# Prove linearly independent

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• Jan 24th 2010, 11:13 AM
450081592
Quote:

Originally Posted by Raoh
yes i am(Nod).

I think he proving by contradiction, suppose L.D then get L.I
• Jan 24th 2010, 11:24 AM
Raoh
Quote:

Originally Posted by 450081592
I think he proving by contradiction, suppose L.D then get L.I

he didn't find any contradiction,besides the proof itself needs to be explained(Rofl).
• Jan 24th 2010, 02:38 PM
Bruno J.
One way to prove that $P \longrightarrow Q$ is to show that $- Q \longrightarrow -P$, which is what I did. If $u,v,w$ are linearly dependent then $S$ must be linearly dependent. Thus if $S$ is linearly independent we must necessarily have that $u,v,w$ are also linearly independent.

The proof is given. Which part don't you understand?
• Jan 25th 2010, 06:02 AM
HallsofIvy
Quote:

Originally Posted by Raoh
he didn't find any contradiction,besides the proof itself needs to be explained(Rofl).

He did find a contradiction- that the set of vectors $S =\{au+bv+cw,\ du+ev+fw,\ gu+hv+iw\}$ was linearly dependent when the hypothesis is that it was independent.

And I thought his proof was very clear.
• Jan 26th 2010, 04:25 AM
Raoh
if $u,v$ and $w$ were dependent, $\text{dim}_{\mathbb{R}}Span(u,v,w)\leq 2$,therefore $u,v$ and $w$ will be in $Span(u,v,w)$ with a dimension less than 2,hence they can't be independent.
is that what you're trying to say(Thinking)