Finding a basis for the span of 4 vectors.

In summary, the conversation discusses different methods for finding a basis for a given set of vectors. These include using the determinant, the gram-schmidt process, and row-reducing a matrix formed from the vectors. The end goal is to determine which vectors are linearly independent and can be used as a basis for the subspace spanned by the given vectors.
  • #1
M55ikael
17
0
Hi!

Homework Statement


I can't for the life of me figure out how to do this. I need to find the basis for the span of these four vectors:

V1= 3, 1, -2, -4
V2 = -5, -3, 5, 9
V3 = 5, -1, 0, -2
V4 = -1, 5 -6 -8

2. The attempt at a solution

I've figured out that the determinant is zero, so in its current form it's definitely not a basis. At least one of the vectors are obsolete, and they don't span R4. I was going to take away a vector at random and see if the remaining vectors were linearly dependant or not, but I'm not sure how to prove that a non square matrix is linearly independent.

Example: Say I take away one vector and use gauss-jordan elimination on the matrix comprised of the remaining vectors. I would eventually end up with all the constants equal to zero, or one or more rows of zeros. But if I ended up with an identity matrix, it wouldn't actually prove that the ONLY solution was that alle the constants are zero, would it?

This is my first post , so please be kind ;-) Thanks for helping me out!

EDIT: Sorry, this might have been better suited for the physics sub-forum.
 
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  • #2
Have you tried the gram-schmidt process? it'll eliminate linearly independent vectors by itself.
 
  • #3
Because they are dependent, there exist numbers, a, b, c, and d, not all 0, such that
a(3, 1, -2, -4)+ b(-5, -3, 5, 9)+ c(5, -1, 0, -2)+ d(-1, 5, -6, -8)= (0, 0, 0, 0).

That is the same as saying 3a- 5b+ 5c- d= 0, a- 3b- c+ 5d= 0, -2a+ 5b- 6d= 0, and -4a+ 9b- 2c- 8d= 0. Solve for a, b, d, and d (there will be an infinite number of solutions). You can drop a vector for which the coefficient is non-zero. Then check the remaining three to see if they are independent. If not, repeat!
 
  • #4
Thanks! It seems the problem was that I don't know how to solve solve equations with indefinite amounts of solutions.

When using Gauss-Jordan elimination I found the following:
a = 7d
b = 4d
c = 0
d = b/4

Am I supposed know that the constants (ie. a,b,c,d) are indefinite when they are dependent on each other in this manner?

I choose a=1 and find that the equations do add up, but the fact that I had to do it illustrates that I'm not used to this kind of thinking.
 
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  • #5
Another way to find a basis for the subspace spanned by the given vectors is to form a matrix with the vectors as columns in the matrix.

After forming the matrix, row-reduce it. If the vectors are linearly independent, the matrix will have no rows that are all zero. If the vectors are linearly dependent, one or more rows in the reduced matrix will consist entirely of zeros. That's what happens in this case: two of the rows have entries that are all zero.

Once the matrix is completely reduced, you can write the system of equations that the reduced matrix represents, and get a basis from the system.

For this problem, the reduced matrix is
[tex]\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right][/tex].

The system that this matrix represents is
[tex]c_1=-5c_3+7c_4[/tex]
[tex]c_2=-2c_3+4c_4[/tex]
[tex]c_3=1c_3 + 0c_4[/tex]
[tex]c_4=0c_3 + 1c_4[/tex]

The two basis vectors are the vectors whose entries are the coefficients of c3 and c4.
 
  • #6
Mark44 said:
For this problem, the reduced matrix is
[tex]\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right][/tex].

The system that this matrix represents is
[tex]c_1=-5c_3+7c_4[/tex]
[tex]c_2=-2c_3+4c_4[/tex]
[tex]c_3=1c_3 + 0c_4[/tex]
[tex]c_4=0c_3 + 1c_4[/tex]

The two basis vectors are the vectors whose entries are the coefficients of c3 and c4.

Thanks! English isn't my first language, so I didn't quite get the final sentence.

