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N term metrix question

  1. Feb 5, 2008 #1
  2. jcsd
  3. Feb 5, 2008 #2

    EnumaElish

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    Why not set it up as |A − λI| = 0 (where I is the identity matrix), then solve for λ?
     
  4. Feb 5, 2008 #3
    it makes no difference

    my gowl was to find the determinat
    and that what i have done
    now what next??
    i encountered some realy tough problems

    ??
     
  5. Feb 5, 2008 #4

    EnumaElish

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    Just start with the formula I stated, then proceed from there.
     
  6. Feb 5, 2008 #5
    i know your formula
    youll noticed that i used it
    but in a different step
    i dont think there is any difference
    i have trouble to finish it
     
  7. Feb 5, 2008 #6

    HallsofIvy

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    That's exactly what he did. Although he miswrote it as det A= ... when he has det A-[itex]\lambda[/itex]I after the "=".


    transgalactic, for [itex]\lambda= 1[/itex] you have [itex]x_2+ x_3+ \cdot\cdot\cdot+ x_n= 0[/itex], [itex]-x_2= x_2[/itex], ..., [itex]-x_n= x_n[/itex]. What does that give you? (It should give you one simple vector.)

    For [itex]\lamba= 0[/itex], you have [itex]x_1+ x_2+ \cdot\cdot\dot+ x_n= 0[/itex]. What does that give you? (n- 1 vectors)
     
  8. Feb 6, 2008 #7
    for L=1
    you got the expression of x2 in "n" terms
    i cant do it indefinetly

    for L=0
    i know that i have n-1 vectors

    what do i do now
    what do i right as the answer of the question
    ??
     
  9. Feb 6, 2008 #8

    HallsofIvy

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    For [itex]\lambda= 1[/itex] you have, as I said before, [itex]-x_2= x_2[/itex],[itex]-x_3= x_3[/itex], etc. That tells you that [itex]x_2= x_3= \cdot\cdot\cdot= 0[/itex]! Then, of course, the first equation, [itex]x_2+ x_1+ \cdot\cdot\cdot+ x_n= 0[/itex] is automatically satisfied. Every number except [itex]x_1[/itex] must be 0. Since [itex]x_1[/itex] does not appear in any equation, it is arbitrary. All eigenvectors corresponding to [itex]\lambda= 1[/itex] are of the form <a, 0, 0, ..., 0> which is spanned, of course, by <1, 0, 0, ..., 0>.

    If [itex]\lambda= 0[/itex] then you have the single equation [itex]x_1+ x_2+ \cdot\cdot\cdot+ \x_n= 0[/itex] which is the same as [itex]x_n= -(x_1+ x_2+ x_3+ \cdot\cdot\cdot+ x_{n-1})[/itex].
    Now do as I have suggested before: take each [itex]x_i[/itex] equal to 0 in turn, the others 0, and solve for [itex]x_n[/itex]. That will give you the n-1 vectors you need.
     
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