# Prove linearly independent

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• Jan 20th 2010, 05:49 PM
450081592
Prove linearly independent
Suppose that u, v, w are vectors in $R^n$ such that
S = {au + bv + cw, du + ev + fw, gu + hv +iw} is linearly independent for some numbers a,b,c,d,e,f,g,h, and i. Prove that u,v,w are linearly independent.
Is the set T = {u + v + w, u + 2v + 3w, 2u + 3v + 4w, u - 4v + 2w } linearly dependent or linearly independent?
• Jan 20th 2010, 06:09 PM
kjchauhan
The set T = {u + v + w, u + 2v + 3w, 2u + 3v + 4w, u - 4v + 2w } is LD,
because 2u+3v+4w = (u+v+w)+(U+2v+3w).
• Jan 21st 2010, 09:45 AM
Runty
Quote:

Originally Posted by kjchauhan
The set T = {u + v + w, u + 2v + 3w, 2u + 3v + 4w, u - 4v + 2w } is LD,
because 2u+3v+4w = (u+v+w)+(U+2v+3w).

Thanks for solving that half of the problem. Now for the harder half...
• Jan 21st 2010, 10:21 AM
Raoh
Quote:

Originally Posted by 450081592
Suppose that u, v, w are vectors in $R^n$ such that
S = {au + bv + cw, du + ev + fw, gu + hv +iw} is linearly independent for some numbers a,b,c,d,e,f,g,h, and i. Prove that u,v,w are linearly independent.
Is the set T = {u + v + w, u + 2v + 3w, 2u + 3v + 4w, u - 4v + 2w } linearly dependent or linearly independent?

$(au+bv+cw)+(du+ev+fw)+(gu+hv+iw)=0$
$(a+d+g)u+(b+e+h)v+(c+f+i)w=0$
$S$ is L.I.
$a+d+g=0,b+e+h=0$ and $c+f+i=0$
therefore u,v,w are L.I.
• Jan 21st 2010, 07:17 PM
450081592
Quote:

Originally Posted by Raoh
$(au+bv+cw)+(du+ev+fw)+(gu+hv+iw)=0$
$(a+d+g)u+(b+e+h)v+(c+f+i)w=0$
$S$ is L.I.
$a+d+g=0,b+e+h=0$ and $c+f+i=0$
therefore u,v,w are L.I.

Here, you only prove that there is a trivil solution for u, v , w, but there are may be other solutions that satisfy this, how do you know this is the only solution?
• Jan 22nd 2010, 01:23 AM
Roam
In that case I would put my vectors into a matrix. Then I will try to determine whether the homogeneous linear system has non-trivial solutions. Keep in mind that a homogeneous linear system has only the trivial solution iff its column vectors are linearly independent. There are ways to know whether the homogeneous system has the trivial solutions.
You can show that the matrix is invertible, therefore the homogeneous linear system has ONLY the trivial solution.
• Jan 22nd 2010, 08:45 AM
450081592
Quote:

Originally Posted by Roam
In that case I would put my vectors into a matrix. Then I will try to determine whether the homogeneous linear system has non-trivial solutions. Keep in mind that a homogeneous linear system has only the trivial solution iff its column vectors are linearly independent. There are ways to know whether the homogeneous system has the trivial solutions.
You can show that the matrix is invertible, therefore the homogeneous linear system has ONLY the trivial solution.

but how do you prove invertiable, it does not specify what u, v ,w are, and we dont know how many entries are in each vectors?
• Jan 22nd 2010, 08:59 AM
Raoh
Quote:

Originally Posted by 450081592
but how do you prove invertiable, it does not specify what u, v ,w are, and we dont know how many entries are in each vectors?

Exactly,the vectors are in $\mathbb{R}^n$.
• Jan 22nd 2010, 09:35 AM
450081592
Quote:

Originally Posted by Raoh
Exactly,the vectors are in $\mathbb{R}^n$.

Right, so what you did above is not enough to prove linearly independent, is it?
• Jan 22nd 2010, 09:55 AM
Raoh
Quote:

Originally Posted by 450081592
Right, so what you did above is not enough to prove linearly independent, is it?

i believe i should let someone else handle this (Wink).
• Jan 22nd 2010, 11:17 AM
Bruno J.
Suppose $u,v,w$ are linearly dependent. Let $W=\mbox{span }\{u,v,w\}$, so $0 \leq \mbox{dim }W < 3$. Let $S =\{au+bv+cw,\ du+ev+fw,\ gu+hv+iw\}$. Since $S \subset W$, we have $0<\mbox{dim span }S < 3 = |S|$ and therefore $S$ must be linearly dependent.
• Jan 23rd 2010, 05:39 AM
Raoh
Quote:

Originally Posted by Bruno J.
Suppose $u,v,w$ are linearly dependent. Let $W=\mbox{span }\{u,v,w\}$, so $0 \leq \mbox{dim }W < 3$. Let $S =\{au+bv+cw,\ du+ev+fw,\ gu+hv+iw\}$. Since $S \subset W$, we have $0<\mbox{dim span }S < 3 = |S|$ and therefore $S$ must be linearly dependent.

would you please explain how did you do that(Rofl) ?
thanks.
• Jan 23rd 2010, 09:20 AM
Bruno J.
Are you familiar with dimension and span?
• Jan 23rd 2010, 11:07 AM
450081592
Quote:

Originally Posted by Bruno J.
Suppose $u,v,w$ are linearly dependent. Let $W=\mbox{span }\{u,v,w\}$, so $0 \leq \mbox{dim }W < 3$. Let $S =\{au+bv+cw,\ du+ev+fw,\ gu+hv+iw\}$. Since $S \subset W$, we have $0<\mbox{dim span }S < 3 = |S|$ and therefore $S$ must be linearly dependent.

did you mean linear dependent or linearly independent, we are trying to prove it independent, so it cannot be linearly dependent
• Jan 23rd 2010, 11:25 AM
Raoh
Quote:

Originally Posted by Bruno J.
Are you familiar with dimension and span?

yes i am(Nod).
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