8. Description of determinants according to Weierstrass

We will now show that Properties I and II, derived in the last section, together with Equation (7.6) are characteristic for a determinant:

Theorem 3

Let f(A) be a polynomial in the n² elements of the matrix A with the properties:

(1) f(A) is linear and homogeneous with respect to the elements of each row of A.
(2) f(A) changes its sign as two rows of A are interchanged.
(3) For the unit matrix E, f(E) = 1.

Then f(A) equals the determinant |A| of the matrix A. (K.Weierstrass 1823-1852)

Proof: According to Property (1), f(A) has the form

where r₁, r₂, ···, r_n , independently of each other, take the numbers 1, 2, ···, n and the are certain coefficients, which do not depend on the a₁₁, ···, a_nn and some of which can vanish. In fact, due to (8.1), every individual term of f(A) has from each row of A exactly one element as factor.

Then there follows immediately from (8.1):

(1a) f(A) = 0, if one row has only zeros, because then every term of f(A) has one factor 0.

Moreover, (8.1) yields:

(1b) f(A) becomes lf(A), as you multiply all elements of any row of A by l, because every term of f(A) contains from the elements of the by l multiplied row exactly one element, that is, every individual term is multiplied by l.

Property (2) yields directly:

(2a) f(A) = 0 if A contains two equal rows, because you can argue as during the derivation of IIa from II in Section 7.

Now add in the matrix A to the elements of the i-th row corresponding elements of another row, multiplied by l - say, of the j-th row. Denote the new matrix by M*. Then

where the matrix A₀ arises from A by replacement of its i-th row by its j-th row. Hence, by (2a), F(A₀) = 0, because A₀ has two equal rows, and

(2b) F(A) does not change as you add in A to one row another row multiplied by l.

Now construct, starting from the matrix A, a finite sequence of matrices

A, A₁, ··· , A_t,

for which

f(A) = f(A₁) = ··· = f(A_t).

The (82) and (8.3) yield f(A) = |A|, and Theorem 3 has been proved.

The construction of this sequence is achieved as follows:

Step 1: View the first elements of all rows of the matrix A. If all of them are zero, at the first step, A is not changed and A₁ = A. Otherwise, let the first element of the i-th row differ from 0. If i > 1, interchange the i-th row with the first row and simultaneously multiply it by -1. Then, according to Ib and II of Section 7 or according to (1b) and (2), respectively, neither the determinant nor the polynomial f change their values. Hence you have arrived at a matrix in which the first element of the first row is 0.

You can now form by addition of suitable multiples of the first row to the second, ···, n-th row a matrix, in which the first elements of the second, ···, n-th row vanish. This concludes the first step.

Hence, after the first step, you obtain in each case a matrix A₁, in which only the top most element 0. However, you can state this also as follows: In the first column of A₁, there are only zeroes below the principal diagonal . Moreover, due to the properties of the above polynomial f and the determinant

f(A₁) = f(A) and |A₁| = |A|.

Step 2: View the second elements of the second to n-th rows of A₁. If all of them are zero, during the second step, A₁ is not changed and you set A₂ = A₁. Otherwise there exists in the interval 2, ···, n at least one j, when the second element of the j-th row differs from zero. If j > 0, you interchange the j-th row with the second one and multiply it simultaneously by -1. Then neither the determinant nor the polynomial change their values. In this way you arrive at a matrix in which the second element in the second row differs from zero. You can now form by addition of suitable multiples of the second row to the third up to the n-th row a matrix A₂, in which the second elements of the third to the n-th row vanish. This concludes the second step. After it, you have obtained in each case a matrix A₂, in which the first two columns below the main diagonal contain only 0. Moreover,

f(A₂) = f(A₁) and |A₂| = |A₁|.

The next steps correspond: At the k-th step, A_k-1 is not changed and you set A_k = A_k-1. Otherwise you make sure that the k-th element in the k-th row differs from 0. You then add suitable multiples of the k-th row to the (k + 1)-th to n-th rows, in order to obtain thus a matrix A_k, in which the k-th elements of the (k+1)-th to n-th row vanish. After the k-th step, you have a matrix A_k, in which the first k columns below the main diagonal contain only 0. Moreover

f(A_k) = f(A_k-1) and |A_k| = |A_k-1|.

After n - 1 steps, you arrive in this manner at a matrix A_n-1, in which there occur only zeroes below the main diagonal. In addition, you have

f(A_n-1) = f(A) and |A_n-1| = |A|.

You now consider the main diagonal elements of A_n-1 and study two cases depending on whether all of them differ from 0 or one of them is 0.

In the first case, you can add in sequence for i = 2, 3, ··· , n suitable multiples of the i-th row to the first up to the (i - 1)-th row, so that all elements above the main diagonal vanish. In this way, you obtain a diagonal matrix A_t with

f(A_t) = f(A) and |A_t| = |A|.

As we have noted at the end of Section 7, |A_t| is equal to the product of the elements in the main diagonal of A_t. From Properties (1b) and (3) of the polynomial f, you now draw the same conclusion as for the determinant in Section 7 that also f(A_t) equals the product of the elements of the main diagonal of A_t, whence

f(A_t) = |A_t|

and Theorem 3 has been proved.

We must still attend to the case that there occurs among the elements of the main diagonal of A_n-1 at least one zero. Let the m-th row of A_n-1 be the row with the largest number in which the main diagonal element vanishes. As in the preceding case, you can in each case arrive at a matrix A_t in which in the (m+1)-th up to n-th columns there are also above the main diagonal only zeroes. Consider the m-th row of A_t. The first m - 1 elements of this row have already vanished at the conclusion of the (m - 1)-th step. The element in the main diagonal vanishes according to our assumption. And moreover the (m+1)-th to n-th elements have also been made to vanish. Hence the m-th row of the matrix A_t contains only zeroes, that is, according to (1a) or Ia, respectively,

f( A_t) = 0 and |A_t| = 0,

and Theorem 3 has been proved completely.

Comments on Theorem 3: If you replace Property (3) of the polynomial f by (3'):

f(E) = g for an arbitrary g, then

f(A) = g ·A_t.

In fact, Property (3) has only been employed once and, in fact, for the evaluation of f( A_t) for a diagonal matrix A_t. If a₁, ···, a_n are the diagonal elements of A_t, you obtain at first from (1b)

f( A_t) = a₁···a_n·f(E)

and then moreover, due to (3'),

f( A_t) = a₁···a_n·g.

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