您所在的位置：
&数学计算器 V1.7.3.0 (支持矩阵运算) 绿色版
点击查看大图
大小：546 KB
语言：简体中文
授权：免费软件
数学计算器软件介绍
数学计算器功能：1、支持矩阵的加、减、乘、除、求秩、求逆、特征值、广义逆、qr分解、lu分解、schur分解、奇异值分解、满秩分解、约当标准化、givens变换、化为hessenberg矩阵……2、支持常规线性方程组的求解以及不确定稀疏矩阵的求解
3、支持一元任意次多项式的加、减、乘、除、求根、变号数……【凡是与多项式相关的命令函数均以poly开头】
4、支持任意一元函数的求根、求定积分、二维空间作图……例如任意给个方程:sin(x)*exp(x)-x^2+8=0,求x的解,本程序可以很轻易解决.
5、支持数据的多项式拟合,数据的lagrange插值,数据的三次样条插值以及函数的最佳平方逼近
6、数学计算器支持大数的加、减、乘、除、求余、求次方(任意次方,包括负数,小数)、求最小公倍数、求最大公约数、求阶乘(50000以内的数)、任意位数的2进制与10进制与16进制的转换……【凡是与大数计算的相关函数均已bg开头】
7、支持空间几何的点与直线、平面求距离,直线与直线、平面求距离和夹角和交点、直线与任意曲面求交点,【凡是空间几何相关的函数均已geo开头】
8、数学计算器支持概率统计里的第一、二类贝塞尔函数,第一、二类艾里函数,贝塔函数,伽马函数,x2分布,正太分布,f分布,t分布求解……
9、支持坐标变换、数独的求解以及构造一些特殊的矩阵(如:幻方、hankel、hurwitz矩阵等)
10、支持常规表达式的计算,例如,当你在界面输入如下一行的内容后,按下键盘的enter键,将得到0.37这个答案
&&& sin(45*cos(36+exp(12))+log(36))*4.59
数学计算器相关下载
数学计算器下载地址
12326次下载
3524次下载
3094次下载
5019次下载
2065次下载
7482次下载
84835次下载
5023次下载
3510次下载
1277次下载
84835次下载
74933次下载
49997次下载
49738次下载
49377次下载
49353次下载
49032次下载
48677次下载
48582次下载
48249次下载
热门关键字
扫红码得红包5亿红包100%中奖 上传我的文档
 下载
 收藏
该文档贡献者很忙，什么也没留下。
 下载此文档
正在努力加载中...
矩阵的概念与基本运算
下载积分：3000
内容提示：矩阵的概念与基本运算，矩阵的概念与
文档格式：PPT|
浏览次数：9|
上传日期： 05:34:27|
文档星级：
该用户还上传了这些文档
矩阵的概念与基本运算
官方公共微信golang go语言的矩阵运算 - 为程序员服务
为程序员服务
go语言的矩阵运算
// Package matrix implements a simple library for creating and
// manipulating matrices, and performing basic linear algebra.
package matrix
type Matrix struct {
rows, columns int
// the number of rows and columns.
[]int // the contents of the matrix as one long slice.
// Set lets you define the value of a matrix at the given row and
// column.
func (A *Matrix) Set(r, c, val int) {
A.data[findIndex(r, c, A)] = val
// Get retrieves the contents of the matrix at the row and column.
func (A *Matrix) Get(r, c int) int {
return A.data[findIndex(r, c, A)]
// Print converts the matrix into a string and then outputs it to fmt.Printf.
func (A *Matrix) Print() {
// Find the width (in characters) that each column needs to be.
We hold these
// widths as strings, not ints, because we're going to use these in a printf
// function.
columnWidths := make([]string, A.columns)
for i := range columnWidths {
var maxLength int
thisColumn := A.Column(i + 1)
for j := range thisColumn {
thisLength := len(strconv.Itoa(thisColumn[j]))
if thisLength & maxLength {
maxLength = thisLength
columnWidths[i] = strconv.Itoa(maxLength)
// We have the widths, so now output each element with the correct column
// width so that they line up properly.
for i := 0; i & A. i++ {
thisRow := A.Row(i + 1)
fmt.Printf(&[&)
for j := range thisRow {
var printFormat string
if j == 0 {
printFormat = &%& + columnWidths[j] + &s&
printFormat = & %& + columnWidths[j] + &s&
fmt.Printf(printFormat, strconv.Itoa(thisRow[j]))
fmt.Printf(&]\\n&)
// Column returns a slice that represents a column from the matrix.
