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Math Help - Determine a basis for a subspace

  1. #1
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    Determine a basis for a subspace

    For the following matrix, determine a basis for each of the subspace R(A^{\top}), N(A), R(A), and N(A^{\top}).

    A=\left[\begin {array}{cc}3&4\\\noalign{\medskip}6&8\end{array}\r  ight]

    So I found the RREF of A to be \left[\begin {array}{cc}1&\dfrac{4}{3}\\\noalign{\medskip}0&0\e  nd{array}\right]
    (1,\dfrac{4}{3}) forms a basis for R(A) and \left[\begin {array}{c}1\\\noalign{\medskip}\dfrac{4}{3}\end{ar  ray}\right] forms a basis for R(A^{\top}).
    x_{1}+\dfrac{4}{3}x_{2}=0
    Setting x_{1}=\alpha and x_{2}=\beta I got
    \textbf{x}=\alpha\left[\begin {array}{c}0\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]+\beta\left[\begin {array}{c}\dfrac{-4}{3}\\\noalign{\medskip}0\end{array}\right]
    So x is a equal to the N(A)? This is where I am stuck, I don't know where to go from here or even if I begun my process correctly. Help would be greatly appreciated, thank you!
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  2. #2
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    What is N(A)? I know R(A) is the row space of A.
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  3. #3
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    Quote Originally Posted by KGene View Post
    What is N(A)? I know R(A) is the row space of A.
    N(A) is the nullspace of A.
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    Quote Originally Posted by pakman View Post
    N(A) is the nullspace of A.
    Suppose x=(x_1,x_2) is in the nullspace of A. By looking at the row space, it must satisfies the equation x_1+4/3x_2=0 as you mentioned earlier. So x_1,x_2 are related as given in the equation. Can you write x in term of x_1 only?
    If you could do so, factor out x_1, then you can easily see a basis element of N(A).

    Hope this helps.
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    Note also that \dim N(A)=2-\dim R(A)=2-1=1.
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  6. #6
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    Hi !

    Quote Originally Posted by KGene View Post
    Suppose x=(x_1,x_2) is in the nullspace of A. By looking at the row space, it must satisfies the equation x_1+4/3x_2=0 as you mentioned earlier. So x_1,x_2 are related as given in the equation. Can you write x in term of x_1 only?
    If you could do so, factor out x_1, then you can easily see a basis element of N(A).

    Hope this helps.
    Or just find a vector x=\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} that satisfies this equality (apart from the null vector )
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  7. #7
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    Quote Originally Posted by KGene View Post
    Suppose x=(x_1,x_2) is in the nullspace of A. By looking at the row space, it must satisfies the equation x_1+4/3x_2=0 as you mentioned earlier. So x_1,x_2 are related as given in the equation. Can you write x in term of x_1 only?
    If you could do so, factor out x_1, then you can easily see a basis element of N(A).

    Hope this helps.
    So x in terms of only x_{1} would be
    \textbf{x}=x1\left[\begin {array}{c}-1\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]

    Did you mean it like that?
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  8. #8
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    Quote Originally Posted by pakman View Post
    So x in terms of only x_{1} would be
    \textbf{x}=x1\left[\begin {array}{c}-1\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]

    Did you mean it like that?
    There is a mistake to be fixed, x_2=-3/4x_1. So it should be x_1\left [\begin{array}{c}1\\ -3/4\end{array}\right ] instead.
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