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Math Help - Simple Matrix Question

  1. #1
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    Simple Matrix Question

    A=  \left( \begin{array}{cccc}<br />
0 & 1 & 1 & 1 \\ \\<br />
0 & 0 & 0 & 0  \\ \\<br />
0 & 1 & 0 & 1<br />
\end{array} \right)

    a) Find the Basis,
    b) the dimension of the row space of A,
    c) the column space of A
    d) the null space of A
    e) the null space of A^T

    Solutions:

    a) Basis = [tex](1,0,1), (1,0,0), (1,0,1) [/Math]
    b) dimension of row space is 3 or 2, that row of 0s is setting me off
    c) not sure if it's 3 or 4
    d) I'm not sure if this would involve the row, the column or both, considering that they are nothing but 0s

    e) A^T=  \left( \begin{array}{ccc}<br />
0 & 0 & 0 \\ \\<br />
1 & 0 & 1 \\ \\<br />
1 & 0 & 0 \\ \\<br />
1 & 0 & 1 <br />
\end{array} \right)

    again those rows and columns of 0 are throwing me off.
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  2. #2
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    Quote Originally Posted by lllll View Post
    A= \left( \begin{array}{cccc}<br />
0 & 1 & 1 & 1 \\ \\<br />
0 & 0 & 0 & 0 \\ \\<br />
0 & 1 & 0 & 1<br />
\end{array} \right)

    a) Find the Basis,
    b) the dimension of the row space of A,
    c) the column space of A
    d) the null space of A
    e) the null space of A^T

    Solutions:

    a) Basis = (1,0,1), (1,0,0), (1,0,1)
    b) dimension of row space is 3 or 2, that row of 0s is setting me off
    c) not sure if it's 3 or 4
    d) I'm not sure if this would involve the row, the column or both, considering that they are nothing but 0s

    e) A^T= \left( \begin{array}{ccc}<br />
0 & 0 & 0 \\ \\<br />
1 & 0 & 1 \\ \\<br />
1 & 0 & 0 \\ \\<br />
1 & 0 & 1 <br />
\end{array} \right)

    again those rows and columns of 0 are throwing me off.
    elemntery row operations do not change the row space of a matrix


    \begin{bmatrix}<br />
0 && 1 && 1 && 1 \\<br />
0 && 0 && 0 && 0 \\<br />
0 && 1 && 0 && 1 \\<br />
\end{bmatrix}

    in reduced row eschelon from we get


    \begin{bmatrix}<br />
0 && 1 && 1 && 1 \\<br />
0 && 0 && 1 && 0 \\<br />
0 && 0 && 0 && 0 \\<br />
\end{bmatrix} \to \begin{bmatrix}<br />
0 && 1 && 0 && 1 \\<br />
0 && 0 && 1 && 0 \\<br />
0 && 0 && 0 && 0 \\<br />
\end{bmatrix}

    so the basis of the row space is (0,1,0,1) and (0,0,1,0) so the dimention of the row space is 2.

    The leading 1's in the matrix above occur in column 2 and 3. The leading 1's let us know what columns are the basis for the column space. So we go back to the original matrix and choose columns 2 and 3.

    so the basis of the column space is (1,0,1) and (1,0,0) so it has dimention 2.

    using the rank nullity theorem?? I'm not sure it that is what it is called

    Rank(A)+Nullity(A)=number of columns of A

    2+Nullity(A)=4 nullity=2

    The columspace of A is the row space of A transpose so

    2+nullity(A trans)=3 nullity=1.

    I hope this helps.
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