// This works by examining each row, and adding the nth element of
// each to the column slice.
func (A *Matrix) Column(n int) []int {
col := make([]int, A.rows)
for i := 1; i &= A. i++ {
col[i-1] = A.Row(i)[n-1]
return col
// Row returns a slice that represents a row from the matrix.
func (A *Matrix) Row(n int) []int {
return A.data[findIndex(n, 1, A):findIndex(n, A.columns+1, A)]
// Multiply multiplies two matrices together and return the resulting matrix.
// For each element of the result matrix, we get the dot product of the
// corresponding row from matrix A and column from matrix B.
func Multiply(A, B Matrix) *Matrix {
C := Zeros(A.rows, B.columns)
for r := 1; r &= C. r++ {
A_row := A.Row(r)
for c := 1; c &= C. c++ {
B_col := B.Column(c)
C.Set(r, c, dotProduct(A_row, B_col))
// Add adds two matrices together and returns the resulting matrix.
// this, we just add together the corresponding elements from each matrix.
func Add(A, B Matrix) Matrix {
C := Zeros(A.rows, A.columns)
for r := 1; r &= A. r++ {
for c := 1; c &= A. c++ {
C.Set(r, c, A.Get(r, c)+B.Get(r, c))
// Identity creates an identity matrix with n rows and n columns.
// multiply any matrix by its corresponding identity matrix, you get the
// original matrix.
The identity matrix looks like a zero-filled matrix with
// a diagonal line of one's starting at the upper left.
func Identity(n int) Matrix {
A := Zeros(n, n)
for i := 0; i & len(A.data); i += (n + 1) {
A.data[i] = 1
// Zeros creates an r x c sized matrix that's filled with zeros.
The initial
// state of an int is 0, so we don't have to do any initialization.
func Zeros(r, c int) Matrix {
return Matrix{r, c, make([]int, r*c)}
// New creates an r x c sized matrix that is filled with the provided data.
// The matrix data is represented as one long slice.
func New(r, c int, data []int) Matrix {
if len(data) != r*c {
panic(&[]int data provided to matrix.New is great than the provided capacity of the matrix!'&)
A := Zeros(r, c)
A.data = data
// findIndex takes a row and column and returns the corresponding index
// from the underlying data slice.
func findIndex(r, c int, A *Matrix) int {
return (r-1)*A.columns + (c - 1)
