How Can You Prove Linear Dependence in Vector Sets?

[vsage]

I've already found the answer to this solution but I want to check my methods because the class is very proof-based and the professor likes to take off points for style in proofs on tests:

5. Is {(1, 4, -6), (1, 5, 8), (2, 1, 1), (0, 1, 0)} a linearly independent subset of R^3? Justify your answer.

Obviously the answer is no because R^3 has a dimension of 3 and if you're given 4 generating vectors then one isn't necessary. However, I tried creating the linear system and got stuck trying to prove that for this system:

a + b + 2c = 0
4a + 5b + c + d = 0
-6a + 8b + c = 0

that at least one of a, b, c and d is nonzero and

a(1, 4, -6) + b(1, 5, 8) + c(2, 1, 1) + d(0, 1, 0) = 0

works for at least one nonzero a, b, c, d. Little help? The professor alluded to the fact that you can use Gaussian Reduction but he said we wouldn't learn how to do that for another chapter but he regularly uses it in class. The only way I know of it is studying ahead a year back or so but how do I reduce this?

 

[vsage]

Nevermind I got it.. Proved that a(1, 4, -6) + b(1, 5, 8) + c(2, 1, 1) = (0, -1, 0)

 

[M98Ranger]

To prove linear dependence, we need to show that at least one of the vectors in the set is a linear combination of the others. In other words, there exists scalars a, b, c, and d (not all zero) such that:

a(1, 4, -6) + b(1, 5, 8) + c(2, 1, 1) + d(0, 1, 0) = 0

To solve for these scalars, we can use Gaussian reduction. The first step is to set up the augmented matrix:

[1 1 2 0 | 0]
[4 5 1 1 | 0]
[-6 8 1 0 | 0]

Next, we can perform row operations to reduce this matrix to row-echelon form. This involves using elementary row operations, such as multiplying a row by a non-zero scalar, adding a multiple of one row to another, or switching the positions of two rows. The goal is to get the matrix in the form:

[1 0 0 a | 0]
[0 1 0 b | 0]
[0 0 1 c | 0]

If we are able to achieve this form, then we have found the values of a, b, and c that satisfy the equation above. If we are not able to achieve this form, then the set is linearly independent.

In this case, we can perform the following row operations to reduce the matrix:

R2 - 4R1 --> R2 (this eliminates the a term in the second row)
R3 + 6R1 --> R3 (this eliminates the a term in the third row)
R2 - 2R3 --> R2 (this eliminates the c term in the second row)

This gives us the following reduced row-echelon form:

[1 0 0 a | 0]
[0 1 0 b | 0]
[0 0 1 c | 0]

Since we are able to achieve this form, we can conclude that the set is linearly dependent. In fact, we can see that a = -2b and c = 6b, which means that for any non-zero value of b, we can find values of a and c that satisfy the equation above. Therefore, we have proven that at least one of