Linear (in)dependence of vector pairs and triples

In summary, if you are seeking a counterexample to a statement "Pairs of vectors are linearly independent if and only if", an example that violates that statement is a proof.
  • #1
ibkev
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Homework Statement


Suppose that pairs of vectors ##v_1, v_2## and ##v_1, v_3## and ##v_2, v_3## are linearly independent. Must ##v_1, v_2, v_3## be linearly independent?

Homework Equations


So this means ##v_1 \neq av_2##, ##v_1 \neq bv_3## and ##v_2 \neq cv_3##

The Attempt at a Solution



For the sake of intuition, I started playing around with vectors in ##\mathbb {R}^3##. And I came up with values for ##v_1, v_2##, and ##v_3## that are independent as pairs but linearly dependent when together:

(1,2,1)
(0,1,1)
(1,1,0)

So I've answered the question by existence but this doesn't feel very satisfying, especially since I only shown this for ##\mathbb {R}^3## and not shown it in general. Do you have suggestions for a better proof?
 
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  • #2
ibkev said:

Homework Statement


Suppose that pairs of vectors ##v_1, v_2## and ##v_1, v_3## and ##v_2, v_3## are linearly independent. Must ##v_1, v_2, v_3## be linearly independent?

Homework Equations


So this means ##v_1 \neq av_2##, ##v_1 \neq bv_3## and ##v_2 \neq cv_3##

The Attempt at a Solution



For the sake of intuition, I started playing around with vectors in ##\mathbb {R}^3##. And I came up with values for ##v_1, v_2##, and ##v_3## that are independent as pairs but linearly dependent when together:

(1,2,1)
(0,1,1)
(1,1,0)

So I've answered the question by existence but this doesn't feel very satisfying, especially since I only shown this for ##\mathbb {R}^3## and not shown it in general. Do you have suggestions for a better proof?
This is a good way to do the problem, although I would have started with three vectors in ##\mathbb{R}^2##. The question "Must ##v_1, v_2, v_3## be linearly independent?" is asking whether three such vectors are always independent. By showing three vectors that are pairwise linearly independent, but not independent as a whole, you have answered the question.

If you want to extend this to a higher dimension space, just do essentially the same thing that you did to come up with your three vectors. Pick two vectors that aren't parallel (neither is a multiple of the other), and for the third vector use the sum of the other two. As long as the third vector isn't a scalar multiple of either of the other two, that's your example. The equation ##c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} = \vec{0}## will have a nonzero solution for the constants; namely ##c_1 = 1, c_2 = 1, c_3 = -1##.
 
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  • #3
Ok that makes a lot of sense - much appreciated Mark!
 
  • #4
ibkev said:

Homework Statement


Suppose that pairs of vectors ##v_1, v_2## and ##v_1, v_3## and ##v_2, v_3## are linearly independent. Must ##v_1, v_2, v_3## be linearly independent?

Homework Equations


So this means ##v_1 \neq av_2##, ##v_1 \neq bv_3## and ##v_2 \neq cv_3##

The Attempt at a Solution



For the sake of intuition, I started playing around with vectors in ##\mathbb {R}^3##. And I came up with values for ##v_1, v_2##, and ##v_3## that are independent as pairs but linearly dependent when together:

(1,2,1)
(0,1,1)
(1,1,0)

So I've answered the question by existence but this doesn't feel very satisfying, especially since I only shown this for ##\mathbb {R}^3## and not shown it in general. Do you have suggestions for a better proof?

Having an example is the best possible proof. You can extend it immediately to ##n## dimensions by appending ##n-3## zero components to the end of each of your vectors.
 
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  • #5
>What is the condition for the independence of three vectors?

The condition for linear independence is that only the trivial linear combination (where all vector coefficients are zero) is zero.

Is that what what you meant by your question?
 
  • #6
ibkev said:
>What is the condition for the independence of three vectors?

The condition for linear independence is that only the trivial linear combination (where all vector coefficients are zero) is zero.

Is that what what you meant by your question?
Yes. As @Mark44 said,
##c_1\vec{v_1} + c_2\vec{v_2} + c_3\vec{v_3} = \vec{0}## has only the trivial (all zero) solution if the vectors are independent. Assume it is not the case, and at least one of the coefficients (say c1) is not zero. Then you can choose it 1, and express ##\vec v_1 ## as linear combination of the other two vectors.
##\vec{v_1} = - c_2\vec{v_2} - c_3\vec{v_3} ##
If both coefficients differ from zero, ##\vec v_1 ## is not the multiple of any of the other two vectors.
 
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  • #7
Ah - thanks for this! I'm still working on the switch to proof oriented math from computation...
 
  • #8
ibkev said:
Ah - thanks for this! I'm still working on the switch to proof oriented math from computation...

I don't think you really understand the issue here. When you seek a counterexample to a statement, just giving an example that violates that statement IS a proof---a 100% mathematically valid proof. There is no need to try anything more "abstract" or general.
 

Related to Linear (in)dependence of vector pairs and triples

1. What is the definition of linear independence in vector pairs and triples?

Linear independence in vector pairs and triples refers to the property of two or three vectors being able to form a unique combination that results in the zero vector. This means that none of the vectors can be expressed as a linear combination of the others.

2. How is the linear independence of vector pairs and triples tested?

The linear independence of vector pairs and triples is tested using the determinant of the matrix formed by the vectors. If the determinant is equal to zero, the vectors are linearly dependent, and if it is non-zero, the vectors are linearly independent.

3. What is the significance of linear independence in vector pairs and triples?

Linear independence is significant in determining the dimension of a vector space. If the vectors are linearly independent, they form a basis for the vector space. It also allows for easier computation and simplification of vector operations.

4. Can a set of linearly dependent vectors be reduced to a linearly independent set?

Yes, a set of linearly dependent vectors can be reduced to a linearly independent set by using the process of Gaussian elimination. This involves transforming the vectors into a row-echelon form and removing any zero rows to obtain a linearly independent set.

5. How is linear independence related to linear transformations?

Linear independence is closely related to linear transformations as it determines the number of dimensions in which a linear transformation can occur. If the vectors are linearly independent, the transformation can occur in the full vector space, but if they are linearly dependent, the transformation will be restricted to a lower-dimensional subspace.

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