我有一个矩阵m
:
(m <- matrix(c(26,14,24,14,20,42,24,42,90), 3))
# [,1] [,2] [,3]
# [1,] 26 14 24
# [2,] 14 20 42
# [3,] 24 42 90
当我运行solve(m)
来计算矩阵的逆矩阵时,我收到以下错误消息:
solve(m)
solve.default(m)出错: 系统在计算上是单数的:倒数条件数= 6.21104e-18
答案 0 :(得分:9)
我们可以看到,必须以多种方式实现这一点,每种方式都意味着不可逆性:
1) m
的决定因素为零:
> det(m)
[1] -2.685852e-12
2) m具有零特征值,即eigen(m)$values[3]
。等效地,m
的零空间是非空的 - 它等于eigen(m)$vectors[, 3]
所跨越的1维空间
> e <- eigen(m); e
$values
[1] 1.180000e+02 1.800000e+01 -6.446353e-15
$vectors
[,1] [,2] [,3]
[1,] -0.2881854 9.486833e-01 0.1301889
[2,] -0.4116935 1.110223e-16 -0.9113224
[3,] -0.8645563 -3.162278e-01 0.3905667
> N <- e$vector[, 3] # nullspace
> m %*% N # shows that N is indeed mapped to zero
[,1]
[1,] 5.329071e-15
[2,] 0.000000e+00
[3,] 0.000000e+00
3) m
的列不线性独立。特别是在其他列上回归m[,1]
给出了完美拟合(即拟合值等于m[, 1]
),因此从线性模型的系数得到7 * m[,2] - 3 * m[, 3]
等于m[, 1]
。
> fm <- lm(m[, 1] ~ m[, 2] + m[, 3] + 0)
> all.equal(fitted(fm), m[, 1]) # perfect fit
[1] TRUE
> coef(fm)
m[, 2] m[, 3]
7 -3
> all.equal(7 * m[, 2] - 3 * m[, 3], m[, 1])
[1] TRUE
4) cholesky分解的对角线为零:
> chol(m, pivot = TRUE)
[,1] [,2] [,3]
[1,] 9.486833 2.529822 4.4271887
[2,] 0.000000 4.427189 0.6324555
[3,] 0.000000 0.000000 0.0000000
attr(,"pivot")
[1] 3 1 2
attr(,"rank")
[1] 2
Warning message:
In chol.default(m, pivot = TRUE) :
the matrix is either rank-deficient or indefinite
5) m
不是满级,即排名小于3:
> attr(chol(m, pivot = TRUE), "rank")
[1] 2
Warning message:
In chol.default(m, pivot = TRUE) :
the matrix is either rank-deficient or indefinite
注意:输入可通过以下方式重复输出:
m <- matrix(c(26, 14, 24, 14, 20, 42, 24, 42, 90), 3)
答案 1 :(得分:8)
问题是列不是线性独立的。
第一列* -1/3 +第二列* 7/3等于第三列。
-m[, 1] * 1/3 + 7/3 * m[, 2]
# [1] 24 42 90