38 Permutation Formula

Existence of the Determinant Learning Goals: students learn that the determinant really exists, and find some formulas for it. So far our formula for the determinant is ±(product of pivots). This isn’t such a good formulas, because for all we know changing the order of the rows might change the pivots, or at least the sign. And it doesn’t tell us how the determinant depends on any particular entry in the matrix. So we need some other ways of finding the determinant. “Other,” not “better.” Why? Because in practice we find the determinant by reducing and multiplying the pivots (if we bother finding it at all—its more theoretically useful than practically useful). In theory, though, other formulas give us more insight. Let’s use the properties we have found to discover a new formula. ⎡ a11 … a1n ⎤ Start with a matrix ⎢ ⎥ . We have the linearity property that says the ⎢ ! " ! ⎥ ⎢am1 # amn ⎥ ⎣ ⎦ a11 … a1n determinant is linear in each row. So we can split up ! " ! in the linear combination of a # a m1 mn 1 0 ! 0 0 1 ! 0 0 0 ! 1 a a ! a a a ! a a a ! a the first row: a 21 22 2n + a 21 22 2n +!+ a 21 22 2n . So 11 " " # " 12 " " # " n1 " " # " a a ! a a a ! a a a ! a n1 n2 nn n1 n2 nn n1 n2 nn there are n terms. Each of these can be split up into the linear combination of the second row, and so on, until we get nn terms which are all possible combinations of one element from each row, times a determinant of a matrix with one 1 in each row. But many—indeed most—of these terms are zero. For the matrices that have ones in the same column will have two identical rows and thus have determinant zero. The only terms remaining are the ones where each 1 is in a different column. Permutations and permutation matrices We recognize these matrices as permutation matrices, the P’s we used in row reduction to swap rows. We will formulize this a little more now: Definition: a permutation σ is a function whose domain and image are both the set of numbers {1, 2, …, n}. The permutation matrix P is the matrix which has one 1 in each row, and the 1 in row k is in column σ(k). The determinant of a permutation matrix is either 1 or –1, because after changing rows around (which changes the sign of the determinant) a permutation matrix becomes I, whose determinant is one. Definition: the sign of a permutation, sgn(σ), is the determinant of the corresponding permutation matrix. Of course, this may not be well defined. How do we know that one way of turning P into I doesn’t require an odd number of row swaps while a different way of doing it might need an even number? We still have to prove that sgn(σ) is uniquely defined. But once that is done we will have proved Theorem: the determinant of a matrix is where the summation is ∑a1σ (1)a2σ (2) !anσ (n) sgn(σ ) σ taken over all possible permutations σ of 1 – n. The text refers to this as the “big formula.” It is quite large—there are n! terms in the sum. This is one reason why determinants are not taken this way—computationally n! compared to n3 operations for row reduction is a nightmare. This does prove something, though. If the sign of a permutation is well-defined, then there is a unique function that satisfies our defining properties for determinants. In other words, we now know that there is at most one determinant. If we show that signs are well-defined and that this formula does actually satisfy the properties, we will have shown existence, and we’re done. Theorem: sgn(σ) is well-defined. Proof: we will find a function whose behavior is very easy to understand why it is well defined and show that it is equivalent to sgn(σ). To this end, given permutation σ, define its “disorderliness” d(σ) (not a standard term—just a convenience for this proof) to be the number of pairs (i, j) with i < j but σ(i) > σ(j). For example, in the permutation (3, 7, 1, 4, 2, 6, 5) (that is, σ(1) = 3, σ(2) = 7 and so forth) the disordered pairs are (1, 3), (1, 7), (2, 3), (2, 4), (2, 7), (4, 7), (5, 6), (5, 7) and (6, 7), so the disorderliness is 9. It is clear that the disorderliness of a permutation is well-defined. We will show that sgn(σ) = 1 if d(σ) is even and –1 if it is odd. First, if any two adjacent values in σ are switched, the disorderliness changes by exactly one. For example, d(3, 7, 4, 1, 2, 6, 5) is 10, while that of (3, 7, 1, 4, 2, 6, 5) is 9. The simple reason for this is that switching two adjacent numbers changes the relative order of only the pair involving those two numbers. Now swapping any two terms in σ will change the disorderliness by an odd number. Why? Because any swap can be achieved by swapping adjacent items k times (if there are k things between the items being swapped) until the two items to be swapped are next to each other, swapping them, and then making k more adjacent swaps until the rest of the terms are back in order. For example, there are two things between the a and the d below, and we can swap them and leave everything else alone, with 2⋅2 + 1 = 5 adjacent swaps: (a, b, c, d, e) → (b, a, c, d, e) → (b, c, a, d, e) → (b, c, d, a, e) → (b, d, c, a, e) → (d, b, c, a, e). Thus any swap will always change the disorderliness by 1 an odd number of times, so the disorderliness changes by an odd number. Now the disorderliness of the identity permutation (corresponding to the identity matrix, of course) is zero. If σ is any permutation of even disorderliness, it always takes an even number of swaps to turn it into the identity. An odd number can’t do it, because changing an even number by an odd number an odd number of times can’t leave you with zero! Similarly, every sequence of swaps that changes a permutation with odd disorderliness into the identity must have an odd number of swaps. Since a swap of terms in a permutation corresponds to a swap of rows in a matrix, the number of swaps to recover the identity will always be either even or odd for a particular permuation. So even (disorderliness) permutations have signs of +1 and odd permutations have signs of –1. Now, we finish by showing that our “big formula” does indeed have the defining properties of a determinant so that, as promised, the determinant exists and is unique. Theorem: the formula satisfies the defining qualities of a determinant. Proof: three of the properties are easy. Namely, det(I) = 1 is trivial, since sgn(I) = 1 and I is the only term in its expansion. The pulling-out-multiples-from-a-row property is similarly easy, since this multiple appears exactly once in every term in the sum, as we dissect the row in which it was multiplied. The swapping rows property is also simple because the same products of aij’s occurs in the determinant of the swapped matrix, only with a permutation with swapped rows, so all of the signs are changed. That leaves the add-a-multiple-of-a-row property. So let’s say r times row i has been added to row j. Then (note carefully the subscripts of the second term in the parentheses!) ∑a1σ (1) !(a jσ ( j) + raiσ ( j) )!anσ (n) sgn(σ ) = ∑a1σ (1) !a jσ ( j) !anσ (n) sgn(σ ) σ σ + . The first of these two sums is det(A). The second, r∑a1σ (1) !aiσ ( j) !anσ (n) sgn(σ ) σ when we sum over all sigmas, cancels itself out. For each pair k and l has exactly the same product of a’s with σ(i) = k and σ(j) = l and another σ with σ(i) = l and σ(j) = k, but all other values the same. These two sigmas have opposite signs, so the terms cancel each other in the sum, and thus the second sum is zero. .

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