Proving Linear Independence of u, v and w

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Homework Help Overview

The discussion revolves around the concept of linear independence in vector spaces, specifically examining the statement that if vectors u, v, and w are linearly independent, then the equation au + bv + cw = 0 holds for some real numbers a, b, and c. Participants are exploring the implications of this statement and how to prove it.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are questioning the correctness of the statement regarding linear independence and discussing how to prove it. There is confusion about whether showing the trivial solution (0u + 0v + 0w = 0) suffices for proving the "for some" aspect of the statement. Additionally, a related question about proving statements involving eigenvalues and elementary row operations is raised, indicating uncertainty about the structure of such proofs.

Discussion Status

Contextual Notes

Participants are grappling with the nuances of proving statements that involve existential quantifiers ("for some") and are considering examples that may or may not align with the definitions they are working with.

annoymage
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Homework Statement




"if u,v,w are linearly independent then au+bv+cw=0 for some a,b,c in R"

first of all is this correct?

and how do i proof if this is correct?

should i show like this,

i know 0u+0v+0w=0

so that imply that statement is true?

because 0 is in R

is this correct? to prove "for some" statement?
 
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annoymage said:

Homework Statement




"if u,v,w are linearly independent then au+bv+cw=0 for some a,b,c in R"
It's true. To show that it's true, look at the definition of linear independence. A similar statement, "if u,v,w are linearly dependent then au+bv+cw=0 for some a,b,c in R", is also true.
annoymage said:
first of all is this correct?

and how do i proof if this is correct?

should i show like this,

i know 0u+0v+0w=0
The equation above is true whether u, v, and w are linearly independent or linearly dependent
annoymage said:
so that imply that statement is true?

because 0 is in R

is this correct? to prove "for some" statement?
 
i see.

but (this is other question), what if it ask to show that this statement is true "If ~~~~ then ~~~~ for some ~~~"

i'm confused

example


"if a is the eigenvalue of matrix A then a is also eigenvalue for matrix B for some elementary row operation on A to B"

i don't know if my grammar is correct. anyway here's example

let A be 2x2 matrices

1. reduce A by changing row 1 to row 2
2. again reduce A by changing row 1 to row 2

that (1) and (2) are ERO so B have the same eigenvalue with A because B=A

so is that implying that "if a is the eigenvalue of matrix A then a is also eigenvalue for matrix B for some elementary row operation on A to B" is true?

is that the way to prove "for some" statement? hep T_T
 
Let's take another look at your original question, "if u,v,w are linearly independent then au+bv+cw=0 for some a,b,c in R"

What this means is that, if u, v, and w are linearly independent, then the equation au + bv + cw = 0 has at least one solution for the constants a, b, and c.

It's also true, but not stated above, that for three linearly independent vectors, there is exactly one such solution; namely, a = b = c = 0.
 

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