A<-matrix(c(4,5,5,6,4,3,6,11,31),nrow=3,ncol=3)
B
B<-cov(A)
[,1] [,2] [,3]
[1,] 0.3333333 -0.8333333 5.0
[2,] -0.8333333 2.3333333 -17.5
[3,] 5.0000000 -17.5000000 175.0
bsvd<-svd(B)
D<-diag(1/bsvd$d)
D
V<-bsvd$v
U<-bsvd$u
a<-V%*%D
a%*%t(U)
a%*%t(U)
[,1] [,2] [,3]
[1,] 3.420461e+15 1.954549e+15 9.772746e+13
[2,] 1.954549e+15 1.116885e+15 5.584426e+13
[3,] 9.772746e+13 5.584426e+13 2.792213e+12
pseudoinverse(B)
[,1] [,2] [,3]
[1,] 0.32090310 -0.5583242 -0.06512408
[2,] -0.55832422 0.9714162 0.11302345
[3,] -0.06512408 0.1130235 0.01887380
I am unsure why this is giving me different answers? The formula for the pseudo inverse is VDU^T from what I have seen online.
Set D[3,3] <- 0, i.e. exclude the zero eigenvalue of B from the inverse, and then recalculate VDU^T and you'll get the same answer as the pseudoinverse function you are using.