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  • Linear Algebra
  • Null Space
  • Echelon Form
  • Solve
    \(A\vec{x}=0\)
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    Null space

    We discussed column space of a matrix
    \(A\)
    as a vector space of linear combination of all of it's column vectors.
    Null space is completely different thing.
    For a matrix say
    \(A\)
    the Null space is the space of all
    \(\vec{x}\)
    that solves
    \(A\vec{x}=\vec{0}\)

    Example in
    \(\mathbb{R}^4\)

    Consider a matrix
    \(A=\begin{bmatrix} 1 & 1 & 2\\ 2 & 4 & 6\\ 3 & 6 & 9\\ 4 & 8 & 12\\ \end{bmatrix}\)

    Null space for
    \(A\)
    is space of all the
    \(\vec{x} = \begin{bmatrix} x_1\\x_2\\x_3\\ \end{bmatrix}\)
    that solves
    \(A\vec{x} = \vec{0}\)

    \(Ax= \begin{bmatrix} 1 & 1 & 2\\ 2 & 4 & 6\\ 3 & 6 & 9\\ 4 & 8 & 12\\ \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ \end{bmatrix}\)

    So you can see our column space is
    \(\mathbb{R}^4\)
    and the space we desire(Null space) is a subspace of
    \(\mathbb{R}^3\)
    .
    What is the obvious value of
    \(\vec{x}\)
    that solves
    \(A\vec{x}=0\)
    ?
    \(\vec{x}=0\)
    What are some other value of
    \(\vec{x}\)
    that solves
    \(A\vec{x}=0\)
    ?
    \(\vec{x}=\begin{bmatrix} -1 \\ -1 \\ 1 \\ \end{bmatrix}\)
    solves
    \(A\vec{x}=0\)
    ,
    and
    \(\vec{x}=\begin{bmatrix} -2 \\ -2 \\ 2 \\ \end{bmatrix}\)
    ,
    \(\vec{x}=\begin{bmatrix} -3 \\ -3 \\ 3 \\ \end{bmatrix}\)
    ,
    \(\vec{x}=\begin{bmatrix} -4 \\ -4 \\ 4 \\ \end{bmatrix},...\)
    also solves
    \(A\vec{x}=0\)
    .
    We can summarize it as,
    \(\vec{x}=a\begin{bmatrix} -1 \\ -1 \\ 1 \\ \end{bmatrix};\quad a\in\mathbb{R}\)
    solves
    \(A\vec{x}=0\)
    .
    What is this form (
    \(a\vec{v}\)
    ) it's a line, so Null space of
    \(A\)
    is a line in
    \(\mathbb{R}^3\)
    .

    How can we assure that the space we get by
    \(A\vec{x}=0\)
    is a vector space?
    We need to show:
    \(1.\)
    Sum of two vectors in Null Space remains inside the null space.
    \(2.\)
    If we multiply a vector of Null space the resultant vector remains in that subspace
    1. Say we have two vectors
      \(\vec{v}\)
      and
      \(\vec{w}\)
      in the Null space.
      So
      \(A\vec{v}=\vec{0}\)
      and
      \(A\vec{w}=\vec{0}\)

      \(\Rightarrow A\vec{v} + A\vec{w}=\vec{0}\)

      \(\Rightarrow A(\vec{v} + \vec{w})=\vec{0}\)

      so
      \(\vec{v}+\vec{w}\)
      is inside the Null space.
    2. Say we have a vectors
      \(\vec{v}\)
      in the Null space.
      So
      \(A\vec{v}=\vec{0}\)

      Say
      \(\vec{w}=c\vec{v};\quad c\in\mathbb{R}\)

      \(A\vec{w}=A(c\vec{v})=cA\vec{v}=c\vec{0}=\vec{0};\quad\{\text{because }A\vec{v}=\vec{0}\)

      So
      \(\vec{w}\)
      is also in Null space.
    This confirms that Null Space is a vector space.