Does that mean that the basis would be:
[tex]\left[ \begin{array}{ccc}-5&7\\-2&4\\1&0\\0&1\end{array}\right][/tex].
?
 
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  • #7
What was your row-reduced matrix? I didn't get c=0.

Ignoring that for a moment, let's take your solution and (arbitrarily) set d=1. Then a=7 and b=4. Plugging this back into your original equation, you have

7v1+4v2+v4=0

or

v4=-7v1-4v2

In other words, v4 can be written as a linear combination of v1 and v2, so you can toss it.
 
  • #8
M55ikael said:
Thanks! English isn't my first language, so I didn't quite get the final sentence.

Does that mean that the basis would be:
[tex]\left[ \begin{array}{cc}-5&7\\-2&4\\1&0\\0&1\end{array}\right][/tex].
?
Sort of. A basis would be those two column vectors (not a 4 x 2 matrix).
 
  • #9
vela said:
What was your row-reduced matrix? I didn't get c=0.

Ignoring that for a moment, let's take your solution and (arbitrarily) set d=1. Then a=7 and b=4. Plugging this back into your original equation, you have

7v1+4v2+v4=0

or

v4=-7v1-4v2

In other words, v4 can be written as a linear combination of v1 and v2, so you can toss it.

Yeah I followed HallsofIvy's suggestion and tossed one of the non-zero vectors, in this case v1. Then I did the elimination again with only three vectors and arrived at this system:
a= 6/5c
b= 7/5 c
c= 5/6a

Then I tossed v2, did the elimination and found that both constants had to be zero.
But the initial elimination was probably flawed as I got c = 0, and the whole thing was just a fluke. However, I do believe the method is valid.
 
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  • #10
Mark44 said:
Sort of. A basis would be those two column vectors (not a 4 x 2 matrix).

Yup, I just don't know how to write things properly in this forum yet. Is there a tutorial btw?
Thanks for helping out everybody.
 
  • #11
You did pretty well. Writing matrices in LaTeX is one of the trickier things to do (in my limited experience at it).

Here's a page with several links: https://www.physicsforums.com/showthread.php?t=386951 [Broken]
 
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  • #12
Mark44 said:
Another way to find a basis for the subspace spanned by the given vectors is to form a matrix with the vectors as columns in the matrix.

After forming the matrix, row-reduce it. If the vectors are linearly independent, the matrix will have no rows that are all zero. If the vectors are linearly dependent, one or more rows in the reduced matrix will consist entirely of zeros. That's what happens in this case: two of the rows have entries that are all zero.

Once the matrix is completely reduced, you can write the system of equations that the reduced matrix represents, and get a basis from the system.

For this problem, the reduced matrix is
[tex]\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right][/tex].

The system that this matrix represents is
[tex]c_1=-5c_3+7c_4[/tex]
[tex]c_2=-2c_3+4c_4[/tex]
[tex]c_3=1c_3 + 0c_4[/tex]
[tex]c_4=0c_3 + 1c_4[/tex]

The two basis vectors are the vectors whose entries are the coefficients of c3 and c4.
Actually, Mark made a slight error here. Those two vectors give you a basis of the null space of the matrix. You're actually looking for a basis of the column space of the matrix. You can identify which of the original vectors form such a basis by locating where the pivots are in the reduced matrix. In this case, they're in the first two columns, so v1 and v2 would be the basis you're looking for.
 
  • #13
vela said:
Actually, Mark made a slight error here. Those two vectors give you a basis of the null space of the matrix. You're actually looking for a basis of the column space of the matrix. You can identify which of the original vectors form such a basis by locating where the pivots are in the reduced matrix. In this case, they're in the first two columns, so v1 and v2 would be the basis you're looking for.

What's a pivot?
 
  • #14
A pivot is the leading 1 in a row after the matrix has been reduced to row-echelon form.
 
  • #15
vela said:
A pivot is the leading 1 in a row after the matrix has been reduced to row-echelon form.

Are there any proofs? It seems so simple I wonder why it's not mentioned in my book.
 
