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Linear operator

  1. Feb 9, 2010 #1
    I have a question about the invertibility of a linear operator T.

    In Friedberg's book, Theorem 6.18 (c) claims that if B is an orthonormal basis for a finite-dimensional inner product space V, then T(B) is an orthonromal basis for V.

    I don't understand the proof, I think the book only prove that T(B) is orthonormal.

    If T is not one-to-one, why T(B) is also linear independent?
     
  2. jcsd
  3. Feb 9, 2010 #2

    quasar987

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    You're right.. T(B) need not be linearly independent if T isn't injective.

    But what hypothesis on T does the book make exactly?

    Because it is not true that any linear operator, even invertible ones send orthonormal basis to orthonormal basis.
     
  4. Feb 10, 2010 #3
    Could you explain why an invertible linear operator can't send orthonormal basis to orthonormal basis?
     
    Last edited: Feb 10, 2010
  5. Feb 10, 2010 #4

    HallsofIvy

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    ?? You started by saying that you understood the proof that an invertible linear operator sends an orthonormal basis into an orthonormal basis. Now you are asking why that can't be true?

    If T maps a finite dimensional space to a finite dimensional space and is invertible, then it maps a basis to a basis. If the spaces are not finite dimensional, that may not be true.
     
  6. Feb 10, 2010 #5

    quasar987

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    I'm not saying an invertible linear operator can't send orthonormal basis to orthonormal basis, I'm saying that it can, but in general, it wont.

    Consider for instance T:R²-->R² given by T(x,y)=(x+y,y). It is an invertible linear operator that sends the orthonormal basis {(1,0), (0,1)} to the non orthonormal basis {(1,0), (1,1)}.
     
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