Thread: Prove linearly independent

1. Prove linearly independent

Suppose that u, v, w are vectors in $\displaystyle R^n$ 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?

2. 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).

3. 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...

4. Originally Posted by 450081592
Suppose that u, v, w are vectors in $\displaystyle R^n$ 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?
$\displaystyle (au+bv+cw)+(du+ev+fw)+(gu+hv+iw)=0$
$\displaystyle (a+d+g)u+(b+e+h)v+(c+f+i)w=0$
$\displaystyle S$ is L.I.
$\displaystyle a+d+g=0,b+e+h=0$ and $\displaystyle c+f+i=0$
therefore u,v,w are L.I.

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

6. 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.

7. 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?

8. 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 $\displaystyle \mathbb{R}^n$.

9. Originally Posted by Raoh
Exactly,the vectors are in $\displaystyle \mathbb{R}^n$.
Right, so what you did above is not enough to prove linearly independent, is it?

10. 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 .

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

12. Originally Posted by Bruno J.
Suppose $\displaystyle u,v,w$ are linearly dependent. Let $\displaystyle W=\mbox{span }\{u,v,w\}$, so $\displaystyle 0 \leq \mbox{dim }W < 3$. Let $\displaystyle S =\{au+bv+cw,\ du+ev+fw,\ gu+hv+iw\}$. Since $\displaystyle S \subset W$, we have $\displaystyle 0<\mbox{dim span }S < 3 = |S|$ and therefore $\displaystyle S$ must be linearly dependent.
would you please explain how did you do that ?
(i don't get your proof)
thanks.

13. Are you familiar with dimension and span?

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

15. Originally Posted by Bruno J.
Are you familiar with dimension and span?
yes i am.

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