r 中对角矩阵的逆矩阵

Inverse of a diagonal matrix in r

我想取这个子集对角矩阵的逆(我希望我已经以正确的方式创建了一个 dput):

dput(DMIdiag[1:20,1:20])
structure(c(581.166666666667, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 5397.42, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1254652.485, 0, 0, 0, 0, 0, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15164.7616666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3721176.8, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1805.11333333333, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 266406189.456667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 478590.468333333, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 508510.586666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 348034.096666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 945785.841666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1056505.53666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1389813.64166667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 28585.2133333333, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 713675.006666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 661456.686666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 53533.9066666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 7291.31666666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 49375.4466666667, 
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 807758.686666667
), .Dim = c(20L, 20L))

我的代码:

DMI <- p + rowSums(Z, na.rm = TRUE)*1 #p is a vector and Z is a matrix
DMIdiag <- diag(DMI) #produces the matrix above
DMIinv <- solve(DMIdiag)

运行 最后一行给我:

Error in solve.default(DMIdiag) : 
  Lapack routine dgesv: system is exactly singular: U[173,173] = 0

我真的不明白这个矩阵怎么是奇异的,如果这就是问题所在?

您可以通过用倒数替换对角线分量来得到对角矩阵的逆矩阵。因此,如果你这样做

DMIinv <- DMIDiag
diag(DMIinv) <- 1/diag(DMIDiag)

然后 DMIinv 将是 DMIDiag 的倒数,我们可以通过这样做来证明:

DMIDiag %*% DMIinv
#>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
#>  [1,]    1    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [2,]    0    1    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [3,]    0    0    1    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [4,]    0    0    0    1    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [5,]    0    0    0    0    1    0    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [6,]    0    0    0    0    0    1    0    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [7,]    0    0    0    0    0    0    1    0    0     0     0     0     0     0     0     0     0     0     0     0
#>  [8,]    0    0    0    0    0    0    0    1    0     0     0     0     0     0     0     0     0     0     0     0
#>  [9,]    0    0    0    0    0    0    0    0    1     0     0     0     0     0     0     0     0     0     0     0
#> [10,]    0    0    0    0    0    0    0    0    0     1     0     0     0     0     0     0     0     0     0     0
#> [11,]    0    0    0    0    0    0    0    0    0     0     1     0     0     0     0     0     0     0     0     0
#> [12,]    0    0    0    0    0    0    0    0    0     0     0     1     0     0     0     0     0     0     0     0
#> [13,]    0    0    0    0    0    0    0    0    0     0     0     0     1     0     0     0     0     0     0     0
#> [14,]    0    0    0    0    0    0    0    0    0     0     0     0     0     1     0     0     0     0     0     0
#> [15,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     1     0     0     0     0     0
#> [16,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     1     0     0     0     0
#> [17,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     1     0     0     0
#> [18,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     1     0     0
#> [19,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     1     0
#> [20,]    0    0    0    0    0    0    0    0    0     0     0     0     0     0     0     0     0     0     0     1

这是 20 x 20 单位矩阵。