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anonymous

  • 5 years ago

hey im struggling with a qustion, can someone help please? if A = the matrices: −1 4 −2 −3 4 0 −3 1 3 Find an invertable matrix P and diagonal matrix D so that AP = PD. thank you for any feedback x

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  1. amistre64
    • 5 years ago
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    greenday will have to help you since they feel noone is capable of helping out ...

  2. amistre64
    • 5 years ago
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    D is just the eugene values on the identity

  3. amistre64
    • 5 years ago
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    I cant recall for the life of me what to do with the P tho

  4. amistre64
    • 5 years ago
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    do you recall how to find the eugene values?

  5. phi
    • 5 years ago
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    It is asking for the eigenvalues (diagonal entries of D) and eigenvectors (cols of P)

  6. amistre64
    • 5 years ago
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    −1-L 4 −2 −3 4-L 0 −3 1 3-L determinate of that and solve for L

  7. amistre64
    • 5 years ago
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    i spose we could echlon it too to find out but i never get a good result that way

  8. amistre64
    • 5 years ago
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    eugene vectors; thats the P .... ill remember that someday lol

  9. amistre64
    • 5 years ago
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    −1 4 −2 −3 4 0 −3 1 3 −1 4 −2 0 -8 6 0 -11 9 −1 4 −2 0 -8 6 0 0 6/8 maybe?

  10. amistre64
    • 5 years ago
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    −1-L 4 −2 −3 4-L 0 −3 1 3-L (-1-L)( (4-L)(3-L)-0) +3 (4(3-L)-1()-2) -3(0-(4-L)(-2) (-1-L)(L^2-7L+12) 36-12L+6 -24+6L -L^3+6L^2-11L+6 with any luck thats the eugene equations

  11. amistre64
    • 5 years ago
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    L=-1 1+6+11+6 not= 0 ugh!!

  12. amistre64
    • 5 years ago
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    maybe the eugenes are the negative diags? or i simply mismathed the whole thing

  13. JamesJ
    • 5 years ago
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    Yes, the matrix D has the eigenvalues of A down it's diagonal. So first you need to find the eigenvalues. Then the matrix P is the matrix of the eigenvectors as column vectors corresponding to those eigenvalues.

  14. amistre64
    • 5 years ago
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    if we echoln the matrix, dont we get the Evalues as well? or is that just me wishful thinking?

  15. JamesJ
    • 5 years ago
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    No, we don't. There's no escaping solving the characteristic equation. @sammy_maths: I'm sure you have a worked example of a problem like this in your class notes, albeit for a 2x2 or 3x3 matrix. I can't find a good one to link to right now. But go back and have a look at it. And then if you're still stuck, as us again.

  16. amistre64
    • 5 years ago
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    bummer... http://www.wolframalpha.com/input/?i=eigenvalues+%7B%7B-1%2C4%2C-2%7D%2C%7B-3%2C4%2C0%7D%2C%7B-3%2C1%2C3%7D%7D you can use this for a check :)

  17. amistre64
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=characteristic+polynomial%7B%7B-1%2C4%2C-2%7D%2C%7B-3%2C4%2C0%7D%2C%7B-3%2C1%2C3%7D%7D yay, my char eq was right lol

  18. JamesJ
    • 5 years ago
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    that's a big help.

  19. amistre64
    • 5 years ago
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    i know L=1 works by sheer luck, so whats left over is a quadratic

  20. anonymous
    • 5 years ago
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    I have been following it, and I'm understanding where you get the values from so far The only notes that may help now is the answer to P = 1 2 1 3 3 1 4 3 1 thank you x

  21. amistre64
    • 5 years ago
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    yeah, ill have to go over finding Evectors again to refresh me cobwebs :)

  22. amistre64
    • 5 years ago
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    it may be from the augmented | A LI| gauss jordon stuff

  23. amistre64
    • 5 years ago
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    (A+LI)X = 0

  24. phi
    • 5 years ago
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    Once you find an eigenvalue, subtract it from the diagonal, and find the null space of the resulting matrix. Use elimination to do this

  25. amistre64
    • 5 years ago
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    or simply −1+L 4 −2 −3 4+L 0 −3 1 3+L row reduced for each L

  26. phi
    • 5 years ago
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    to find the evals solve −1-L 4 −2 det −3 4-L 0 = 0 −3 1 3-L as previously posted

  27. amistre64
    • 5 years ago
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    http://www.sosmath.com/matrix/eigen2/eigen2.html might be the same, hard for me to tell

  28. phi
    • 5 years ago
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    it is easy with matlab or wolfram, and tedious by hand (-1-L)(4-L)(3-L) - 4(-3(3-L)) -2(-3 +3(4-L)) = 0

  29. amistre64
    • 5 years ago
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    i see, AX = LX AX - LX = 0 (A-LI)X = 0

  30. amistre64
    • 5 years ago
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    they had a L = -4 which threw me for a loop :)

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