Say I have the matrix $\begin{pmatrix}16&16&8\\8&6&2\\3&4&2\end{pmatrix}$. To give a structure for a module homomorphism whose representation is given by this matrix, I must reduce the matrix to a Smith normal form. Now, in order to calculate the Smith normal form, I have to calculate the elementary divisors. Are they $2,8$ in this case?
I think no. This is because, though the first elementary divisor is given by $\frac{d_1}{d_0}=\frac{2}{1}=2$, but the second elementary divisor is just $\frac{d_2}{d_1}=\frac{2}{2}=1$ in my opinion. This is because, the gcd of all the $2\times 2$ minors of the matrix is $2$. Should we only consider the principal minors while calculating the second elementary divisor? Kindly elaborate. Thanks beforehand.
Reduce the matrix to its normal form by following the usual algorithm. In this case, we would go through the following steps: $$ \pmatrix{16&16&8\\8&6&2\\3&4&2} \to \pmatrix{3&4&2\\8&6&2\\16&16&8} \to \\ \pmatrix{1&4&2\\6&6&2\\8&16&8} \to \pmatrix{1&0&0\\6&-18&-10\\8&-16&-8}\to\\ \pmatrix{1&0&0\\0&-18&-10\\0&-16&-8} \to \pmatrix{1&0&0\\0&18&10\\0&16&8} \to\\ \pmatrix{1&0&0\\0&10&18\\0&8&16} \to \pmatrix{1&0&0\\0&2&2\\0&8&16} \to\\ \pmatrix{1&0&0\\0&2&2\\0&0&8} \to \pmatrix{1&0&0\\0&2&0\\0&0&8}. $$ Consequently, conclude that the invariant factors are $2,8$.
If you insist on using the formula with the minors, note that we have $d_1 = 1 \neq 2$. Indeed, the $3,1$ entry and $3,3$ entries are relatively prime, which means that the entries (the "first order minors") are relatively prime.
For the second factor, we compute the cofactor matrix to be $$ \pmatrix{4&-10&14\\0&8&-16\\-16&32&-32}. $$ The greatest common factor of these entries is $2$, which gives us the invariant factor $2/1 = 2$.
Finally, we compute the matrix of the whole matrix to be $-16$, which gives us the invariant factor $-16/2 = -8$. We can equivalently say that the invariant factor is simply $8$.