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Eigenvector proof

  1. Jun 19, 2009 #1
    1. The problem statement, all variables and given/known data
    If v and w are eigenvectors with different (nonzero) eigenvalues, prove that they are
    linearly independent.


    2. Relevant equations



    3. The attempt at a solution
    Define an operator A such that a is an nxn matrix, and Av=cIv with
    c an eigenvalue and v and eigenvector. Define a basis
    <v1...vn> in that v=vi and w=vk 1<=k<=n and 1<=i<=n,
    and let ci,i, be an element in A. I is the identity matrix.

    Consider I*v, a 1xn column matrix with its lonely nonzero (1) at position 1,i.
    Let the value at ci,i=c.Multiplying I*v by A gives c*Iv . If I*w (another 1xn column
    matrix) had 1 at position 1,i, it would
    correspond with ci,i on A and we would get c*Iw. But we assume w has a different
    eigenvalue. Therefore I*w must have its 1 at a different position to correspond with
    a different value on A (call it k). Since I*v must have 1 at a row different from I*w,
    let c*Ia1v+k*Ia2w=0, and since 1 is at different rows, and c and I and k are not zero,
    a1 and a2 must be 0, so we have c*I*0*v+I*k*0*w=0*v+0*w=0, thus
    a1 and a2 are trivial so v and w are linearly independent.

    I kind of have a gut feeling that this may be too wordy.
     
    Last edited: Jun 19, 2009
  2. jcsd
  3. Jun 20, 2009 #2

    CompuChip

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    Homework Helper

    What you have looks correct, but - as you said - perhaps you are overdoing a little :)

    I would just start with stating that v and w are eigenvectors:
    (1a) A v = c v
    (1b) A w = d w
    for some numbers c and d. We know that c is not equal to d and neither is equal to 0.

    Now not being linearly independent means that there does not exist a number k such that w = k v. This is not pleasant to work with, so a proof by contradiction suggests itself. Suppose that there does exist a k such that
    (2) w = k v.

    Now can you derive a contradiction?
    (Note: the rest of the proof is rather straightforward, because all you have to work with are equations (1a), (1b) and (2)).
     
  4. Jun 20, 2009 #3

    Dick

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    This is a special case of the last problem you posted. You don't need a basis and you don't need a matrix. If Av=av and Au=bu (a not equal b) you want to show that if cv+du=0 then both c and d are zero. Hit that equation with (A-aI).
     
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