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Thread: 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 $\displaystyle R(A^{\top})$, $\displaystyle N(A)$, $\displaystyle R(A)$, and $\displaystyle N(A^{\top})$.

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

    So I found the RREF of A to be $\displaystyle \left[\begin {array}{cc}1&\dfrac{4}{3}\\\noalign{\medskip}0&0\e nd{array}\right]$
    $\displaystyle (1,\dfrac{4}{3})$ forms a basis for $\displaystyle R(A)$ and $\displaystyle \left[\begin {array}{c}1\\\noalign{\medskip}\dfrac{4}{3}\end{ar ray}\right]$ forms a basis for $\displaystyle R(A^{\top})$.
    $\displaystyle x_{1}+\dfrac{4}{3}x_{2}=0$
    Setting $\displaystyle x_{1}=\alpha$ and $\displaystyle x_{2}=\beta$ I got
    $\displaystyle \textbf{x}=\alpha\left[\begin {array}{c}0\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]+\beta\left[\begin {array}{c}\dfrac{-4}{3}\\\noalign{\medskip}0\end{array}\right]$
    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 $\displaystyle N(A)$? I know $\displaystyle R(A)$ is the row space of $\displaystyle A$.
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    Quote Originally Posted by KGene View Post
    What is $\displaystyle N(A)$? I know $\displaystyle R(A)$ is the row space of $\displaystyle A$.
    $\displaystyle N(A)$ is the nullspace of A.
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    Quote Originally Posted by pakman View Post
    $\displaystyle N(A)$ is the nullspace of A.
    Suppose $\displaystyle x=(x_1,x_2)$ is in the nullspace of $\displaystyle A$. By looking at the row space, it must satisfies the equation $\displaystyle x_1+4/3x_2=0$ as you mentioned earlier. So $\displaystyle x_1,x_2$ are related as given in the equation. Can you write $\displaystyle x$ in term of $\displaystyle x_1$ only?
    If you could do so, factor out $\displaystyle x_1$, then you can easily see a basis element of $\displaystyle N(A)$.

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

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

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

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

    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 $\displaystyle x_{1}$ would be
    $\displaystyle \textbf{x}=x1\left[\begin {array}{c}-1\\\noalign{\medskip}\dfrac{-3}{4}\end{array}\right]$

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