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Math Help - Two basic questions!

  1. #1
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    Two basic questions!

    Ok so I have two quick questions.

    The problem I'm trying to solve is as follows:

    Find a so that the null space of A^{T} has dimension 0. Find a basis for the column space of A^{T} once a has been determined.


    we know A =  \begin{bmatrix} a &1 &-1 &3 \\ 0 &a &-1 &0 \\ 1 &-1 &0 &a \end{bmatrix}

    so I went ahead and set a = 0 so that the dimension of A^{T} = 0.

    With that, I was wondering if I would instantly be able to say:

    a basis for the column space of A^{T} = all 3 columns of A^{T} which, when reduced, would be =
     SP{\begin{bmatrix} 1\\0\\0\\0 \end{bmatrix} , \begin{bmatrix} 0\\1\\0\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\1\\0 \end{bmatrix} }




    For my second question, I was wondering if there's a generalized way to find a. For this problem, I pretty much did trial and error and looked at what value I could give a so that, when I put A^{T} into R.E.F. it would be linearly independent. Is that the only way to find a value for a?
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  2. #2
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    Since no one has answered this yet, perhaps I need to make it more clear what I was thinking.

    I thought that if I can conclude that dim(N(A)) = 0, then the rank = 3. As in an M x N matrix, N = Rank + Nullity. Therefore, since the rank is 3 and n = 3, a basis for the column space would have to be the 3 columns.


    Am I correct in thinking this?
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by addieandjack View Post
    Since no one has answered this yet, perhaps I need to make it more clear what I was thinking.

    I thought that if I can conclude that dim(N(A)) = 0, then the rank = 3. As in an M x N matrix, N = Rank + Nullity. Therefore, since the rank is 3 and n = 3, a basis for the column space would have to be the 3 columns.


    Am I correct in thinking this?
    Why don't you just apply A to basis elements of the domain space and pick a intelligently so that the image of those basis elements are linearly independent?
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  4. #4
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    Yes. If a=0, then the only solution to A^Tx=0 is x=0, so then \mathrm{dim} \mathrm{null} A=0.


    In this case, as you said, \mathrm{rank} A^T=3-0=3 (I use 3 here because A^T has 3 columns; you have to be careful since A^T is not square.)

    Since the 3 columns generate the column space, this forces them to be a basis for it. (The rank of a matrix is equal to its column rank.)
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