[線代] diagonally dominant

[handsomepow (handsomepow)]

 Matrix A=[a_ij] 屬於 R^n*n is called strictly column

 diagonally dominant or diagonally dominant for short, if

                         n
              |a_ii| > sigma |a_ji|.
                      i=1,j=1

 Assume after one step of Gaussian elimination,

 the matrix A has the following form [a_11 (a_1)^T]
                                     [ 0     A_2  ]
                                    wwwwwwwwwwwwwwww
                               (這裡(a_1)^T 看不懂是啥東西)

(a) Show that A is nonsingular. Hint: Use Gershgorin's theorem.

(b) Show that matrix A_2 is still strictly column diagonally dominant,

    i.e.,Gaussian elimination with partial pivoting does not actually

    permute any rows.

  請求高手教我解這題





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※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 111.242.6.101
推 TWN2        :(a)直接用定理秒殺 (b)利用|x|-|y|<=|x-y|<=|x|+|y|
→ TWN2        :跟定義秒殺 然後你diagonally dominant寫錯 請見wiki
不太懂怎使用定理跟定義 可以請T大講解嗎 謝謝
※ 編輯: handsomepow     來自: 111.242.6.101        (11/25 08:02)
推 TWN2        :不如你說你哪裡卡住好了
→ handsomepow :(a)我知道Gershgorin's定理 但是跟A有啥關係
→ handsomepow :還有wwwwwww上面的矩陣那個(a_1)^T 我不太懂
→ TWN2        :(a)eigenvalue非0 (a_1)^T不重要 只是個vector
→ handsomepow :(b)還是不懂@@
→ TWN2        :你diagonally dominant當然永遠做不出來
→ handsomepow :diagonally dominan |對角線|>|該列其他元素合|
→ handsomepow :大於才對  打錯!
→ TWN2        :你去查一下wiki好嗎= = 不覺得你那行很怪嗎...
http://en.wikipedia.org/wiki/Diagonally_dominant_matrix
網路是大於等於  題目只有大於 =.=
※ 編輯: handsomepow     來自: 140.123.63.51        (11/25 11:30)