// dotProduct calculates the algebraic dot product of two slices.
This is just
// the sum
of the products of corresponding elements in the slices.
// this when we multiply matrices together.
func dotProduct(a, b []int) int {
var total int
for i := 0; i & len(a); i++ {
total += a[i] * b[i]
return total
//该片段来自于
您可能的代码
相关聚客文章
相关专栏文章R中的矩阵运算 - R爱好者的日志,人人网,R爱好者的公共主页
想用R软件做计量的——请在自己学校的图书馆搜索论文《运用r进行计量经济学教学》学习
R中的矩阵运算
矩阵计算R笔记
1 创建一个向量在R中可以用函数c()来创建一个向量，例如：& x=c(1,2,3,4)& x[1] 1 2 3 42 创建一个矩阵在R中可以用函数matrix()来创建一个矩阵，应用该函数时需要输入必要的参数值。& args(matrix)function (data = NA, nrow = 1, ncol = 1, byrow = FALSE, dimnames= NULL)data 项为必要的矩阵元素，nrow 为行数，ncol 为列数，注意nrow 与ncol 的乘积应为矩阵元素个数，byrow 项控制排列元素时是否按行进行，dimnames 给定行和列的名称。例如：& matrix(1:12,nrow=3,ncol=4)[,1] [,2] [,3] [,4][1,] 1 4 7 10[2,] 2 5 8 11[3,] 3 6 9 12& matrix(1:12,nrow=4,ncol=3)[,1] [,2] [,3][1,] 1 5 9[2,] 2 6 10[3,] 3 7 11[4,] 4 8 12& matrix(1:12,nrow=4,ncol=3,byrow=T)[,1] [,2] [,3][1,] 1 2 3[2,] 4 5 6[3,] 7 8 9[4,] 10 11 12& rowname[1] "r1" "r2" "r3"& colname=c("c1","c2","c3","c4")& colname[1] "c1" "c2" "c3" "c4"& matrix(1:12,nrow=3,ncol=4,dimnames=list(rowname,colname))c1 c2 c3 c4r1 1 4 7 10r2 2 5 8 113 矩阵转置A 为m&n 矩阵，求A'在R中可用函数t()，例如：& A=matrix(1:12,nrow=3,ncol=4)& A[,1] [,2] [,3] [,4][1,] 1 4 7 10[2,] 2 5 8 11[3,] 3 6 9 12& t(A)[,1] [,2] [,3][1,] 1 2 3[2,] 4 5 6[3,] 7 8 9[4,] 10 11 12若将函数t()作用于一个向量x，则R默认x为列向量，返回结果为一个行向量，例如：& x[1] 1 2 3 4 5 6 7 8 9 10& t(x)[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10][1,] 1 2 3 4 5 6 7 8 9 10& class(x)[1] "integer"& class(t(x))[1] "matrix"若想得到一个列向量，可用t(t(x))，例如：& x[1] 1 2 3 4 5 6 7 8 9 10& t(t(x))[,1][1,] 1[2,] 2[3,] 3[4,] 4[5,] 5[6,] 6[7,] 7[8,] 8[9,] 9[10,] 10& y=t(t(x))& t(t(y))[,1][1,] 1[2,] 2[3,] 3[4,] 4[5,] 5[6,] 6[7,] 7[8,] 8[9,] 9[10,] 104 矩阵相加减在R中对同行同列矩阵相加减，可用符号：&＋&、&－&，例如：& A=B=matrix(1:12,nrow=3,ncol=4)& A+B[,1] [,2] [,3] [,4][1,] 2 8 14 20[2,] 4 10 16 22[3,] 6 12 18 24& A-B[,1] [,2] [,3] [,4][1,] 0 0 0 0[2,] 0 0 0 0[3,] 0 0 0 05 数与矩阵相乘A 为m&n 矩阵，c&0，在R中求cA 可用符号：&*&，例如：& c=2& c*A[,1] [,2] [,3] [,4][1,] 2 8 14 20[2,] 4 10 16 22[3,] 6 12 18 246 矩阵相乘A 为m&n 矩阵，B为n&k 矩阵，在R中求AB可用符号：&％*％&，例如：& A=matrix(1:12,nrow=3,ncol=4)& B=matrix(1:12,nrow=4,ncol=3)& A%*%B[,1] [,2] [,3][1,] 70 158 246[2,] 80 184 288[3,] 90 210 330若A为n&m矩阵，要得到A'B，可用函数crossprod()，该函数计算结果与t(A)%*%B相同，但是效率更高。