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Math Help - linear independent/vectors

  1. #1
    MHF Contributor arbolis's Avatar
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    linear independent/vectors

    Show wether the vectors are linear independent and if they generate \mathbb{R}^3.
    a) (2,-2,0),(1,-2,1),(-1,4,-3)
    My attempt : I don't understand anything about my teacher's notes. So I've read on wikipedia that in order to determine if some vectors are l.i. I could calculate the determinant of a matrix formed by them. That's what I did here and I found it to be different from 0, so the 3 vectors are l.i.
    Now how can I show that they generate \mathbb{R}^3 or not? I guess I must find a basis (but I don't know how to procede and what it is) and check if it generates \mathbb{R}^3.
    Thanks in advance.
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    Quote Originally Posted by arbolis View Post
    Show wether the vectors are linear independent and if they generate \mathbb{R}^3.
    a) (2,-2,0),(1,-2,1),(-1,4,-3)
    My attempt : I don't understand anything about my teacher's notes. So I've read on wikipedia that in order to determine if some vectors are l.i. I could calculate the determinant of a matrix formed by them. That's what I did here and I found it to be different from 0, so the 3 vectors are l.i.
    Now how can I show that they generate \mathbb{R}^3 or not? I guess I must find a basis (but I don't know how to procede and what it is) and check if it generates \mathbb{R}^3.
    Thanks in advance.
    If you can show these vectors are linearly independent then they would generate \mathbb{R}^3 - why?
    Thus, we just need to check if they are linearly indepedent.

    Say a\begin{bmatrix} 2\\-2\\0 \end{bmatrix} + b\begin{bmatrix}1\\-2\\1 \end{bmatrix} + c \begin{bmatrix}-1\\4\\-3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

    This gives us,
    2a + b - c = 0
    -2a - 2b + 4c = 0
    0a  + b - 3c = 0

    We need to show this only have a trivial solution i.e. a=b=c=0.

    This is equivalent to showing,
    \det \begin{bmatrix}2& 1 & -1 \\ -2&-2&4 \\ 0 & 1 & - 3 \end{bmatrix} \not = 0
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    <2, -2, 0> - 3<1, -2, 1> = <2, -2, 0> + <-3, 6, -3> = <-1, 4, -3>

    The vectors are not linearly independent.
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    MHF Contributor arbolis's Avatar
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    This is equivalent to showing,
    This is what I've done.
    \det A=2(6+4)-(6+0)-(-2+0)=22-6\neq 0 thus they are linear independent. Am I calculating it badly?
    And about
    If you can show these vectors are linearly independent then they would generate - why?
    I asked it myself : is that possible that they generate for example \mathbb{R}^2? And I thought not, because it would mean that 2 of them are linear dependent and the remaining vector is linear independent from the 2 firsts, but I don't know how to show this.
    I probably did an error since icemanfan claims that they are linear dependent...
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    The determinant is (2)(-2)(-3) + (1)(4)(0) + (-1)(-2)(1) - (-1)(-2)(0) - (2)(4)(1) - (1)(-2)(-3) = 12 + 0 + 2 - 0 - 8 - 6 = 0.
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  6. #6
    MHF Contributor arbolis's Avatar
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    Sorry, I made an error of sign. It was \det A= 2(6-4)+...
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  7. #7
    MHF Contributor arbolis's Avatar
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    Sorry not to start it in a new thread, but the entire question states "If they are not l.i., characterize implicitly the subspace generated and give a basis of this subspace".
    I've really no idea about how to do that. Maybe they generate \mathbb{R}^2, but I don't know how to show it...nor how to pick off a basis.
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  8. #8
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    you can do both parts of your problem at the same time without even using determinant: first let A=\begin{bmatrix}2 & 1 & -1 \\ -2 & -2 & 4 \\ 0 & 1 & -3 \end{bmatrix}, \ \bold{b}=\begin{bmatrix}b_1 \\ b_2 \\ b_3 \end{bmatrix}. then \bold{b} is in the subspace spanned by the columns of A,

    if and only if the equation A \bold{x}=\bold{b} has a solution. so you need to consider the augmented matrix [A \mid \bold{b}]=\begin{bmatrix} 2 & 1 & -1 & b_1 \\ -2 & -2 & 4 & b_2 \\ 0 & 1 & -3 & b_3 \end{bmatrix} \sim \begin{bmatrix}2 & 1 & -1 & b_1 \\ 0 & -1 & 3 & b_1+b_2 \\ 0 & 0 & 0 & b_1+b_2+b_3 \end{bmatrix}.

    now it's clear that the columns of A are linearly dependent because the third row of the echelon form of A is all 0. this answers the first part of your question. also it's clear that A \bold{x}=\bold{b} is

    consistent iff b_1+b_2+b_3=0, which gives us b_3=-b_1-b_2. hence \bold{b}=\begin{bmatrix}b_1 \\ b_2 \\ -b_1-b_2 \end{bmatrix}=b_1 \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} + b_2 \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix}. so the subspace generated by the columns of A is a plane which goes through

    the origin and is spanned by the vectors v_1=\begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \ v_2=\begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix}. clearly \{v_1, v_2 \} is a basis for the subspace. this answers the second part of your question.
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  9. #9
    MHF Contributor arbolis's Avatar
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    Thank you very much NonCommAlg, I could follow you entirely despite the fact that all this stuff is new to me.
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