Finding a basis for the span of 4 vectors.

[M55ikael]

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.

 

[zcd]

Have you tried the gram-schmidt process? it'll eliminate linearly independent vectors by itself.

 

[HallsofIvy]

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!

 

[M55ikael]

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.

 

[Mark44]

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
\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right].

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

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

 

[M55ikael]

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

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

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:
\left[ \begin{array}{ccc}-5&7\\-2&4\\1&0\\0&1\end{array}\right].
?

 

[vela]

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.

 

[Mark44]

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

Does that mean that the basis would be:
\left[ \begin{array}{cc}-5&7\\-2&4\\1&0\\0&1\end{array}\right].
?

Sort of. A basis would be those two column vectors (not a 4 x 2 matrix).

 

[M55ikael]

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.

 

[M55ikael]

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.

 

[Mark44]

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

 

[vela]

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
\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right].

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

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.

 

[M55ikael]

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?

 

[vela]

A pivot is the leading 1 in a row after the matrix has been reduced to row-echelon form.

 

[M55ikael]

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.

 

[M55ikael]

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.

 

[Mark44]

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
\left[ \begin{array}{cccc}1&0&5&-7\\0&1&2&-4\\0&0&0&0\\0&0&0&0\end{array}\right].

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

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.

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 c_i'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 v_i'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 c_i 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.