例如：& A=matrix(1:12,nrow=4,ncol=3)& B=matrix(1:12,nrow=4,ncol=3)& t(A)%*%B[,1] [,2] [,3][1,] 30 70 110[2,] 70 174 278[3,] 110 278 446& crossprod(A,B)[,1] [,2] [,3][1,] 30 70 110[2,] 70 174 278[3,] 110 278 4467 矩阵对角元素相关运算例如要取一个方阵的对角元素，& A=matrix(1:16,nrow=4,ncol=4)& A[,1] [,2] [,3] [,4][1,] 1 5 9 13[2,] 2 6 10 14[3,] 3 7 11 15[4,] 4 8 12 16& diag(A)[1] 1 6 11 16对一个向量应用diag()函数将产生以这个向量为对角元素的对角矩阵，例如：& diag(diag(A))[,1] [,2] [,3] [,4][1,] 1 0 0 0[2,] 0 6 0 0[3,] 0 0 11 0[4,] 0 0 0 16对一个正整数z应用diag()函数将产生以z维单位矩阵，例如：& diag(3)[,1] [,2] [,3][1,] 1 0 0[2,] 0 1 0[3,] 0 0 18 矩阵求逆矩阵求逆可用函数solve()，应用solve(a, b)运算结果是解线性方程组ax = b，若b缺省，则系统默认为单位矩阵，因此可用其进行矩阵求逆，例如：& a=matrix(rnorm(16),4,4)& a[,1] [,2] [,3] [,4][1,] 1....3174184[2,] -0....8096514[3,] -0....3679963[4,] -0....2424030& solve(a)[,1] [,2] [,3] [,4][1,] 0....3813059[2,] -0....6957775[3,] -0....1046207[4,] -0....6607851& solve (a) %*%a[,1] [,2] [,3] [,4][1,] 1. 2. -2. -8.[2,] 1. 1. -4. 6.[3,] 2. -4. 1. 6.[4,] 1. 1. 1. 1.9 矩阵的特征值与特征向量矩阵A 的谱分解为A=UΛU',其中Λ 是由A 的特征值组成的对角矩阵，U 的列为A 的特征值对应的特征向量，在R中可以用函数eigen()函数得到U和Λ，& args(eigen)function (x, symmetric, only.values = FALSE, EISPACK = FALSE)其中：x为矩阵，symmetric项指定矩阵x是否为对称矩阵，若不指定，系统将自动检测x是否为对称矩阵。例如：& A=diag(4)+1& A[,1] [,2] [,3] [,4][1,] 2 1 1 1[2,] 1 2 1 1[3,] 1 1 2 1[4,] 1 1 1 2& A.eigen=eigen(A,symmetric=T)& A.eigen$values[1] 5 1 1 1$vectors[,1] [,2] [,3] [,4][1,] 0.5 0.. 0.0000000[2,] 0.5 -0.. 0.8164966[3,] 0.5 -0.. -0.4082483[4,] 0.5 -0.. -0.4082483& A.eigen$vectors%*%diag(A.eigen$values)%*%t(A.eigen$vectors)[,1] [,2] [,3] [,4][1,] 2 1 1 1[2,] 1 2 1 1[3,] 1 1 2 1[4,] 1 1 1 2& t(A.eigen$vectors)%*%A.eigen$vectors[,1] [,2] [,3] [,4][1,] 1. 4. 1. -5.[2,] 4. 1. -1. 6.[3,] 1. -1. 1. -1.[4,] -5. 6. -1. 1.10 矩阵的Choleskey分解对于正定矩阵A，可对其进行Choleskey分解，即：A=P'P，其中P为上三角矩阵，在R中可以用函数chol()进行Choleskey分解，例如：& A[,1] [,2] [,3] [,4][1,] 2 1 1 1[2,] 1 2 1 1[3,] 1 1 2 1[4,] 1 1 1 2& chol(A)[,1] [,2] [,3] [,4][1,] 1....7071068[2,] 0....4082483[3,] 0....2886751[4,] 0....1180340& t(chol(A))%*%chol(A)[,1] [,2] [,3] [,4][1,] 2 1 1 1[2,] 1 2 1 1[3,] 1 1 2 1[4,] 1 1 1 2& crossprod(chol(A),chol(A))[,1] [,2] [,3] [,4][1,] 2 1 1 1[2,] 1 2 1 1[3,] 1 1 2 1[4,] 1 1 1 2若矩阵为对称正定矩阵，可以利用Choleskey分解求行列式的值，如：& prod(diag(chol(A))^2)[1] 5& det(A)[1] 5若矩阵为对称正定矩阵，可以利用Choleskey分解求矩阵的逆，这时用函数chol2inv()，这种用法更有效。