Quod Erat Demonstrandum



Filed under: Pure Mathematics — johnmayhk @ 11:17 下午

5 則迴響 »

  1. Extended question: Let A be the 7 by 7 matrix in your exercise.

    (a) What are the eigenvalues of A.

    (b) What are the eigenvalues of A^{2010}.

    迴響 由 apook — 2010/11/19 @ 2:26 下午 | 回覆

  2. FYI(for other readers who have not heard of eigenvalues):

    t is an eigenvalue of A iff det(A – tI) = 0.

    迴響 由 apook — 2010/11/19 @ 2:28 下午 | 回覆

  3. 複平面上,單位圓內接正七邊形之頂點(其中一頂點為 z=1)。

    迴響 由 johnmayhk — 2010/11/19 @ 5:44 下午 | 回覆

  4. I am more interested in how you get your answer? Did you compute det(A – tI)?

    迴響 由 apook — 2010/11/22 @ 10:42 上午 | 回覆

    • I guess there are many ways to obtain the result

      \det(A - \lambda I) = 1 - \lambda^7

      For me, I just observe that every column in A contains 2 non-zero entries (1 and -\lambda).

      Just consider the definition of deteminant.

      When taking the entry -\lambda in C_1, then we must also take -\lambda‘s from other columns, obtaining (-\lambda)^7 as one of the non-zero terms in sum of \det A.

      When taking the entry 1 in C_1, it is easy to see that it is a must to take 1’s from other columns, obataining the term 1^7 as one of the non-zero terms.

      And all other terms in the sum are zeros, hence result follows.


      For not computing \det(A - \lambda I), how about:

      A\overrightarrow{v} = \lambda \overrightarrow{v} (where \overrightarrow{v} is an eigenvector corresponding to \lambda)


      \overrightarrow{v} = A^7\overrightarrow{v} = \lambda^7v

      (1 - \lambda^7)\overrightarrow{v} = \overrightarrow{0}


      1 - \lambda^7 = 0

      迴響 由 johnmayhk — 2010/11/22 @ 4:21 下午 | 回覆

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