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  • #16
Khanacademy teaches the same method as Vela described here:

I was wondering though, if it's ok for the columns to have constants other than zero above the 1, meaning you would have a row-echelon matrix as opposed to a reduced row-echelon matrix. My logic tells me that the vectors would be linearly independent anyway, so I guess it's ok.
 
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  • #17
Mark44 said:
Another way to find a basis for the subspace spanned by the given vectors is to form a matrix with the vectors as columns in the matrix.

After forming the matrix, row-reduce it. If the vectors are linearly independent, the matrix will have no rows that are all zero. If the vectors are linearly dependent, one or more rows in the reduced matrix will consist entirely of zeros. That's what happens in this case: two of the rows have entries that are all zero.

Once the matrix is completely reduced, you can write the system of equations that the reduced matrix represents, and get a basis from the system.

For this problem, the reduced matrix is
[tex]\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right][/tex].

The system that this matrix represents is
[tex]c_1=-5c_3+7c_4[/tex]
[tex]c_2=-2c_3+4c_4[/tex]
[tex]c_3=1c_3 + 0c_4[/tex]
[tex]c_4=0c_3 + 1c_4[/tex]

The two basis vectors are the vectors whose entries are the coefficients of c3 and c4.

vela said:
Actually, Mark made a slight error here. Those two vectors give you a basis of the null space of the matrix. You're actually looking for a basis of the column space of the matrix. You can identify which of the original vectors form such a basis by locating where the pivots are in the reduced matrix. In this case, they're in the first two columns, so v1 and v2 would be the basis you're looking for.

Vela is right. The two vectors I showed are a basis for the nullspace, which isn't what you're after.

However, the system of equations above (with the ci's) can be used to get basis vectors for the spanning set. In essence, the row reduction goes back to finding solutions to the equation c1v1 + c2v2 + c3v3 + c4v4 = 0, which is what HallsofIvy was saying back in post #3 or so. The vi's are the vectors we started with.

Since we ended up with a couple of rows of zeros in the row-reduced matrix, the four original vectors aren't linearly independent. Earlier, vela showed that two vectors would suffice to span the subspace spanned by the four given vectors.

If you look at the equations for ci that I gave, you can see that all four constants depend on c3 and/or c4. If you arbitrarily set c3 to 1 and c4 to 0, you can see that c1 = -5 and c2 = -2. This means that -5v1 -2v2 + 1v3 = 0, so you can solve for v3 as a linear combination of v1 and v2.

Now, if you set c3 to 0 and c4 to 1, you can see that c1 = 7 and c2 = 4. This means that 7v1 + 4v2 + 1v4 = 0, so you can solve for v4 as a linear combination of v1 and v2.

Since both v3 and v4 are linear combinations of the other two vectors, we can discard them, leaving us with v1 and v2. These vectors span the space spanned by all four vectors, and they are linearly independent (by inspection - neither is a multiple of the other), so they are a basis for the subset of R4 spanned by the original four vectors.
 

1. What is a basis for the span of 4 vectors?

A basis for the span of 4 vectors is a set of 4 linearly independent vectors that can be used to represent any vector in the same vector space.

2. How do you find a basis for the span of 4 vectors?

To find a basis for the span of 4 vectors, you can use the method of Gaussian elimination to reduce the vectors to their row echelon form. The pivot columns of the reduced matrix will form the basis for the span.

3. Can a set of 4 vectors always form a basis for their span?

No, a set of 4 vectors may not always form a basis for their span. They must be linearly independent in order to form a basis. If they are not, it is possible to reduce the set to a smaller number of linearly independent vectors that will still span the same space.

4. What is the importance of finding a basis for the span of 4 vectors?

Finding a basis for the span of 4 vectors is important because it allows us to represent any vector in the same vector space using a smaller set of linearly independent vectors. This can make calculations and operations easier and more efficient.

5. Are there different methods for finding a basis for the span of 4 vectors?

Yes, there are different methods for finding a basis for the span of 4 vectors. Some other methods include using determinants or finding the null space of the matrix formed by the vectors. However, Gaussian elimination is often the most efficient method.

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