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Range of a linear transformation to power n

  1. Aug 9, 2011 #1
    1. The problem statement, all variables and given/known data

    How to prove that: the range of a square matrix A (linear transformation) to the power of n+1 is a subspace of the Range of A to the power n, for all n >= 1?

    i.e. Range (A^(n+1)) is a subspace of Range (A^n)

    2. Relevant equations



    3. The attempt at a solution

    I can prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)). Not sure if this is the basis of the prove.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 9, 2011 #2

    jbunniii

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    Hint:

    [tex]A^{n+1}x = A^n(Ax)[/tex]
     
  4. Aug 10, 2011 #3
    Thank you jbunniii. Using the hint you gave me I could prove that Kernal (A^n) is a subspace of Kernal (A^(n+1)), because A^n(A(0)) = 0. but what about range? I think if i could prove the range of A is always a subspace of domain of A, then i am done. How to prove this? Thanks.
     
  5. Aug 10, 2011 #4

    jbunniii

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    If y is in the range of [itex]A^{n+1}[/itex], then by definition,

    [tex]y = A^{n+1}x[/tex]

    for some x.

    But

    [tex]y = A^n(Ax)[/tex]

    So what does that imply?
     
  6. Aug 10, 2011 #5
    Got it. Thanks so much.
     
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