# Pascal matrix

In mathematics, particularly matrix theory and combinatorics, a Pascal matrix is a matrix (possibly infinite) containing the binomial coefficients as its elements. It is thus an encoding of Pascal's triangle in matrix form. There are three natural ways to achieve this: as a lower-triangular matrix, an upper-triangular matrix, or a symmetric matrix. For example, the 5 × 5 matrices are:

${\displaystyle L_{5}={\begin{pmatrix}1&0&0&0&0\\1&1&0&0&0\\1&2&1&0&0\\1&3&3&1&0\\1&4&6&4&1\end{pmatrix}}\,\,\,}$
${\displaystyle U_{5}={\begin{pmatrix}1&1&1&1&1\\0&1&2&3&4\\0&0&1&3&6\\0&0&0&1&4\\0&0&0&0&1\end{pmatrix}}\,\,\,}$
${\displaystyle S_{5}={\begin{pmatrix}1&1&1&1&1\\1&2&3&4&5\\1&3&6&10&15\\1&4&10&20&35\\1&5&15&35&70\end{pmatrix}}=L_{5}\times U_{5}}$
There are other ways in which Pascal's triangle can be put into matrix form, but these are not easily extended to infinity.[1]

## Definition

The non-zero elements of a Pascal matrix are given by the binomial coefficients:

${\displaystyle L_{ij}={i \choose j}={\frac {i!}{j!(i-j)!}},j\leq i}$
${\displaystyle U_{ij}={j \choose i}={\frac {j!}{i!(j-i)!}},i\leq j}$
${\displaystyle S_{ij}={i+j \choose i}={i+j \choose j}={\frac {(i+j)!}{i!j!}}}$

such that the indices i, j start at 0, and ! denotes the factorial.

## Properties

The matrices have the pleasing relationship Sn = LnUn. 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 Un. In other words, matrices Sn, Ln, and Un are unimodular, with Ln and Un having trace n.

The trace of Sn is given by

${\displaystyle {\text{tr}}(S_{n})=\sum _{i=1}^{n}{\frac {[2(i-1)]!}{[(i-1)!]^{2}}}=\sum _{k=0}^{n-1}{\frac {(2k)!}{(k!)^{2}}}}$

with the first few terms given by the sequence 1, 3, 9, 29, 99, 351, 1275, ... (sequence A006134 in the OEIS).

## Construction

A Pascal matrix can actually be constructed by taking the matrix exponential of a special subdiagonal or superdiagonal matrix. The example below constructs a 7 × 7 Pascal matrix, but the method works for any desired n × n Pascal matrices. The dots in the following matrices represent zero elements.

${\displaystyle {\begin{array}{lll}&L_{7}=\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.\\1&.&.&.&.&.&.\\.&2&.&.&.&.&.\\.&.&3&.&.&.&.\\.&.&.&4&.&.&.\\.&.&.&.&5&.&.\\.&.&.&.&.&6&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&.&.&.&.&.&.\\1&1&.&.&.&.&.\\1&2&1&.&.&.&.\\1&3&3&1&.&.&.\\1&4&6&4&1&.&.\\1&5&10&10&5&1&.\\1&6&15&20&15&6&1\end{smallmatrix}}\right];\quad \\\\&U_{7}=\exp \left(\left[{\begin{smallmatrix}{\color {white}1}.&1&.&.&.&.&.\\{\color {white}1}.&.&2&.&.&.&.\\{\color {white}1}.&.&.&3&.&.&.\\{\color {white}1}.&.&.&.&4&.&.\\{\color {white}1}.&.&.&.&.&5&.\\{\color {white}1}.&.&.&.&.&.&6\\{\color {white}1}.&.&.&.&.&.&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&1&1&1&1&1&1\\.&1&2&3&4&5&6\\.&.&1&3&6&10&15\\.&.&.&1&4&10&20\\.&.&.&.&1&5&15\\.&.&.&.&.&1&6\\.&.&.&.&.&.&1\end{smallmatrix}}\right];\\\\\therefore &S_{7}=\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.\\1&.&.&.&.&.&.\\.&2&.&.&.&.&.\\.&.&3&.&.&.&.\\.&.&.&4&.&.&.\\.&.&.&.&5&.&.\\.&.&.&.&.&6&.\end{smallmatrix}}\right]\right)\exp \left(\left[{\begin{smallmatrix}{\color {white}i}.&1&.&.&.&.&.\\{\color {white}i}.&.&2&.&.&.&.\\{\color {white}i}.&.&.&3&.&.&.\\{\color {white}i}.&.&.&.&4&.&.\\{\color {white}i}.&.&.&.&.&5&.\\{\color {white}i}.&.&.&.&.&.&6\\{\color {white}i}.&.&.&.&.&.&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&1&1&1&1&1&1\\1&2&3&4&5&6&7\\1&3&6&10&15&21&28\\1&4&10&20&35&56&84\\1&5&15&35&70&126&210\\1&6&21&56&126&252&462\\1&7&28&84&210&462&924\end{smallmatrix}}\right].\end{array}}}$

It is important to note that one cannot simply assume exp(A) exp(B) = exp(A + B), for n × n matrices A and B; this equality is only true 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 for the construction is that both are nilpotent; that is, when raised to a sufficiently great 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.