如：& chol2inv(chol(A))[,1] [,2] [,3] [,4][1,] 0.8 -0.2 -0.2 -0.2[2,] -0.2 0.8 -0.2 -0.2[3,] -0.2 -0.2 0.8 -0.2[4,] -0.2 -0.2 -0.2 0.8& solve(A)[,1] [,2] [,3] [,4][1,] 0.8 -0.2 -0.2 -0.2[2,] -0.2 0.8 -0.2 -0.2[3,] -0.2 -0.2 0.8 -0.2[4,] -0.2 -0.2 -0.2 0.811 矩阵奇异值分解A为m&n矩阵，rank(A)= r, 可以分解为：A=UDV',其中U'U=V'V=I。在R中可以用函数scd()进行奇异值分解，例如：& A=matrix(1:18,3,6)& A[,1] [,2] [,3] [,4] [,5] [,6][1,] 1 4 7 10 13 16[2,] 2 5 8 11 14 17[3,] 3 6 9 12 15 18& svd(A)$d[1] 4. 1. 3.$u[,1] [,2] [,3][1,] -0...4082483[2,] -0...8164966[3,] -0...4082483$v[,1] [,2] [,3][1,] -0...[2,] -0...[3,] -0...[4,] -0...[5,] -0...[6,] -0...& A.svd=svd(A)& A.svd$u%*%diag(A.svd$d)%*%t(A.svd$v)[,1] [,2] [,3] [,4] [,5] [,6][1,] 1 4 7 10 13 16[2,] 2 5 8 11 14 17[3,] 3 6 9 12 15 18& t(A.svd$u)%*%A.svd$u[,1] [,2] [,3][1,] 1. -1. -3.[2,] -1. 1. -3.[3,] -3. -3. 1.& t(A.svd$v)%*%A.svd$v[,1] [,2] [,3][1,] 1. 8. -3.[2,] 8. 1. -2.[3,] -3. -2. 1.12 矩阵QR分解A为m&n矩阵可以进行QR分解，A=QR，其中：Q'Q＝I，在R中可以用函数qr()进行QR分解，例如：& A=matrix(1:16,4,4)& qr(A)$qr[,1] [,2] [,3] [,4][1,] -5... -2.[2,] 0... -9.[3,] 0... 2.[4,] 0... -2.$rank[1] 2$qraux[1] 1. 1. 1. 2.$pivot[1] 1 2 3 4attr(,"class")[1] "qr"rank项返回矩阵的秩，qr项包含了矩阵Q和R的信息，要得到矩阵Q和R，可以用函数qr.Q()和qr.R()作用qr()的返回结果，例如：& qr.R(qr(A))[,1] [,2] [,3] [,4][1,] -5... -2.[2,] 0... -9.[3,] 0... 2.[4,] 0... -2.& qr.Q(qr(A))[,1] [,2] [,3] [,4][1,] -0.. -0..[2,] -0.. 0..[3,] -0.. 0..[4,] -0.. -0..& qr.Q(qr(A))%*%qr.R(qr(A))[,1] [,2] [,3] [,4][1,] 1 5 9 13[2,] 2 6 10 14[3,] 3 7 11 15[4,] 4 8 12 16& t(qr.Q(qr(A)))%*%qr.Q(qr(A))[,1] [,2] [,3] [,4][1,] 1. -1. -6. -7.[2,] -1. 1. -4. 7.[3,] -6. -4. 1. -1.[4,] -7. 7. -1. 1.& qr.X(qr(A))[,1] [,2] [,3] [,4][1,] 1 5 9 13[2,] 2 6 10 14[3,] 3 7 11 15[4,] 4 8 12 1613 矩阵广义逆(Moore-Penrose)n&m矩阵A+称为m&n矩阵A的Moore-Penrose逆，如果它满足下列条件：① A A+A=A；②A+A A+= A+；③(A A+)H=A A+；④(A+A)H= A+A在R的MASS包中的函数ginv()可计算矩阵A的Moore-Penrose逆，例如：library(&MASS&)& A[,1] [,2] [,3] [,4][1,] 1 5 9 13[2,] 2 6 10 14[3,] 3 7 11 15[4,] 4 8 12 16& ginv(A)[,1] [,2] [,3] [,4][1,] -0.285 -0. 0.2475[2,] -0.145 -0. 0.1325[3,] -0.005 0. 