R可以求解欠定线性系统：
A = matrix((1:12)^2,3,4,T)
B = 1:3
qr(A)$rank # 3
qr.solve(A, B) # solutions will have one zero, not necessarily the same one
# 0.1875 -0.5000 0.3125 0.0000
solve(qr(A, LAPACK = TRUE), B)
# 0.08333333 -0.18750000 0.00000000 0.10416667
(它给出了无穷多个解中的一个解(。
然而，如果秩(这里是2(低于行数(这里是3(，它就不起作用：
A = matrix(c((1:8)^2,0,0,0,0),3,4,T)
B = c(1,2,0)
A
# [,1] [,2] [,3] [,4]
# [1,] 1 4 9 16
# [2,] 25 36 49 64
# [3,] 0 0 0 0
qr.solve(A, B) # Error in qr.solve(A, B) : singular matrix
solve(qr(A, LAPACK = TRUE), B) # Error in qr.coef(a, b) : error code 3
但这个系统确实有解决方案！
我知道一般的解决方案是使用SVD或A的广义/伪逆(见这个问题及其答案(，但是：
对于solve或qr.solve，是否存在将系统AX=B自动缩减为仅列(a(行的等效系统CX=D的平均值，对于该系统，qr.solve(C, D)将简单地开箱即用？
示例：
C = matrix(c((1:8)^2),2,4,T)
D = c(1,2)
qr.solve(C, D)
# -0.437500 0.359375 0.000000 0.000000
qr.coef和qr似乎起到了作用：
(A <- matrix(c((1:8)^2, 0, 0, 0, 0), nrow = 3, ncol = 4, byrow = TRUE))
# [,1] [,2] [,3] [,4]
# [1,] 1 4 9 16
# [2,] 25 36 49 64
# [3,] 0 0 0 0
(B <- c(1, 2, 0))
# [1] 1 2 0
(X0 <- qr.coef(qr(A), B))
# [1] -0.437500 0.359375 NA NA
X0[is.na(X0)] <- 0
X0
# [1] -0.437500 0.359375 0.000000 0.000000
# Verification:
A %*% X0
# [,1]
# [1,] 1
# [2,] 2
# [3,] 0
第二个例子：
(A<-matrix(c(1, 2, 0, 0, 1, 2, 0, 0, 1, 2, 1, 0), nrow = 3, ncol = 4, byrow = TRUE))
# [,1] [,2] [,3] [,4]
# [1,] 1 2 0 0
# [2,] 1 2 0 0
# [3,] 1 2 1 0
(B<-c(1, 1, 2))
# [1] 1 1 2
qr.solve(A, B)
# Error in qr.solve(A, B) : singular matrix 'a' in solve
(X0 <- qr.coef(qr(A), B))
# [1] 1 NA 1 NA
X0[is.na(X0)] <- 0
X0
# [1] 1 0 1 0
A %*% X0
# [,1]
# [1,] 1
# [2,] 1
# [3,] 2