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Math Help - Prove linearly independent

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    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?
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    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).
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    Quote Originally Posted by kjchauhan View Post
    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...
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    Quote Originally Posted by 450081592 View Post
    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.
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    Quote Originally Posted by Raoh View Post
    (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?
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    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.
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    Quote Originally Posted by Roam View Post
    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?
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    Quote Originally Posted by 450081592 View Post
    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.
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    Quote Originally Posted by Raoh View Post
    Exactly,the vectors are in \mathbb{R}^n.
    Right, so what you did above is not enough to prove linearly independent, is it?
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    Quote Originally Posted by 450081592 View Post
    Right, so what you did above is not enough to prove linearly independent, is it?
    i believe i should let someone else handle this .
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    MHF Contributor Bruno J.'s Avatar
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    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.
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    Quote Originally Posted by Bruno J. View Post
    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 ?
    (i don't get your proof)
    thanks.
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    MHF Contributor Bruno J.'s Avatar
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    Are you familiar with dimension and span?
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    Quote Originally Posted by Bruno J. View Post
    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
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    Quote Originally Posted by Bruno J. View Post
    Are you familiar with dimension and span?
    yes i am.
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