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Another question about projections.

  1. Jul 8, 2009 #1
    Say we have a transformation T[tex]\in[/tex]L(V). Now suppose a subspace of V (U) is in the rangespace of T. Now suppose PUv=u with u=a1u1+...+amvm.
    Now apply T to u to get T(u)=b1u1+...+bmum=/=a1u1+...+amvm. What would happen
    if we apply PUto T(u)? In other words, what would we end up with after computing PUT(u)?
    I'm just wondering whether or not applying a transformation (a non-projection in this case) after a projection would result
    in a different output after applying the projection to the image of T? In other words, would
    PUT(u) map to a1u1+..+amum or would it map to b1u1+..+bmum?
    Thank you for your response!
     
  2. jcsd
  3. Jul 8, 2009 #2
    Em..It's a bit hard to understand your post, could you explain what do these mean:
    T [tex]
    \in
    [/tex] L(V),
    V (U) (what does U mean)
    PU (projection?)
    T(u)=b1u1+...+bmum=/=a1u1+...+amvm
    It would be helpful if you can clarify these
     
  4. Jul 9, 2009 #3
    T is a transformation that sends a vector from V to V.
    U is a subspace of V. PU is a projection onto U. And T(u)=b1u1+..+bmum=/=a1u1+...+amum means that T is not
    an identity operator, since u=a1u1+...+amum.
     
    Last edited: Jul 9, 2009
  5. Jul 9, 2009 #4
    I think in general cases they wouldn't be same, just take an exmple
    projection
    [tex]
    P = \left( {\begin{array}{*{20}c}
    1 & 0 \\
    0 & 0 \\
    \end{array}} \right)
    [/tex]
    transformation(I take a rotation)
    [tex]
    T = \left( {\begin{array}{*{20}c}
    0 & { - 1} \\
    1 & 0 \\
    \end{array}} \right)
    [/tex]
    take a vector
    [tex]
    \overrightarrow v = \left[ \begin{array}{l}
    1 \\
    1 \\
    \end{array} \right]
    [/tex]
    then compute as what you said
    [tex]
    P\overrightarrow v = \left[ \begin{array}{l}
    1 \\
    0 \\
    \end{array} \right]
    [/tex]
    [tex]
    PT\overrightarrow v = \left[ \begin{array}{l}
    - 1 \\
    0 \\
    \end{array} \right]
    [/tex]
     
  6. Jul 9, 2009 #5
    That answers it! Thank you!
     
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