensional vector? How many are in the spanning set?
 
  • #15
ykaire
15
0
Now you should look at your row reduced matrix and assess whether this set of spanning vectors allows you to descrive all possible vectors in R^4.

Hint: What's the minimum number of vectors in a basis to describe a four-dimensional vector? How many are in the spanning set?

the minimum number of vectors would be 4?
and there are 4 (?) vectors in my spanning set, i believe?
so, does that mean that R4 is indeed a span {a1, a2, a3, a4}?
 
  • #16
36,702
8,694
To help you think about this from a geometric perspective, given any four vectors a1, a2, a3, and a4 in R4, Span{a1, a2, a3, a4 } is some subset of R4.

If all four vectors are the zero vector, the span of this set will be the zero vector.
If all four vectors are nonzero but are nonzero multiples of each other, the span of this set will be a one dimensional subset (subspace actually) of R4. IOW, the span will be a line in R4 that goes through the origin.

If two of the vectors are linear combinations of the other two, and the other two are not multiples of each other, then the span of the set will be a two-dimensional subset - a plane in R4.

If there are three vectors that don't lie in the same plane, and the fourth is a linear combination of the others, then the span of the set is a three-dimensional subset of R4.

Finally, if none of the vectors is a linear combination of any of the remaining vectors, then the span of the set is R4 itself.
 

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