Some logic

1. Sep 26, 2010

annoymage

1. The problem statement, all variables and given/known data

"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?

2. Sep 26, 2010

Staff: Mentor

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.
The equation above is true whether u, v, and w are linearly independent or linearly dependent

3. Sep 26, 2010

annoymage

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

4. Sep 26, 2010

Staff: Mentor

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.