hey there and thanks for the replies to my earlier question, here is the next one: i have defined both the (*) and (/) operators for my new type data Matrix a = Matr {unMatr :: [[a]]} | Scalar a deriving (Show, Eq) (*) being matrix multiplication and (/) being multiplication with inverse i also defined them for scalars this allows me to use the (^^) powering operator, which works fine with non-zero exponents. however to my surprise when i try (^^ 0) the answer is (Scalar 1), and not the identity matrix as expected. does this mean that (a ^^ 0) is not defined as (a ^^ 1 * a ^^ (-1)) (or better yet (a / a)) in the prelude? if so, can i redefine it so that it gives the right answer? i am also very interested in how ghci got the answer (Scalar 1), it seems quite magical:) here are my operator definitions: instance RealFrac a => Num (Matrix a) where (Matr as) * (Matr bs) = Matr $ timesH as bs (Scalar a) * (Matr b) = Matr $ timesSH a b (Matr a) * (Scalar b) = Matr $ timesSH b a (Scalar a) * (Scalar b) = Scalar (a * b) instance RealFrac a => Fractional (Matrix a) where (Scalar a) / (Scalar b) = Scalar (a / b) (Matr a) / (Scalar b) = Matr $ map (map (/ b)) a a / b | a == b = idM a | otherwise = a * (inv b) liftMatr = (Matr .) . (. unMatr) transH = foldr (zipWith (:)) (repeat []) timesH :: RealFrac a => [[a]] -> [[a]] -> [[a]] timesH as bs = map (\p -> map (foldr1 (+) . zipWith (*) p) (transH bs)) as timesSH :: RealFrac a => a -> [[a]] -> [[a]] timesSH a = map (map (a *)) idM :: RealFrac a => Matrix a -> Matrix a idM = liftMatr idMH idMH :: RealFrac a => [[a]] -> [[a]] idMH as = let n = length as in take n . map (take n) . iterate (0 :) $ 1 :repeat 0 -- View this message in context: http://old.nabble.com/powering-of-new-types-tp26891202p26891202.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.