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Thread: Finding the span in v3

  1. #1
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    Finding the span in v3

    For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer

    v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]
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    Re: Finding the span in v3

    Do you have any idea how to solve the problem?

    Have you tried anything so far?
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    Re: Finding the span in v3

    Yes, I've tried reducing it down to echelon form, but I have
    {1 0 0}
    {3 0 0}
    {5 8 10+h}

    (Sorry, I don't know how to format the matricies)
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    Re: Finding the span in v3

    \begin{bmatrix}
    1 & -3 & 5 \\
    0 & 0 & 8 \\
    0 & 0 & 10+h\\
    \end{bmatrix}

    ^^ this
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    Re: Finding the span in v3

    Quote Originally Posted by Cammy View Post
    For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer
    v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]
    Because $-3v_1=v_2$ there is no value of $h$ for which $v_3$ is a multiple of $v_1$.
    Please review your post. Have you made a typo?
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    Re: Finding the span in v3

    Quote Originally Posted by Plato View Post
    Because $-3v_1=v_2$ there is no value of $h$ for which $v_3$ is a multiple of $v_1$.
    Please review your post. Have you made a typo?
    That's how my book states it
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    Re: Finding the span in v3

    Quote Originally Posted by Cammy View Post
    That's how my book states it
    Plato is correct. There is no $h$ that will give you a set of linearly independent vectors.

    $\left |\begin{pmatrix}1 &-3 &2 \\-3 &9 &-6 \\5 &-7 &h \end{pmatrix}\right | = 0,~\forall h$
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    Re: Finding the span in v3

    Quote Originally Posted by Cammy View Post
    For what values of h is v3 in Span(v1, v2) and for what values of h is {v1, v2, v3} linearly dependent? Justify each answer v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h]

    Quote Originally Posted by romsek View Post
    Plato is correct. There is no $h$ that will give you a set of linearly independent vectors.
    Actually the question asks about linearly dependent vectors and not independent.

    But because $-3v_1=v_2$ it follows that $Span(v_1,v_2)=Span(v_1)$.
    There is no value of $h$ for which $v_3\in Span(v_1)$
    Therefore for every value of $h$ the set $\{v_1,v_2,v_3\}$ is dependent.
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  9. #9
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    Re: Finding the span in v3

    More fundamentally, vectors v1, v2, and v3 are "linearly independent" if the only way av1+ bv2+cv3= 0 is to have a= b= c. They are "linearly dependent" if there exist values of a, b, and c other than a= b= c= 0 that make that true.

    With v1 = [1 -3 2], v2 = [-3 9 -6], v3 = [5 -7 h], that becomes a[1 -3 2]+ b[-3 9 -6]+ c[5 -7 h]= [a- 3b+ 5c -3a+ 9b- 7c 2a- 6b+ hc]= [0 0 0].

    So we must have a- 3b+ 5c= 0, -3a+ 9b- 7c= 0, 2a- 6b+ hc= 0.

    If we multiply the first equation by 3 and add to the second equation, we get (3- 3)a+ (9- 3)b- (7- 5)c= 6b- 2c= 0. If we multiply the first equation by 2 and subtract from the third equation, we get (2- 2)a- (6- 6)b+ (h- 10)c= (h- 10)c= 0. If h is any number other than 10, h- 10 is not 0 so we can divide by it to get c= 0. Then 6b- 2c= 6b= 0 so b= 0 and the first equation becomes a- 3b+ 5c= a= 0. If h= 10 then c can be any number and there are "non-trivial" solutions.
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