如何使用代码在R或Python中求解矩阵方程PAP^T=I

Question

x=matrix(c(rnorm(10,1,3), rnorm(10,7,2), rnorm(10,15,4)), byrow=FALSE, nrow=10)
xc=x
for (k in 1:3) {
  xc[,k]=xc[,k]-mean(xc[,k])
}
A=var(xc)

出去

> A
          [,1]      [,2]      [,3]
[1,] 10.166746 -2.619763 -6.717475
[2,] -2.619763  3.291888  3.052898
[3,] -6.717475  3.052898 22.313351

我想解PAP^T=I，其中p是3x3，p^T是p的转置，I是3x3单位矩阵。解算（A）解出PA=I，但我想要PAP^T=I。我找不到答案，将矩阵相乘将花费太长时间。有人知道如何在R或Python中实现这一点吗？我也不想使用P也可以是^（-1/2）的事实。通过代码的解决方案会很好

library(stringr)
A=matrix(c(letters[1:4], letters[6:8], letters[10:11]), nrow=3)
mult=function(A, B) {
  AB=matrix(nrow=3, ncol=3)
  for (r in 1:3) {
    for (c in 1:3) {
      AB[r,c]=str_c("(", A[r,1], ")*(", B[1,c], ")+(", A[r,2], ")*(", B[2,c], ")+(", A[r,3], ")*(", B[3,c], ")")
    }
  }
  return(AB)
}
AB=mult(A,B)
ABAT=mult(AB, t(A))

seteq=function(A, B) {
  E=matrix(nrow=3, ncol=3)
  for (r in 1:3) {
    for (c in 1:3) {
      E[r,c]=str_c(A[r,c], "=", B[r, c])
    }
  }
  dim(E)=c(9,1)
  return(E[,1])
}
Eq=seteq(ABAT, diag(rep(1, 3)))
res=""
for (i in 1:9) {
  res=str_c(res, Eq[i])
  if (i!=9) res=str_c(res, ",")
}
res

给我方程式

((a)*(18.9052758456522)+(d)*(0.834836569187845)+(h)*(9.76575358263375))*(a)+((a)*(0.834836569187845)+(d)*(0.555656932954595)+(h)*(1.03585433030833))*(d)+((a)*(9.76575358263375)+(d)*(1.03585433030833)+(h)*(6.84209765783974))*(h)=1,((b)*(18.9052758456522)+(f)*(0.834836569187845)+(j)*(9.76575358263375))*(a)+((b)*(0.834836569187845)+(f)*(0.555656932954595)+(j)*(1.03585433030833))*(d)+((b)*(9.76575358263375)+(f)*(1.03585433030833)+(j)*(6.84209765783974))*(h)=0,((c)*(18.9052758456522)+(g)*(0.834836569187845)+(k)*(9.76575358263375))*(a)+((c)*(0.834836569187845)+(g)*(0.555656932954595)+(k)*(1.03585433030833))*(d)+((c)*(9.76575358263375)+(g)*(1.03585433030833)+(k)*(6.84209765783974))*(h)=0,((a)*(18.9052758456522)+(d)*(0.834836569187845)+(h)*(9.76575358263375))*(b)+((a)*(0.834836569187845)+(d)*(0.555656932954595)+(h)*(1.03585433030833))*(f)+((a)*(9.76575358263375)+(d)*(1.03585433030833)+(h)*(6.84209765783974))*(j)=0,((b)*(18.9052758456522)+(f)*(0.834836569187845)+(j)*(9.76575358263375))*(b)+((b)*(0.834836569187845)+(f)*(0.555656932954595)+(j)*(1.03585433030833))*(f)+((b)*(9.76575358263375)+(f)*(1.03585433030833)+(j)*(6.84209765783974))*(j)=1,((c)*(18.9052758456522)+(g)*(0.834836569187845)+(k)*(9.76575358263375))*(b)+((c)*(0.834836569187845)+(g)*(0.555656932954595)+(k)*(1.03585433030833))*(f)+((c)*(9.76575358263375)+(g)*(1.03585433030833)+(k)*(6.84209765783974))*(j)=0,((a)*(18.9052758456522)+(d)*(0.834836569187845)+(h)*(9.76575358263375))*(c)+((a)*(0.834836569187845)+(d)*(0.555656932954595)+(h)*(1.03585433030833))*(g)+((a)*(9.76575358263375)+(d)*(1.03585433030833)+(h)*(6.84209765783974))*(k)=0,((b)*(18.9052758456522)+(f)*(0.834836569187845)+(j)*(9.76575358263375))*(c)+((b)*(0.834836569187845)+(f)*(0.555656932954595)+(j)*(1.03585433030833))*(g)+((b)*(9.76575358263375)+(f)*(1.03585433030833)+(j)*(6.84209765783974))*(k)=0,((c)*(18.9052758456522)+(g)*(0.834836569187845)+(k)*(9.76575358263375))*(c)+((c)*(0.834836569187845)+(g)*(0.555656932954595)+(k)*(1.03585433030833))*(g)+((c)*(9.76575358263375)+(g)*(1.03585433030833)+(k)*(6.84209765783974))*(k)=1

但是Wolfram Alpha没有识别出这个输入

Answer 1

正如你所注意到的，如果你用代数“平方根”分解A

A = L * L^T

那么问题就变成了：

(P*L) * (P*L)^T = Id

这意味着P*L是酉矩阵https://en.wikipedia.org/wiki/Unitary_matrix

因此，该解由任何酉矩阵U（U * U^T=Id）形成

P=inv(L)*U

有无限多的解决方案

您明确地声明不希望使用U=Id的平凡解决方案P=inv(L)，但您没有说明原因，并且您没有提供关于p

请注意，@G.Grothendieck的好答案中的第一个解决方案就是：用特征向量V以对角形式D（特征值）减少A

inv(V)*A*V = D
D = sqrt(D)*sqrt(D)
P = inv(sqrt(D))*V