求矩阵特征值的pascal代码。

Question

求矩阵特征值的pascal代码。
多谢了。
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Answer 1

这个好象有点难度,一定要用PASCAL吗?MATLAB不行吗?

Answer 2

我记得OpenGL单元里有好多Matrix的函数，你找找看

Answer 3

按照求Eigenvalue的方法算就是了，貌似求Eigenvalue很简单的样子

Answer 4

这是以前解线性方程组写的一段代码，也许你可以参考一下：procedure a_equation(var A:matrx2;N:integer;var B:array of real);
//AX=B,结果 A=A逆，B=A逆*B
var
  IPIV,INDXR,INDXC:array[1..50] of integer;
  I,J,K,L,LL:integer;
  BIG,PIVINV,DUM:real;
  IROW,ICOL:integer;
  temp:array[1..50] of real;
begin
  for J:=1 to N do
    temp[J]:=B[J-1];  for J:=1 to N do
    IPIV[J]:=0;  for I:=1 tO N do
    BEGIN
      BIG:=0;
      for j:=1 to N do
        begin
          if IPIV[J] <> 1 then
            BEGIN
               for K:=1 to N do
                 begin
                   if IPIV[K] = 0 then
                     BEGIN
                       IF Abs(A[J,K]) >= BIG then
                         begin
                           BIG:=Abs(A[J,K]);
                           IROW:=J;
                           ICOL:=K;
                         end;
                     END
                   ELSE
                     BEGIN
                       if IPIV[K] > 1 then
                         begin
                            ShowMessage('系数矩阵不可逆！');
                            exit;
                          end;
                     END;                 end;
            END;
        end;
      IPIV[ICOL]:=IPIV[ICOL]+1;
      If IROW <> ICOL then
        begin
          for L:=1 to N do
            BEGIN
              DUM:=A[IROW,L];
              A[IROW,L]:=A[ICOL,L];
              A[ICOL,L]:=DUM;
            END;
          DUM:=B[IROW];
          B[IROW]:=B[ICOL];
          B[ICOL]:=DUM;
        end;
      INDXR[I]:=IROW;
      INDXC[I]:=ICOL;
      if A[ICOL,ICOL] = 0 then
        begin
          ShowMessage('系数矩阵不可逆！');
          exit;
        end;
      PIVINV:=1/A[ICOL,ICOL];
      A[ICOL,ICOL]:=1;
      for L:=1 to N do
        A[ICOL,L]:=A[ICOL,L]*PIVINV;
      B[ICOL]:=B[ICOL]*PIVINV;
      for LL:=1 tO N do
        begin
          if LL <>  ICOL then
            BEGIN
              DUM:=A[LL,ICOL];
              A[LL,ICOL]:=0;
              for L:=1 to N DO
                A[LL,L]:=A[LL,L]-A[ICOL,L]*DUM;
              B[LL]:=B[LL]-B[ICOL]*DUM;
            END;
        end;
    END;
   for L:=N Downto 1 do
     begin
       if INDXR[L] <> INDXC[L] then
         BEGIN
           for K:=1 TO N do
             begin
               DUM:=A[K,INDXR[L]];
               A[K,INDXR[L]]:=A[K,INDXC[L]];
               A[K,INDXC[L] ]:=DUM;
             end;
         END;
     end;  for i:=1 to N do
    begin
      B[i-1]:=0;
      for j:=1 to N do
        B[i-1]:=B[i-1]+A[i,j]*temp[j];
    end;
end;