A fact about the invariant factors of a matrix

Let A be an n \times n matrix over a field F.

Let 1 \leq k \leq n, and let s,t : k \rightarrow n be injective, monotone functions. (That is, 1 < s_1 < s_2 < \ldots < s_k < n.) The (s,t)-minor of A, denoted (A)_{s,t}, is the k \times k matrix whose (i,j) entry is the (s_i,t_j) entry of A. We will call such s and t crunchy.

Let d_k(xI-A) be the unique monic generator in F[x] of the ideal H_k(xI-A) = (\mathsf{det}\ (xI-A)_{s,t} \ |\ s,t\ \mathrm{crunchy\ on}\ k).

Now suppose S is the Smith normal form of xI-A, and that the diagonal entries of S are D_i. Prove that D_i = d_i/d_{i-1}, taking d_0 = 1.

(Setting d_0 = 1 is in fact the proper thing to do. Think of a matrix as a function from n^2 to F; then if n = 0, this function is empty. Now S_0 = 1, and in the formula \mathsf{det}\ A = \sum_{\sigma \in S_0} \varepsilon(\sigma) \prod_{i=1}^0 a_{\sigma(i),i}, the empty product is 1, and we have one summand. So the ideal in F[x] is (1).)

[I consulted these notes by Gregg Musiker for this problem.]

We begin by arguing that d_k(A) = d_k(EA), where E is an elementary matrix. (We discussed elementary matrices previously.) Recall that left multiplication by an elementary matrix corresponds to one of the three elementary row operations. Let s and t be crunchy, and let S = \mathsf{im}\ s and T = \mathsf{im}\ t.

Suppose E interchanges rows i and j.

  1. If i,j \notin S, then (EA)_{s,t} = (A)_{s,t}, and so \mathsf{det}\ (EA)_{s,t} = \mathsf{det}\ (A)_{s,t}.
  2. If i,j \in S, then (EA)_{s,t} is obtained from (A)_{s,t} by interchanging rows s^{-1}(i) and s^{-1}(j). So we have (EA)_{s,t} = E^\prime (A)_{s,t} for some row-swapping elementary matrix E^\prime, and thus \mathsf{det}\ (EA)_{s,t} = -\mathsf{det}\ \mathsf{det}\ (A)_{s,t}.
  3. Suppose i \in S and j \notin S. Now let s^\prime be the (unique) crunchy function whose image is S with i replaced by j. Now (EA)_{s,t} is obtained from (A)_{s^\prime,t} by swapping some rows. So (EA)_{s,t} = E^\prime(A)_{s^\prime,t}, where E^\prime is some product of row-swapping elementary matrices, and we have \mathsf{det}\ (EA)_{s,t} = (-1)^p \mathsf{det}\ (A)_{s^\prime,t} for some p.

Thus we have H_k(A) = H_k(EA), and so d_k(A) = d_k(EA).

Now suppose E multiplies row i by a field element \alpha.

  1. If i \notin S, then (EA)_{s,t} = (A)_{s,t}, and so \mathsf{det}\ (EA)_{s,t} = \mathsf{det}\ (A)_{s,t}.
  2. If i \in S, then (EA)_{s,t} is obtained from (A)_{s,t} by multiplying row s^{-1}(i) by \alpha. So we have \mathsf{det}\ (EA)_{s,t} = \alpha \mathsf{det}\ (A)_{s,t}.

Thus we have H_k(A) = H_k(EA), and so d_k(A) = d_k(EA).

Finally, suppose E adds \alpha times row j to row i.

  1. If i \notin S, then (EA)_{s,t} = (A)_{s,t}, and so \mathsf{det}\ (EA)_{s,t} = \mathsf{det}\ (A)_{s,t}.
  2. If i \in S and j \in S, then (EA)_{s,t} is obtained from (A)_{s,t} by adding a multiple of row s^{-1}(j) to row s^{-1}(i). So (EA)_{s,t} = E^\prime(A)_{s,t} for some row-adding elementary matrix E^\prime, and we have \mathsf{det}\ (EA)_{s,t} = \mathsf{det}\ (A)_{s,t}.
  3. Suppose i \in S and j \notin S. Now (EA)_{s,t} is obtained from (A)_{s,t} by adding some unrelated row vector to row s^{-1}(i). In particular, we have \mathsf{det}\ (EA)_{s,t} = \sum_{\sigma \in S_k} \varepsilon(\sigma) \prod_{\ell=1}^k (a_{s(\ell), \sigma^{-1}(s(\ell))} + \delta_{i,s(\ell)} \alpha a_{j,\sigma^{-1}(s(\ell)}) = \mathsf{det}\ (A)_{s,t} + \mathsf{det}\ (E^\prime A)_{s,t}, where E^\prime is a product of two elementary matrices; one which swaps rows i and j, and one which multiplies the (new) ith row by \alpha. We’ve already seen that \mathsf{det}\ (E^\prime A)_{s,t} is a unit multiple of \mathsf{det}\ (A)_{s,t}.

Thus we have H_k(A) = H_k(EA), and so d_k(A) = d_k(EA).

So for any elementary matrix E, d_k(EA) = d_k(A). Now note that (A)_{s,t}^\mathsf{T} = (A^\mathsf{T})_{t,s}, so that d_k(A^\mathsf{T}) = d_k(A). So we also have d_k(AE) = d_k(A) for elementary matrices E.

By Theorem 21 in D&F, we have P(xI-A)Q = S in Smith Normal Form, where P and Q are products of elementary matrices. In particular, d_k(xI-A) = d_k(S).

Now we claim that d_k(S)/d_{k-1}(S) = D_k(S).

Claim: If t \neq s, then \mathsf{det}\ (S)_{s,t} = 0. To see this, Note that if we remove some row from a diagonal matrix, then the result has a zero column. Unless we also remove the corresponding column, the determinant is 0.

Now suppose s is crunchy on k. By the divisibility condition on the diagonal entries of S, we have S_{i,i} | S_{s(i),s(i)} for all i. In particular, the determinant of the k \times k minor with s(i) = i for 1 \leq i \leq k divides the determinant of every other k \times k minor. So d_k(S) = \prod_{i=1}^k D_i(S), and d_k/d_{k-1} = D_k follows.

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