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Math Help - Basis and Dimension of the solution space

  1. #1
    Jav
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    Question Basis and Dimension of the solution space

    I have a more simple problem, but my answer is wrong according to my book.

    Find a basis for, and the dimension of, the solution space of Ax=0.

    A is a matrix of rows:
    1 2 -3
    2 -1 4
    4 3 -2

    I have reduced this to row-echelon form and I get:
    1 2 -3
    0 1-2
    0 0 0
    Then, I know that the Rank(A) is 2.
    columns=rank+nullity. Then, the nullity is 1.

    Now, how can I have a nullity or dimension of the Nullspace to be 1, when I have two rows that are the basis. Namely, (1,2,-3) and (0,1,-2).

    Thank you.
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  2. #2
    Jav
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    Re: Vector Spaces

    Ok, I made many mistakes. I know the answer now.

    I end up with a reduced row echelon:
    1 0 1
    0 1 -2
    0 0 0

    (Numbers to the right of "x" represent sub-numbers)
    x1=- x3
    x2=2 x3
    x3=t (all real)

    Then, the basis is:
    -1
    2
    1.

    I have more questions. How do I know that I'm being asked to find a basis in this form rather than one for the row or column space?
    Thank you for your patience.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Re: Vector Spaces

    Quote Originally Posted by Jav View Post
    Now, how can I have a nullity or dimension of the Nullspace to be 1, when I have two rows that are the basis. Namely, (1,2,-3) and (0,1,-2).
    You are confusing the null space N with the row space. Take into account that N is the set of (x,y,z)\in\mathbb{R}^3 such that \begin{Bmatrix} x+2y-3z=0\\y-2z=0\end{matrix} Now, use the method we commented you.


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  4. #4
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    Re: Vector Spaces

    A basis of what? Not the kernel certainly, since \begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}1 \\ 2 \\ -3 \end{bmatrix}= \begin{bmatrix}14 \\ -12 \\ 10\end{bmatrix} and \begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}0 \\ 1 \\ -2 \end{bmatrix}= \begin{bmatrix} 8 \\ 8 \\  7\end{bmatrix}, not the 0 vector.
    The kernel is given by the solutions to
    ] \begin{bmatrix}1 & 2 & -3 \\ 2 & -1 & 4 \\ 4 & 3 & -2\end{bmatrix}\begin{bmatrix}x \\ y\\ v \end{bmatrix}= \begin{bmatrix}0\\ 0 \\ 0\end{bmatrix} which can be got by row reducing the augmented matrix
    \begin{bmatrix}1 & 2 & -3 & 0 \\ 2 & -1 & 4 & 0 \\ 4 & 3 & -2 & 0\end{bmatrix}

    Yes, that reduces to much what you have:
    \begin{bmatrix}1 & 2 & 3 & 0 \\ 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0\end{bmatrix}
    But now that is equivalent to the equations x+ 2y+ 3z= 0, y- 2z= 0. From the second equation, y= 2z and from the first, x+ 2y+ 3z= x+ 2(2z)+ 3z= x+ 7z= 0 from which x= -7z. That is, any vector in the kernel is of the form <x, y, z>= <-7z, 2z, z>= z<-7, 2, 1>. A basis for the kernel is the singleton set {<-7, 2, 1>}.
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