Linear Independence: Is the set LI?

[Saladsamurai]

Homework Statement

Give that u and v are LI and that u and w are LI and that v and w are LI, is the set {u,v,w} LI ? Prove or disprove.

The Attempt at a Solution

I know that this can be done by providing a counterexample. But I wanted to know if there is a way to prove it generally? That is, without guessing and checking to see if a counterexample works?

And if not, is there a smart way to guess? That is, is there a way to narrow the scope of my guessing? I am thinking that letting u, v, w be vectors with lots of 0's and 1's would help, but I am not sure why.

Any thoughts? Just looking to improve my approach to these.

 

[LCKurtz]

I would think about what if the vectors are in a 2 dimensional space like R²?

 

[Saladsamurai]

I would think about what if the vectors are in a 2 dimensional space like R²?

Hmmm ... how about:

u = (0,1)
v = (1,0)
w = (1,1)

Then u and v are LI, u and w are LI, v and w are LI, but u + v = w so {u,v,w} are LD.

Is that what you were thinking?

 

[hgfalling]

You might consider trying to prove that the statement is true and see what kind of problems you run into. Or consider what kinds of cool things you could prove if that were true?

It's hard to say how to guess about these things; it's not that easy to do them systematically, and really not that fruitful in terms of developing your mathematical intuition. You want to be able to look at this problem and just jump to "oh, u,v,u+v."

 

[Saladsamurai]

You might consider trying to prove that the statement is true and see what kind of problems you run into. Or consider what kinds of cool things you could prove if that were true?

It's hard to say how to guess about these things; it's not that easy to do them systematically, and really not that fruitful in terms of developing your mathematical intuition. You want to be able to look at this problem and just jump to "oh, u,v,u+v."

I'm sorry hgfalling, but are you referring to "the statement" in the OP or in post #3 ? I can't tell if you posted during or after I replied

 

[hgfalling]

I meant the original post (ie, "pairwise LI implies the union is LI")

 

[LCKurtz]

Hmmm ... how about:

u = (0,1)
v = (1,0)
w = (1,1)

Then u and v are LI, u and w are LI, v and w are LI, but u + v = w so {u,v,w} are LD.

Is that what you were thinking?

That's the idea, but I was thinking a bit more generally. If you already know that a 2 dimensional space has a maximal linear independent set of 2 vectors, there isn't much left to do.

 

[Saladsamurai]

That's the idea, but I was thinking a bit more generally. If you already know that a 2 dimensional space has a maximal linear independent set of 2 vectors, there isn't much left to do.

Hi LCKurtz But I don't already know this Sounds kind of "mathy" to me. I don't know that I would just know that unless I actually read somewhere explicitly.