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Math Help - Dimension of the column space of a matrix

  1. #1
    Member
    Joined
    Mar 2009
    From
    Gold Coast, Queensland, Australia
    Posts
    105

    Dimension of the column space of a matrix

    I have this problem that I am doing in a book, and well, the answer in the back says the dimension of the column space and the row space is two.



    \left(<br />
\begin{matrix}<br />
1 & 1 & 0\\<br />
1  & 3 & 1\\<br />
3 & 1 & -1\\<br />
\end{matrix}<br />
\right)<br />

    I row reduced it to:

    \left(<br />
\begin{matrix}<br />
1 & 0 & 0\\<br />
0 & 1 & 0\\<br />
0 & 0 & 1\\<br />
\end{matrix}<br />
\right)<br />


    So I thought this would mean that they are independent columns and the space it spans has a dimension of 3.

    Am I missing something here?

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  2. #2
    Senior Member Twig's Avatar
    Joined
    Mar 2008
    From
    Gothenburg
    Posts
    396

    recheck

    Hi

    Recheck your calculations.

    I get after row-operations:

     \left[\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{matrix}\right]

    Which shows that the matrix has two pivot columns and thus rank = 2.
    Equivalently, dim Nul (A) = number of free variables, and
    n = dim Nul(A) + rank (A)

    And we see here that we have one free variable, so rank (A) = n-dim Nul(A) = 2
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  3. #3
    Member
    Joined
    Mar 2009
    From
    Gold Coast, Queensland, Australia
    Posts
    105

    you're right

    Thanks mate.
    I put
    <br />
a_{33} = 1<br />
    instead of -1 in the original matrix before I reduced it.

    cheers.
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