Quod Erat Demonstrandum

2010/09/14

唔打:錯在哪裡之方程組

Filed under: Pure Mathematics — johnmayhk @ 8:59 下午

2 則迴響 »

  1. 可否這樣解釋:
    將(*)看成
    [p q 0][x] [A]
    [r s 0][y]=[B]
    [t u 0][z] [C]
    則乘以transpose matrix一步相當於
    [p r t][p q 0][x] [A]
    [q s u][r s 0][y]=[B]
    [0 0 0][t u 0][z] [C]
    借row operations的概念可知此舉相當於
    將3rd row清零,亦即將第三條等式變成0+0+0=0,或無視第三條等式。
    故乘以transpose matrix一步沒有考慮到第三條等式的存在,
    其結論自然只為必要條件而非充份條件。

    迴響 由 Ivan — 2010/09/15 @ 8:39 下午 | 回覆

    • Thank you Ivan.

      I just think about that

      for non-square matrix A,

      (A^tA)u = (A^t)v

      does not imply

      Au = v

      in general.

      迴響 由 johnmayhk — 2010/09/16 @ 6:17 下午 | 回覆


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