## Variants

Interesting variants can be obtained by obvious modification of the matrix-logarithm PL7 and then application of the matrix exponential.

The first example below uses the squares of the values of the log-matrix and constructs a 7 × 7 "Laguerre"- matrix (or matrix of coefficients of Laguerre polynomials

${\displaystyle {\begin{array}{lll}&LAG_{7}=\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.\\1&.&.&.&.&.&.\\.&4&.&.&.&.&.\\.&.&9&.&.&.&.\\.&.&.&16&.&.&.\\.&.&.&.&25&.&.\\.&.&.&.&.&36&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&.&.&.&.&.&.\\1&1&.&.&.&.&.\\2&4&1&.&.&.&.\\6&18&9&1&.&.&.\\24&96&72&16&1&.&.\\120&600&600&200&25&1&.\\720&4320&5400&2400&450&36&1\end{smallmatrix}}\right];\quad \end{array}}}$

The Laguerre-matrix is actually used with some other scaling and/or the scheme of alternating signs. (Literature about generalizations to higher powers is not found yet)

The second example below uses the products v(v + 1) of the values of the log-matrix and constructs a 7 × 7 "Lah"- matrix (or matrix of coefficients of Lah numbers)

${\displaystyle {\begin{array}{lll}&LAH_{7}=\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.\\2&.&.&.&.&.&.\\.&6&.&.&.&.&.\\.&.&12&.&.&.&.\\.&.&.&20&.&.&.\\.&.&.&.&30&.&.\\.&.&.&.&.&42&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&.&.&.&.&.&.&.\\2&1&.&.&.&.&.&.\\6&6&1&.&.&.&.&.\\24&36&12&1&.&.&.&.\\120&240&120&20&1&.&.&.\\720&1800&1200&300&30&1&.&.\\5040&15120&12600&4200&630&42&1&.\\40320&141120&141120&58800&11760&1176&56&1\end{smallmatrix}}\right];\quad \end{array}}}$

Using v(v − 1) instead provides a diagonal shifting to bottom-right.

The third example below uses the square of the original PL7-matrix, divided by 2, in other words: the first-order binomials (binomial(k, 2)) in the second subdiagonal and constructs a matrix, which occurs in context of the derivatives and integrals of the Gaussian error function:

${\displaystyle {\begin{array}{lll}&GS_{7}=\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.\\.&.&.&.&.&.&.\\1&.&.&.&.&.&.\\.&3&.&.&.&.&.\\.&.&6&.&.&.&.\\.&.&.&10&.&.&.\\.&.&.&.&15&.&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&.&.&.&.&.&.\\.&1&.&.&.&.&.\\1&.&1&.&.&.&.\\.&3&.&1&.&.&.\\3&.&6&.&1&.&.\\.&15&.&10&.&1&.\\15&.&45&.&15&.&1\end{smallmatrix}}\right];\quad \end{array}}}$

If this matrix is inverted (using, for instance, the negative matrix-logarithm), then this matrix has alternating signs and gives the coefficients of the derivatives (and by extension the integrals) of Gauss' error-function. (Literature about generalizations to greater powers is not found yet.)

Another variant can be obtained by extending the original matrix to negative values:

${\displaystyle {\begin{array}{lll}&\exp \left(\left[{\begin{smallmatrix}.&.&.&.&.&.&.&.&.&.&.&.\\-5&.&.&.&.&.&.&.&.&.&.&.\\.&-4&.&.&.&.&.&.&.&.&.&.\\.&.&-3&.&.&.&.&.&.&.&.&.\\.&.&.&-2&.&.&.&.&.&.&.&.\\.&.&.&.&-1&.&.&.&.&.&.&.\\.&.&.&.&.&0&.&.&.&.&.&.\\.&.&.&.&.&.&1&.&.&.&.&.\\.&.&.&.&.&.&.&2&.&.&.&.\\.&.&.&.&.&.&.&.&3&.&.&.\\.&.&.&.&.&.&.&.&.&4&.&.\\.&.&.&.&.&.&.&.&.&.&5&.\end{smallmatrix}}\right]\right)=\left[{\begin{smallmatrix}1&.&.&.&.&.&.&.&.&.&.&.\\-5&1&.&.&.&.&.&.&.&.&.&.\\10&-4&1&.&.&.&.&.&.&.&.&.\\-10&6&-3&1&.&.&.&.&.&.&.&.\\5&-4&3&-2&1&.&.&.&.&.&.&.\\-1&1&-1&1&-1&1&.&.&.&.&.&.\\.&.&.&.&.&0&1&.&.&.&.&.\\.&.&.&.&.&.&1&1&.&.&.&.\\.&.&.&.&.&.&1&2&1&.&.&.\\.&.&.&.&.&.&1&3&3&1&.&.\\.&.&.&.&.&.&1&4&6&4&1&.\\.&.&.&.&.&.&1&5&10&10&5&1\end{smallmatrix}}\right].\end{array}}}$