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The answer above by @A5C1D2H2I1M1N2O1R2T1 is not entirely correct due to index order in the upper.tri assignment. The answer by @Valentin_Stefan is, using the outer product. You will only see this when using four elements.

df <- data.frame(list(
    "Group" = c("A", "B", "C", "D"),
    "v.1" =   c( 1,   1,   1,   3 ),
    "v.2" =   c( 1,   1,   1,   2 )
    ))

m <- matrix(0, nrow = nrow(df), ncol = nrow(df),
            dimnames=list(df$Group, df$Group))

m[lower.tri(m)] <- combn(df$v.1, 2, FUN=diff)
m[upper.tri(m)] <- combn(df$v.2, 2, FUN=diff)
m
#   A B C D
# A 0 0 0 0
# B 0 0 1 1
# C 0 0 0 1
# D 2 2 2 0

(see the m[2,3] element which should be at m[1,4] instead)

I would transpose to correct; maybe there is an easier way?

n <- matrix(0, nrow = nrow(df), ncol = nrow(df),
            dimnames=list(df$Group, df$Group))

n[lower.tri(n)] <- combn(df$v.2, 2, FUN=diff)
n <- t(n)
n[lower.tri(n)] <- combn(df$v.1, 2, FUN=diff)

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