<s>[INST] Smith normal form of matrix over $Z$? I was wondering if someone could help me find the Smith normal form of the matrix A over $Z$ defined as follows:
$A =
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & 2 & 4 & 8 & 16\\
1 & 3 & 9 & 27 & 81\\
1 & 4 & 16 & 64 & 256\\
1 & 5 & 25 & 125 & 625\\
\end{bmatrix}$
Clearly this is the same as:
$\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & 2 & 2^2 & 2^3 & 12^4\\
1 & 3 & 3^2 & 3^3 & 3^4\\
1 & 4 & 4^2 & 4^3 & 4^5\\
1 & 5 & 5^2 & 5^3 & 5^4\\
\end{bmatrix}$
Is there some "trick" I am unaware of? Through row and column operations I just get stuck at a matrix looking like this:
$\begin{bmatrix}
1 & 0 & 0 & 0 & 0\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
0 & 1 & 1 & 1 & 1\\
\end{bmatrix}$
 [/INST] Define
$$P := \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 \\ -1 & 3 & -3 & 1 & 0 \\ 1 & -4 & 6 & -4 & 1 \end{bmatrix}.$$
Then
$$P A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 2 & 12 & 50 \\ 0 & 0 & 0 & 6 & 60 \\ 0 & 0 & 0 & 0 & 24\end{bmatrix} =
\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & 0 & 24\end{bmatrix} \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 1 & 6 & 25 \\ 0 & 0 & 0 & 1 & 10 \\ 0 & 0 & 0 & 0 & 1\end{bmatrix} =: DQ.$$
Now since $P$ and $Q$ are unimodular matrices by inspection, it follows that $D$ is the Smith normal form of $A$.
This should generalize nicely to higher values of $n$: if $P$ is the matrix whose $i,j$ element is $(-1)^{i+j} \binom{i-1}{j-1}$, then the $i,j$ element of $PA$ will be $(\Delta^{i-1} x^{j-1}) |_{x = 1}$, which is equal to 0 if $i > j$; equal to $(i-1)!$ if $i = j$; and divisible by $(i-1)!$ if $i < j$.  Therefore, we will again be able to get a similar factorization $PA = DQ$ where $D$ is diagonal with $D_{ii} = (i-1)!$; and $Q$ is upper triangular with 1 on the diagonal and therefore unimodular.
</s>