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Math Help - Basis and dimension

  1. #1
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    Basis and dimension

    1) Find a basis for, and the dimension of, col(A), where

    A=\begin{pmatrix}1&1&0&2&1\\0&0&-1&-2&2\\-1&-1&1&4&-1\\1&1&0&4&2\end{pmatrix}

    I'm don't really know what col(A) means. But would you start of this question by:

    1) Reducing into reduced row-echelon form.
    2) Then looking at the leading columns (or pivot columns) and take those to the basis.
    3) Dimension can then be determined

    So the reduced row-echelon form is:

    A=\begin{pmatrix}1&1&0&0&0\\0&0&1&0&-3\\0&0&0&1&0.5\\0&0&0&0&0\end{pmatrix}

    I just need someone to verify whether what I'm doing is correct here.
    Last edited by acevipa; October 3rd 2010 at 06:04 PM.
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  2. #2
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    Quote Originally Posted by acevipa View Post
    1) Find a basis for, and the dimension of, col(A), where

    A=\begin{pmatrix}1&1&0&2&1\\0&0&-1&-2&2\\-1&-1&1&4&-1\\1&1&0&4&2\end{pmatrix}

    I'm don't really know what col(A) means. But would you start of this question by:

    1) Reducing into reduced row-echelon form.
    2) Then looking at the leading columns (or pivot columns) and take those to the basis.
    3) Dimension can then be determined

    So the reduced row-echelon form is:

    A=\begin{pmatrix}1&1&0&0&0\\0&0&1&0&-3\\0&0&0&1&0.5\\0&0&0&0&0\end{pmatrix}

    I just need someone to verify whether what I'm doing is correct here.
    Your "second" 'A' might better be denoted 'U'.

    So you've found a basis for the column space of U (i.e., col(U)); they are columns 1, 3 and 4 of U.

    But what you want is a basis for the column space of A (i.e., col(A)).

    It's important to understand that A doesn't have the same column space as U.
    The elimination process will leave some stuff unchanged (e.g., nullspace), but the columns will be entirely different.

    Nevertheless, there's an important theorem telling us that when certain columns of U form a basis for col(U), the corresponding columns of A form a basis for col(A).
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  3. #3
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    Quote Originally Posted by PiperAlpha167 View Post
    Your "second" 'A' might better be denoted 'U'.

    So you've found a basis for the column space of U (i.e., col(U)); they are columns 1, 3 and 4 of U.

    But what you want is a basis for the column space of A (i.e., col(A)).

    It's important to understand that A doesn't have the same column space as U.
    The elimination process will leave some stuff unchanged (e.g., nullspace), but the columns will be entirely different.

    Nevertheless, there's an important theorem telling us that when certain columns of U form a basis for col(U), the corresponding columns of A form a basis for col(A).
    Okay, for U, the pivot columns are columns 1, 3, 4. So wouldn't the basis for A be columns 1, 3 and 4 of A.

    What is col(A)?
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  4. #4
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    Quote Originally Posted by acevipa View Post
    Okay, for U, the pivot columns are columns 1, 3, 4. So wouldn't the basis for A be columns 1, 3 and 4 of A.

    What is col(A)?
    There are four fundamental subspaces in linear algebra. The column space is one of them.
    Some authors consider these spaces so important that they state the fundamental theorem of linear algebra completely in terms of these subspaces.

    The column space of A is often abbreviated, col(A). Sometimes it's abbreviated, R(A), and called the range of A. The word 'range' as used here is consistent with the usual idea of the range of a function, say f, as the set of all possible values f(x). In the case of A, col(A) is simply all possible vectors Ax.
    With other words, it is all possible b for which Ax=b can be solved.

    Those pivot columns of U form a basis for the column space of U, col(U).
    You want a basis for col(A). You've already done all the work.
    Just pick out the columns of A that correspond to the pivot columns of U.
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  5. #5
    Senior Member MacstersUndead's Avatar
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    Let {u_n} be the column vectors in U.
    {u_1} = {u_2}
    and
    {u_5} = -3{u_3} + 0.5{u_4}

    With these linear combinations in mind, what can you say about the {a_2}? {a_5}?
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