Linear Independence: Homework Equations & Solutions

In summary, the conversation discusses two parts (a and b) of a homework problem involving linear combinations and linear independence. In part (a), the equation a*1 + b*√2 + c*√3 = 0 is analyzed and it is shown that a, b, and c must all be zero for the equation to be true. In part (b), a similar argument is made with the set {1, 1 + √5, (1 + √5)2} and it is shown that the third "vector" is a linear combination of the first two. The issue of division by zero is also briefly mentioned.
  • #1
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Homework Statement



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Homework Equations





The Attempt at a Solution



For part (a):

a*1 + b*√2 + c*√3 = 0
assume a, b, c not all zero
a + b√2 = -c√3
a2 + 2b2 + 2ab√2 = 3c2
a2 + 2b2 - 3c2 = -2ab√2
(a2 + 2b2 - 3c2)/(-2ab) = √2

which is not possible since we take a, b, c to be rational, and √2 is irrational.

thus our assumption of a, b, c not all zero was false and we must have a=b=c=0.

For part (b), a similar argument, but easier:

{1, 1 + √5, (1 + √5)2} = {1, 1 + √5, 1 + 2√5 + 5}

a + b(1 + √5) + c(1 + 2√5 + 5) = 0
assume a, b, c not all zero
1 + b + b√5 + c + 2c√5 + 5c = 0
1 + b + c + 5c = -b√5 - 2c√5
1 + b + 6c = (-b - 2c)√5
(1 + b + 6c)/(-b - 2c) = √5

same story as before.

Now that was super easy. And the assignment says to be careful with the structure of my argument. And I hate denominators because they can't be zero. If a = b = c = 0, then we have 0/0, which is of indeterminate form, which is ok! but I'm getting the idea that my argument is flawed because of division by zero...
 

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  • #2
Looks OK to me ... a,b,c not all zero is required for linear independence anyway.
 
  • #3
For part b, note that your set is {1, 1 + √5, 6 + 2√5}. Clearly the third "vector" is a linear combination of the first two in the list.
 
  • #4
yeah i was so caught up in things that i forgot to combine the 1 and 5, thanks everyone.
 

1. What is the concept of linear independence?

Linear independence refers to a set of vectors in a vector space that cannot be represented as a linear combination of other vectors in the same space. In other words, no vector in the set can be expressed as a linear combination of the other vectors in the set.

2. Why is linear independence important in mathematics and science?

Linear independence is a fundamental concept in mathematics and science, especially in fields such as linear algebra, physics, and engineering. It allows us to understand and solve systems of equations, analyze data, and determine the relationship between different variables.

3. How can I determine if a set of vectors is linearly independent?

A set of vectors is linearly independent if the only solution to the equation c1u1 + c2u2 + ... + cnun = 0, where ci are scalars and ui are the vectors, is when c1 = c2 = ... = cn = 0. In other words, if the only way to make the linear combination of the vectors equal to zero is by setting all the coefficients to zero, then the set is linearly independent.

4. What are some real-life applications of linear independence?

Linear independence has many applications in real-life situations. For example, in physics, it is used to determine the independence of forces acting on an object. In economics, it is used to analyze market trends and identify factors that contribute to economic growth. In data analysis, it is used to identify redundant variables and simplify models.

5. How can I use linear independence to solve a system of equations?

If a set of vectors is linearly independent, it means that none of the vectors can be expressed as a linear combination of the others. This property can be used to solve systems of equations by expressing each variable as a linear combination of the independent vectors. This leads to a unique solution for the system of equations.

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