0.0175[4,] 0.135 0. -0.0975验证性质1：& A%*%ginv(A)%*%A[,1] [,2] [,3] [,4][1,] 1 5 9 13[2,] 2 6 10 14[3,] 3 7 11 15[4,] 4 8 12 16验证性质2：& ginv(A)%*%A%*%ginv(A)[,1] [,2] [,3] [,4][1,] -0.285 -0. 0.2475[2,] -0.145 -0. 0.1325[3,] -0.005 0. 0.0175[4,] 0.135 0. -0.0975验证性质3:& t(A%*%ginv(A))[,1] [,2] [,3] [,4][1,] 0.7 0.4 0.1 -0.2[2,] 0.4 0.3 0.2 0.1[3,] 0.1 0.2 0.3 0.4[4,] -0.2 0.1 0.4 0.7& A%*%ginv(A)[,1] [,2] [,3] [,4][1,] 0.7 0.4 0.1 -0.2[2,] 0.4 0.3 0.2 0.1[3,] 0.1 0.2 0.3 0.4[4,] -0.2 0.1 0.4 0.7验证性质4:& t(ginv(A)%*%A)[,1] [,2] [,3] [,4][1,] 0.7 0.4 0.1 -0.2[2,] 0.4 0.3 0.2 0.1[3,] 0.1 0.2 0.3 0.4[4,] -0.2 0.1 0.4 0.7& ginv(A)%*%A[,1] [,2] [,3] [,4][1,] 0.7 0.4 0.1 -0.2[2,] 0.4 0.3 0.2 0.1[3,] 0.1 0.2 0.3 0.4[4,] -0.2 0.1 0.4 0.714 矩阵Kronecker积n&m矩阵A与h&k矩阵B的kronecker积为一个nh&mk维矩阵，公式为：11 11nmn h km mn mh nka aa a& &&& &&A = & & & && && &B BA BB B?? ? ??在R中kronecker积可以用函数kronecker()来计算，例如：& A=matrix(1:4,2,2)& B=matrix(rep(1,4),2,2)& A[,1] [,2][1,] 1 3[2,] 2 4& B[,1] [,2][1,] 1 1[2,] 1 1& kronecker(A,B)[,1] [,2] [,3] [,4][1,] 1 1 3 3[2,] 1 1 3 3[3,] 2 2 4 4[4,] 2 2 4 415 矩阵的维数在R中很容易得到一个矩阵的维数，函数dim()将返回一个矩阵的维数，nrow()返回行数，ncol()返回列数，例如：& A=matrix(1:12,3,4)& A[,1] [,2] [,3] [,4][1,] 1 4 7 10[2,] 2 5 8 11[3,] 3 6 9 12& nrow(A)[1] 3& ncol(A)[1] 416 矩阵的行和、列和、行平均与列平均在R中很容易求得一个矩阵的各行的和、平均数与列的和、平均数，例如：& A[,1] [,2] [,3] [,4][1,] 1 4 7 10[2,] 2 5 8 11[3,] 3 6 9 12& rowSums(A)[1] 22 26 30& rowMeans(A)[1] 5.5 6.5 7.5& colSums(A)[1] 6 15 24 33& colMeans(A)[1] 2 5 8 1117 矩阵X'X的逆在统计计算中，我们常常需要计算这样矩阵的逆，如OLS估计中求系数矩阵。R中的包&strucchange&提供了有效的计算方法。& args(solveCrossprod)function (X, method = c("qr", "chol", "solve"))其中：method指定求逆方法，选用&qr&效率最高，选用&chol&精度最高，选用&slove&与slove(crossprod(x,x))效果相同，例如：& A=matrix(rnorm(16),4,4)& solveCrossprod(A,method="qr")[,1] [,2] [,3] [,4][1,] 0....2054730[2,] -0....2097302[3,] -0....3162961[4,] 0....3447627& solveCrossprod(A,method="chol")[,1] [,2] [,3] [,4][1,] 0....2054730[2,] -0....2097302[3,] -0....3162961[4,] 0....3447627& solveCrossprod(A,method="solve")[,1] [,2] [,3] [,4][1,] 0....2054730[2,] -0.1543...
阅读(2207)|
人人移动客户端下载 上传我的文档
 下载
 收藏
该文档贡献者很忙，什么也没留下。
 下载此文档
正在努力加载中...
矩阵运算技巧
下载积分：100
内容提示：矩阵运算技巧
文档格式：PDF|
浏览次数：777|
上传日期： 01:09:37|
文档星级：
该用户还上传了这些文档
矩阵运算技巧
官方公共微信