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Math Help - Help figuring out how to find this vector

  1. #1
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    Help figuring out how to find this vector

    I have a matrix A

    2 4 1 1
    3 4 0 2
    1 2 1 2

    Now im trying to find a basis for the nullspace(A)

    After applying GJ to A i get

    1 0 0 4
    0 1 0 0
    0 0 1 3

    Now how would i go from this matrix to finding the single basis vector

    b4 =

    1
    0
    2
    1

    I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by BigBoss22 View Post
    I have a matrix A

    2 4 1 1
    3 4 0 2
    1 2 1 2

    Now im trying to find a basis for the nullspace(A)

    After applying GJ to A i get

    1 0 0 4
    0 1 0 0
    0 0 1 3

    Now how would i go from this matrix to finding the single basis vector

    b4 =

    1
    0
    2
    1

    I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated

    I really don't understand what you did: first, I interchanged the 3rd and 1st rows in the original matrix and brought it to echelon form and got:

    \left(\begin{array}{cccc}1&2&1&2\\0&2&3&4\\0&0&1&3  \end{array}\right)

    Using this matrix a coefficient matrix for a homogeneous system (i.e., equating every row to zero), the general solution is

    \left(\begin{array}{c}\!\!\!\!-8w\\5w\\\!\!\!\!-6w\\2w\end{array}\!\!\right)\,,\,\;with\;\;w\in \mathbb{F}=\mbox{ the field of de{}finition}

    You can now choose any non-zero value for w and you get a basis for the null space of A. The vector you mention above is NOT in the null space of A, as you can easily check.

    Tonio
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  3. #3
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    Ok i'll give you the full example from the notes that i'm looking at.

    Let W be a subspace of Kn and
    v is also contained in Kn

    Need to determine if v is contained in W

    and v =
    3
    1
    2
    3

    W = rowspace(A)

    A =

    2411
    3402
    1212

    over Z5

    According to the notes we will find another matrix B where:
    W = nullspace(B)

    Then after GJ A =

    1004
    0100
    0013

    And then aparantlly

    From this information you can get a single bias vector
     b4 =
    1
    0
    2
    1

    But i dont see how??
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  4. #4
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    Joined
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    153
    Quote Originally Posted by BigBoss22 View Post
    I have a matrix A

    2 4 1 1
    3 4 0 2
    1 2 1 2

    Now im trying to find a basis for the nullspace(A)

    After applying GJ to A i get

    1 0 0 4
    0 1 0 0
    0 0 1 3

    Now how would i go from this matrix to finding the single basis vector

    b4 =

    1
    0
    2
    1

    I don't understand how they went from the matrix to this vector. Any help would be greatly appreciated
    rref(A)=\begin{bmatrix} 1&0&0&4\\0&-2&0&5\\0&0&1&3 \end{bmatrix},\ker(A)=span(\begin{bmatrix} -4\\ \frac{5}{2} \\-3 \\1\end{bmatrix})=span(\begin{bmatrix} -8\\5 \\-6 \\2\end{bmatrix})
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