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Linear algebra

  1. Nov 26, 2006 #1
    can anyone help with this problem?

    Consider a diagonalizable nxn matrix A with m distinct eigenvalues L1,....,Lm show that (A-L1*I)(A-L2*I)****(A-Lm*I)=0
     
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  3. Nov 26, 2006 #2

    0rthodontist

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    Do you recognize the form (A-LI)? Have you ever solved something like (A-LI)x=0?
     
  4. Nov 26, 2006 #3
    isn't that just used to find the eigenvalues? how would you get rid of x?
     
  5. Nov 26, 2006 #4

    0rthodontist

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    Don't think in terms of "getting rid of x." Think about the problem conceptually. What ways do you know to show that some matrix is the zero matrix?
     
  6. Nov 26, 2006 #5
    what do you mean by showing that a matrix is the zero matrix? A matrix is a zero matrix if anything multiplied by it is zero? I know that A is similar do a diagonal matrix D with the eigienvalues down the diagonal. I'm not sure if that would help though?
     
  7. Nov 26, 2006 #6

    0rthodontist

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    Right, a matrix is a zero matrix if any vector multiplied by it is zero. Write the vector as a linear combination of eigenvectors.
     
  8. Nov 26, 2006 #7
    So since A is diagonizable the eigenvectors form an eigenbasis for A. Thus any vector is a linear combination of these eigenvectors. Does that mean that is x is any linear combination of these eigenvectors that (A-L1*I)***(A-Lm*I)*x=0? and if so does it mean since any vector mulitplied by this matrix is zero that this matrix is zero then?
     
  9. Nov 27, 2006 #8

    0rthodontist

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    Can you think of a reason? Try actually writing x as a linear combination rather than just saying that it is.
     
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