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liliy

  • 2 years ago

Show that if A is nxn and has all 0’s on and below the diagonal then An = 0. Hint: do not at first be too ambitious. First find A2 and observe something useful about it. What about A3?

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  1. myininaya
    • 2 years ago
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    What is An?

  2. liliy
    • 2 years ago
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    the matrix

  3. liliy
    • 2 years ago
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    commonly seen as Ax=b. this is An=0.

  4. myininaya
    • 2 years ago
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    Oh I thought there might have been a difference because you called the matrix A then you called it An

  5. myininaya
    • 2 years ago
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    Are you saying n is an eigenvector ?

  6. liliy
    • 2 years ago
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    honesly i dont know wat to do. this is what the teacher asked us

  7. myininaya
    • 2 years ago
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    Or are you say An is the matrix?

  8. liliy
    • 2 years ago
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    idk..lol

  9. myininaya
    • 2 years ago
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    Ok I think |dw:1340595284758:dw| and |dw:1340595304816:dw|

  10. myininaya
    • 2 years ago
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    Is that what you think?

  11. myininaya
    • 2 years ago
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    But that doesn't make since that A_n would be the matrix with nothing but zero entries

  12. myininaya
    • 2 years ago
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    do you mean the determinant is 0?

  13. myininaya
    • 2 years ago
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    |A_n|=0?

  14. myininaya
    • 2 years ago
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    that would make since

  15. myininaya
    • 2 years ago
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    sense*

  16. myininaya
    • 2 years ago
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    I think that is what you mean

  17. myininaya
    • 2 years ago
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    so do you know how to find the determinant of a matrix?

  18. liliy
    • 2 years ago
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    ya ad-bc

  19. myininaya
    • 2 years ago
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    Try finding the determinant of A_2 ? What do you get?

  20. liliy
    • 2 years ago
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    zero

  21. myininaya
    • 2 years ago
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    Ok what about A_3

  22. liliy
    • 2 years ago
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    same

  23. myininaya
    • 2 years ago
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    Ok so we have convinced ourselves that |A_n|=0 But we must prove it

  24. myininaya
    • 2 years ago
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    |dw:1340595692432:dw|

  25. myininaya
    • 2 years ago
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    I would just show a little work for this show a pattern you know

  26. myininaya
    • 2 years ago
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    like how you did for A_3 and then do the nth term you know what I mean?

  27. liliy
    • 2 years ago
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    how do u find determinant for 3x3 or bigger matrix?

  28. myininaya
    • 2 years ago
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    oh ok for an A_3 |dw:1340595913293:dw|

  29. myininaya
    • 2 years ago
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    |dw:1340595950507:dw|

  30. myininaya
    • 2 years ago
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    The signs alternate

  31. myininaya
    • 2 years ago
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    Like you take top entries

  32. liliy
    • 2 years ago
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    only the first row?

  33. myininaya
    • 2 years ago
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    And take everything that isn't below that entry

  34. liliy
    • 2 years ago
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    like if its 10x10 u still only do the first row /?

  35. myininaya
    • 2 years ago
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    For |A_4| |dw:1340596049002:dw|

  36. myininaya
    • 2 years ago
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    |dw:1340596102188:dw|

  37. myininaya
    • 2 years ago
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    and you already know how to find the determinant for a 3 by 3

  38. myininaya
    • 2 years ago
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    Same thing just take the top entries and do the signs alternating

  39. eliassaab
    • 2 years ago
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    \[ A=\left( \begin{array}{cccc} 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^2=\left( \begin{array}{cccc} 0 & 0 & 1 & 10 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^3=\left( \begin{array}{cccc} 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)\\ A^4=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \]

  40. liliy
    • 2 years ago
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    but the determinant of the new 3x3 is gonna also be broken down right? .. im ur case its zero bec the coefficient is zero so it odsnt really mater

  41. myininaya
    • 2 years ago
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    yes just like i did above for the 3 by 3

  42. myininaya
    • 2 years ago
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    But not all the top entries are 0

  43. liliy
    • 2 years ago
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    @eliassaab i dont really understnad wat u wrote

  44. liliy
    • 2 years ago
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    right...

  45. eliassaab
    • 2 years ago
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    See my example below and examine what is going on?

  46. liliy
    • 2 years ago
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    you have zeros. but what are you doing to the matrix?

  47. myininaya
    • 2 years ago
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    Do you think he means to raise A to a power @eliassaab ?

  48. myininaya
    • 2 years ago
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    Instead of finding the determinant ?

  49. eliassaab
    • 2 years ago
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    You raise it to the power 2, then 3, then 4.

  50. myininaya
    • 2 years ago
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    Ok I'm sorry @liliy I don't know what your question is asking anymore.

  51. eliassaab
    • 2 years ago
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    You do not need to deal with determina

  52. eliassaab
    • 2 years ago
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    determinant

  53. liliy
    • 2 years ago
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    so can you start over with me?

  54. liliy
    • 2 years ago
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    what does a^n=0 even mean?

  55. eliassaab
    • 2 years ago
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    Any matrix like yours, when you raise it to the power 2, you get what is first above the diagonal is zero When you raise it to the power 2, you get the first and the second above the diagonal to be zero. When you raise it to the power 3, you get the first and the second and third above the diagonal to be zero. When you raise it to the power 4, you get everything zero.

  56. eliassaab
    • 2 years ago
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    Look at A^4 in my example above to see that A^4=0, this means all the entries of the matrix A^4 are zeros.

  57. myininaya
    • 2 years ago
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    So you are just giving another way right @eliassaab Do you think I interpreted is question correctly?

  58. eliassaab
    • 2 years ago
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    @myininaya, you do not need determinant to do that,

  59. myininaya
    • 2 years ago
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    Yeah I know, but I'm asking you if I interpreted it correctly?

  60. liliy
    • 2 years ago
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    @eliassaab i dont undesrtnad how you started to do the problem. my teacher said start with a^2 .. and move to bigger ones... so wat is a= to a 4x4 and then writing a^2... a^3..

  61. eliassaab
    • 2 years ago
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    Here is a quick proof using the characteristic polynomial f(x) of the matrix A that says the f(A)=0. Our matrix has\( f(x)=x^n\), hence \(f(A)=A^n=0\)

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