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.

  • $\begingroup$ It sounds like you're doing things backwards. Typically, one finds the elementary divisors by first computing the Smith normal form via the standard algorithm. $\endgroup$ – Ben Grossmann Dec 21 '20 at 23:25

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 $8$.

  • $\begingroup$ thanks, but could you elaborate the usual algorithm, as the wiki page does not explain it clearly, and previous para is quite confusing $\endgroup$ – vidyarthi Dec 22 '20 at 7:02
  • 1
    $\begingroup$ @vidyarthi I don't see it explained this way, but here's an equivalent "recursive" characterization. Step 1: using linear combinations if necessary, get at least one entry equal to the greatest common divisor of the matrix entries. Step 2: switch the common divisor to the upper-left entry (with row/column permutations). Step 3: use row/column operations to produce zeroes in the opposite row. Step 4: put the submatrix (excluding the first row/column) into Smith normal form. The steps I follow in this answer are like this except I switch steps 1 and 2. $\endgroup$ – Ben Grossmann Dec 22 '20 at 15:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.