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Linear algebra proof

  1. Apr 6, 2006 #1
    Let T: V->V be a linear transformation where V is finite dimensional. Show ath exactly one of (i) an (ii) holds
    i) T(v) = 0 for some v not zero in V
    ii) T(x) = v has a solution x in V for every v in V


    do they mean that if i holds then ii cannot hold?
    Ok suppose i holds
    T(v) = 0 for some v in V, v not zero
    then T(T(v)) = T(0) = 0
    let T(v) = x
    then T(x) = 0
    only solution here is x = v
    So T(x) = 0 for all x. ANd thus is not possible for T(x) = v if T(v) = 0
     
  2. jcsd
  3. Apr 6, 2006 #2

    Hurkyl

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    There's more to it than that...


    Anyways, your proof doesn't make any sense to me: here are the problems I see:

    Okay. I don't see the point of defining x to be zero, though.

    (1) Solution to what?
    (2) For what are you solving? You haven't written down any unknowns!
    (3) Why is it a solution?
    (4) Why is it the only solution?


    You've defined x to be T(v) (which is zero) -- so it doesn't make sense to say "for all x".
     
  4. Apr 6, 2006 #3
    Clearly i missed the point of the question

    what are they asking for actually?
    i imples that ii cannot h0old and vice versa?? Is that what im aiming for here?
     
  5. Apr 7, 2006 #4

    HallsofIvy

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    Let T: V->V be a linear transformation where V is finite dimensional. Show that exactly one of (i) an (ii) holds
    i) T(v) = 0 for some v not zero in V
    ii) T(x) = v has a solution x in V for every v in V


    "Show that exactly one holds" means "one of these and only one" You must show that for any vector v either (i) is true or (ii) is true and also show that they can't both be true.

    What do you know about T(0)? If (i) is true, if T(v)= 0 for non-zero v, then what is the solution to T(x)= 0?
     
  6. Apr 7, 2006 #5
    '
    isnt T(0) = 0
    suppose T(v) = 0, then if T(x) = 0, then x = v
     
  7. Apr 7, 2006 #6

    HallsofIvy

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    Did you notice the emphasis in the solution?

    Yes, T(0)= 0. Don't you see a problem with that and "T(v)= 0 for some no-zero v"?
     
  8. Apr 7, 2006 #7
    so
    suppose T(v) = 0 for v not zero
    then suppose T(x) = 0, for some x in V then x =v.
    But T(0) = 0 , so x must be zero? But v is non zero. SO the second one cannot hold? Is it like that?
     
  9. Apr 7, 2006 #8

    Hurkyl

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    In general, T(x) = T(y) does not imply x = y.
     
  10. Apr 7, 2006 #9
    right, T in this case is not specified to be one to one
    so for T(x) = 0 ,x MUST be v because of the assertion of the first condition.
     
  11. Apr 7, 2006 #10

    Hurkyl

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    Why do you think (i) tells you x = v?
     
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