【数学】矩阵与矩阵运算

一、矩阵：

1.定义：由 m × n m times n m×n个数 a i j ( i = 1 , 2 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) a_{ij} (i = 1, 2,cdots,m; j = 1, 2,cdots ,n) aij​(i=1,2,⋯,m;j=1,2,⋯,n)排成的 m m m行 n n n列的数表，称为 m × n m times n m×n矩阵，记为：
A = ( a i j ) m × n A = left(a_{ij}right)_{m times n} A=(aij​)m×n​
在C++中，我们一般用结构体表示矩阵:

typedef struct {
	int li[101][101], m, n;
} matrix;

2.分类
(1).行矩阵： 1 × n 1 times n 1×n矩阵
(2).列矩阵： m × 1 m times 1 m×1矩阵
(3). n n n阶矩阵： n × n ntimes n n×n矩阵，记作 A n A_n An​
(4).单位矩阵：对角线为1，其余为0的n阶矩阵，记作 E n E_n En​
(5).零矩阵：全为0的矩阵，记作 O m × n O_{m times n} Om×n​。
二、矩阵运算：
1.加法：
设两个 m × n m times n m×n矩阵 A , B A, B A,B，则 A + B = ( a i j + b i j ) m × n ( i = 1 , 2 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) A + B = left( a_{ij} + b_{ij}right)_{m times n} (i = 1, 2,cdots,m; j = 1, 2,cdots ,n) A+B=(aij​+bij​)m×n​(i=1,2,⋯,m;j=1,2,⋯,n)
代码如下：

matrix operator + (matrix a, matrix b) {
    matrix I;
    I.m = a.m, I.n = a.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < a.n; j++)
            I.li[i][j] = a.li[i][j] + b.li[i][j];
    return I;
}

2.减法：
设两个 m × n m times n m×n矩阵 A , B A, B A,B，则 A − B = ( a i j − b i j ) m × n ( i = 1 , 2 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) A - B = left( a_{ij} - b_{ij}right)_{m times n} (i = 1, 2,cdots,m; j = 1, 2,cdots ,n) A−B=(aij​−bij​)m×n​(i=1,2,⋯,m;j=1,2,⋯,n)
代码如下：

matrix operator - (matrix a, matrix b) {
  matrix I;
  I.m = a.m, I.n = a.n;
  for (int i = 0; i < a.m; i++)
      for (int j = 0; j < a.n; j++)
          I.li[i][j] = a.li[i][j] - b.li[i][j];
  return I;
}

3.数乘：设 m × n m times n m×n矩阵 A A A，有 λ ∈ R lambda in mathbb {R} λ∈R，则 λ A = ( λ a i j ) m × n ( i = 1 , 2 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) lambda A = left(lambda a_{ij}right)_{m times n} (i = 1, 2,cdots,m; j = 1, 2,cdots ,n) λA=(λaij​)m×n​(i=1,2,⋯,m;j=1,2,⋯,n)
代码如下：

matrix numMul(int lambda, matrix a) {
    matrix I;
    I.m = a.m, I.n = a.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < a.n; j++)
            I.li[i][j] = a.li[i][j] * lambda;
    return I;
}

4.矩阵乘法：
设有 m × s m times s m×s矩阵 A A A， s × n s times n s×n矩阵 B B B，定义 A , B A, B A,B的乘积 C = ( c i j ) m × n C = (c_{ij})_{m times n} C=(cij​)m×n​，其中 c i j = ∑ k = 1 s a i k b k j ( i = 1 , 2 , ⋯   , m ; j = 1 , 2 , ⋯   , n ) c_{ij} = sum_{k = 1}^sa_{ik}b_{kj}(i = 1, 2,cdots,m; j = 1, 2,cdots ,n) cij​=k=1∑s​aik​bkj​(i=1,2,⋯,m;j=1,2,⋯,n)记为 C = A B C = AB C=AB，读作 A A A右乘 B B B。
代码如下：

