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In mathematics, particularly matrix theory and combinatorics, the Pascal matrix is an infinite matrix containing the binomial coefficients as its elements. There are three ways to achieve this: as either an upper-triangular matrix, a lower-triangular matrix, or a symmetric matrix. The 5×5 truncations of these are shown below. Upper triangular: lower triangular: symmetric: These matrices have the pleasing relationship ''S''''n'' = ''L''''n''''U''''n''. From this it is easily seen that all three matrices have determinant 1, as the determinant of a triangular matrix is simply the product of its diagonal elements, which are all 1 for both ''Ln'' and ''U''''n''. In other words, matrices ''S''''n'', ''L''''n'', and ''U''''n'' are unimodular, with ''L''''n'' and ''U''''n'' having trace ''n''. The elements of the symmetric Pascal matrix are the binomial coefficients, i.e. : In other words, : Thus the trace of ''Sn'' is given by : with the first few terms given by the sequence 1, 3, 9, 29, 99, 351, 1275, … . ==Construction== The Pascal matrix can actually be constructed by taking the matrix exponential of a special subdiagonal or superdiagonal matrix. The example below constructs a 7-by-7 Pascal matrix, but the method works for any desired ''n''×''n'' Pascal matrices. (Note that dots in the following matrices represent zero elements.) : It is important to note that one cannot simply assume exp(''A'')exp(''B'') = exp(''A'' + ''B''), for ''A'' and ''B'' ''n''×''n'' matrices. Such an identity only holds when ''AB'' = ''BA'' (i.e. when the matrices ''A'' and ''B'' commute). In the construction of symmetric Pascal matrices like that above, the sub- and superdiagonal matrices do not commute, so the (perhaps) tempting simplification involving the addition of the matrices cannot be made. A useful property of the sub- and superdiagonal matrices used in the construction is that both are nilpotent; that is, when raised to a sufficiently high integer power, they degenerate into the zero matrix. (See shift matrix for further details.) As the ''n''×''n'' generalised shift matrices we are using become zero when raised to power ''n'', when calculating the matrix exponential we need only consider the first ''n'' + 1 terms of the infinite series to obtain an exact result. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Pascal matrix」の詳細全文を読む スポンサード リンク
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