矩阵的左邻空间怎么求
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A化成最简型：
1 0 2 0 1
0 1 3 0 0
0 0 0 1 4
所以Ax=0的解是：
[x1,x2,x3,x4,x5]' = x3[-2 -3 1 0 0]' + x5[-1 0 0 -4 1]'
所以零空间为这两个向量张成的空间：
span([-2 -3 1 0 0]',[-1 0 0 -4 1]')
绛旓細闆剁┖闂村氨鏄眰Ax=0鐨勮В A鍖栨垚鏈绠鍨嬶細1 0 2 0 1 0 1 3 0 0 0 0 0 1 4 鎵浠x=0鐨勮В鏄細[x1,x2,x3,x4,x5]' = x3[-2 -3 1 0 0]' + x5[-1 0 0 -4 1]'鎵浠ラ浂绌洪棿涓鸿繖涓や釜鍚戦噺寮犳垚鐨勭┖闂锛歴pan([-2 -3 1 0 0]',[-1 0 0 -4 1]')
绛旓細4. 浜岀淮FFT FFT绠楁硶鍙敤浜庤绠椾簩缁村彉鎹,杩欐椂鍙鍏姹傜煩闃姣忎竴琛岀殑鍙樻崲,鐒跺悗鍐嶆眰鎵鏈夊垪鐨勫彉鎹.鍏剁畻娉曚负:1) for(j=0;j { x1[j]=x[i][j]; y1[j]=y[i][j]; } ;灏嗙i琛屾暟鎹嫹鍏ヤ复鏃舵暟缁,x,y鍒嗗埆涓哄疄閮ㄥ拰铏氶儴 2) FFT1(x1,y1,M,f); ;瀵圭i琛屾暟鎹綔涓缁碏FT 3) for(j=0;j {...
绛旓細瀵逛簬k=3, 姹傚緱{a}=[6/24, 12/24, 6/24,0]', 鍥犳S(n)=n^4/4+n^3/2+n^2/4+0+0=(n(n+1)/2)^2 瀵逛簬k=5, 姹傚緱{a}=[1/6, 1/2, 5/12, 0, -1/12, 0]',鍥犳S(n)=n^6/6+n^5/2+5*n^4/12-n^2/12 鐜板湪姹俷^5/4+(3/4)n^3+n鐨勫墠n椤瑰拰, 鍒...
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