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    I'm struggling on a column space question:

    We have A =

    (1, 3, -4
    2, -1 -1
    -1, -3, 4)

    I found the nullspace to be lamda (1;1;1) but not sure how to go about finding the column space ...
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    Split up the three columns of the matrix, and the column space is given by:

    a*col1 + b*col2 + c*col3, where a, b and c are scalars.
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    (Original post by marcusmerehay)
    Split up the three columns of the matrix, and the column space is given by:

    a*col1 + b*col2 + c*col3, where a, b and c are scalars.
    Still struggling.

    Do I need it in reduced row echelon form?
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    (Original post by Cggh90)
    Still struggling.

    Do I need it in reduced row echelon form?
    You just need the original matrix. Essentially what you're doing is multiplying the matrix by an arbitrary vector (a,b,c)^T.

    The column space is the image of the matrix transformation i.e. all points the matrix maps to. It is clear to see from this that when a=b=c and you multiply out you obtain the zero vector, which is verification of your nullspace.
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    (Original post by marcusmerehay)
    You just need the original matrix. Essentially what you're doing is multiplying the matrix by an arbitrary vector (a,b,c)^T.

    The column space is the image of the matrix transformation i.e. all points the matrix maps to. It is clear to see from this that when a=b=c and you multiply out you obtain the zero vector, which is verification of your nullspace.
    Oh so due to their linear dependancy, (1,1,1) is the column space of A?
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    (Original post by Cggh90)
    Oh so due to their linear dependancy, (1,1,1) is the column space of A?
    No, no, no :p:

    Just keep the general vector and multiply out and you get the following column space:

    \begin{pmatrix} a+3b-4c \\ 2a-b-c \\ -a-3b+4c \end{pmatrix}
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    (Original post by marcusmerehay)
    No, no, no :p:

    Just keep the general vector and multiply out and you get the following column space:

    \begin{pmatrix} a+3b-4c \\ 2a-b-c \\ -a-3b+4c \end{pmatrix}
    So is that the answer?

    I put that answer initially, but wasn't sure it was correct as the three terms aren't linearly independent..
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    (Original post by Cggh90)
    So is that the answer?

    I put that answer initially, but wasn't sure it was correct as the three terms aren't linearly independent..
    The terms are abitrary, and only give a zero vector if they are all the same scalar (you may wish to state this for completeness).
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    (Original post by marcusmerehay)
    The terms are abitrary, and only give a zero vector if they are all the same scalar (you may wish to state this for completeness).
    Ook cheers, so the next part asks "is this spanning set for col (A) minimal?

    It isn't minimal is it, as one vector can be expressed in terms of the other two vectors?

    So a minimal set would be span {1, 3 ; 2, -1} ?


    (1,3
    2,-1)
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    (Original post by Cggh90)
    Ook cheers, so the next part asks "is this spanning set for col (A) minimal?

    It isn't minimal is it, as one vector can be expressed in terms of the other two vectors?

    So a minimal set would be span {1, 3 ; 2, -1} ?


    (1,3
    2,-1)
    (1,3) and (2,1) aren't vectors in the 3-dimensional plane, so they can't span col(A).

    Since your third column of A is a linear combination of the first two (check this), the spanning set is the first two columns.

    Hence a spanning set is given by the first two columns.
 
 
 
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