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Bases of a Linear transformation (Kernel, Image and Union ?

  1. May 20, 2012 #1
    Bases of a Linear transformation (Kernel, Image and Union !?!

    http://dl.dropbox.com/u/33103477/1linear%20tran.png [Broken]

    For the kernel/null space
    [tex] \begin{bmatrix}
    3 & 1 & 2 & -1\\
    2 & 4 & 1 & -1
    \end{bmatrix} = [0]_v [/tex]

    Row reducing I get

    [tex] \begin{bmatrix}
    1 & 0 & \frac{7}{9} & \frac{-2}{9}\\
    0 & 1 & \frac{-1}{10} & \frac{-1}{10}
    \end{bmatrix} = [0]_v [/tex]

    So the basis of the kernel U1 is:
    [tex] \begin{pmatrix}
    \frac{-7}{9}\\
    \frac{1}{10}\\
    1\\
    0
    \end{pmatrix}, \begin{pmatrix}
    \frac{2}{9}\\
    \frac{1}{10}\\
    0\\
    1
    \end{pmatrix} [/tex]

    Now, for the image/range.

    I can write the transformation as:

    [tex] S=\begin{pmatrix}
    1 \\
    1 \\
    2 \\
    3
    \end{pmatrix}, \begin{pmatrix}
    -1 \\
    -3 \\
    -8 \\
    -27
    \end{pmatrix}
    [/tex]

    So we proceed to find the basis of Span S

    [tex] \begin{bmatrix}
    1 &-1 \\
    1 &-3 \\
    2 &-8 \\
    3 &-27
    \end{bmatrix} [/tex]

    Row reducing I get:

    [tex] \begin{bmatrix}
    1 &0 \\
    0 &1 \\
    0 &0 \\
    0 &0
    \end{bmatrix} [/tex]

    Which implies S is linearly independant.

    So the basis is:


    [tex] S=\begin{pmatrix}
    1 \\
    1 \\
    2 \\
    3
    \end{pmatrix}, \begin{pmatrix}
    -1 \\
    -3 \\
    -8 \\
    -27
    \end{pmatrix}
    [/tex]

    Now for U1 union U2

    Should I just put both the basis together. (After checking if they are all independant ?)

    I have no idea about the addition any idea's ??
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 20, 2012 #2
    Re: Bases of a Linear transformation (Kernel, Image and Union !?!

    Any one ?
     
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