#include 
matrix operator * (matrix a, matrix b) {
  matrix I;
  memset(I.li, 0, sizeof I.li);
  I.m = a.m, I.n = b.n;
  for (int i = 0; i < a.m; i++)
      for (int j = 0; j < b.n; j++)
          for (int k = 1; k <= a.n; k++)
              I.li[i][j] += a.li[i][j] * b.li[j][j];
  return I;
}

4.幂运算：
设 A A A是 n n n阶矩阵，定义 A A A的正整数幂为：
A 1 = A ， A k = A k − 1 A A^1 = A， A^k = A^{k - 1}A A1=A，Ak=Ak−1A
又定义 A 0 = E n A^0 = E_n A0=En​
故可以幂运算用矩阵乘法实现，代码如下。

#include 
matrix pows(matrix a, int n) {
    matrix I;
    I.m = I.n = a.n;
    memset(I.li, 0, sizeof I.li);
    for (int i = 0; i < I.n; i++) I.li[i][i] = 1;
    while (n--) I = I * a;
    return I;
}

亦可用快速幂：

#include 
matrix pows(matrix a, int n) {
    matrix I;
    I.m = I.n = a.n;
    memset(I.li, 0, sizeof I.li);
    for (int i = 0; i < I.n; i++) I.li[i][i] = 1;
    while (n > 1) {
        if (n & 1) I = I * a;
        n >>= 1, a = a * a;
    }
    return I * a;
}

如果需要取模的话可以进行取模：

#include 
const int mod = 10000 //此处10000为取模数，可以改变。
matrix operator % (matrix a, matrix b) {
    matrix I;
    memset(I.li, 0, sizeof I.li);
    I.m = a.m, I.n = b.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < b.n; j++)
            for (int k = 1; k <= a.n; k++)
                I.li[i][j] += a.li[i][j] * b.li[j][j] % mod;
    return I;
}

综上，矩阵及运算代码可归为以下代码：

#include 
typedef struct {
    int li[101][101], m, n;
} matrix;
matrix operator + (matrix a, matrix b) {
    matrix I;
    I.m = a.m, I.n = a.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < a.n; j++)
            I.li[i][j] = a.li[i][j] + b.li[i][j];
    return I;
}
matrix operator - (matrix a, matrix b) {
    matrix I;
    I.m = a.m, I.n = a.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < a.n; j++)
            I.li[i][j] = a.li[i][j] - b.li[i][j];
    return I;
}
matrix numMul(int lambda, matrix a) {
    matrix I;
    I.m = a.m, I.n = a.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < a.n; j++)
            I.li[i][j] = a.li[i][j] * lambda;
    return I;
}
matrix operator * (matrix a, matrix b) {
    matrix I;
    memset(I.li, 0, sizeof I.li);
    I.m = a.m, I.n = b.n;
    for (int i = 0; i < a.m; i++)
        for (int j = 0; j < b.n; j++)
            for (int k = 1; k <= a.n; k++)
                I.li[i][j] += a.li[i][j] * b.li[j][j];
    return I;
}
#include 
const int mod = 10000 //此处10000为取模数，可以改变。
matrix operator % (matrix a, matrix b) {
      matrix I;
      memset(I.li, 0, sizeof I.li);
      I.m = a.m, I.n = b.n;
      for (int i = 0; i < a.m; i++)
          for (int j = 0; j < b.n; j++)
              for (int k = 1; k <= a.n; k++)
	              I.li[i][j] += a.li[i][j] * b.li[j][j] % mod;
      return I;
matrix pows(matrix a, int n) {
    matrix I;
    I.m = I.n = a.n;
    memset(I.li, 0, sizeof I.li);
    for (int i = 0; i < I.n; i++) I.li[i][i] = 1;
    while (n > 1) {
        if (n & 1) I = I * a;
        n >>= 1, a = a * a;
    }
    return I * a;
}
matrix modpows(matrix a, int n) {
    matrix I;
    I.m = I.n = a.n;
    memset(I.li, 0, sizeof I.li);
    for (int i = 0; i < I.n; i++) I.li[i][i] = 1;
    while (n > 1) {
        if (n & 1) I = I % a;
        n >>= 1, a = a % a;
    }
    return I % a;
}