Solution of the algebraic problem

Let (K(s,t)), (f(s)), (\varphi(s)) be the functions given the meaning introduced at the beginning; however, for the sake of simplicity, let us take the interval of the variables (s,t) as the interval (0) to (1); furthermore, let (K(s,t)) be a symmetric function in (s,t). We further understand by (n) a positive integer and use for the following proofs the abbreviations:

[ K_{pq} = K\left(\frac{p}{n}, \frac{q}{n}\right) \quad (p, q = 1, 2, \dots, n), ] [ Kxy = K_{11}x_1y_1 + K_{12}x_1y_2 + K_{21}x_2y_1 + \dots + K_{nn}x_ny_n ] [ = \sum_{p,q} K_{pq} x_p y_q, \quad \text{where } K_{pq} = K_{qp}, ] [ \varphi_p = \varphi\left(\frac{p}{n}\right), \quad f_p = f\left(\frac{p}{n}\right), \quad (p = 1, 2, \dots, n), ] [ Kx_1 = K_{11}x_1 + K_{12}x_2 + \dots + K_{1n}x_n, ] [ Kx_2 = K_{21}x_1 + K_{22}x_2 + \dots + K_{2n}x_n, ] [ \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ] [ Kx_n = K_{n1}x_1 + K_{n2}x_2 + \dots + K_{nn}x_n, ] [ [x, y] = x_1y_1 + x_2y_2 + \dots + x_ny_n. ]

It is evidently [ Kxy = [Kx, y] = [Ky, x]. ]

We now lay the algebraic problem at the foundation: let there be given the (n) linear equations

(1) [ f_1 - \varphi_1 = l(K_{11}\varphi_1 + \dots + K_{1n}\varphi_n), ] [ f_2 - \varphi_2 = l(K_{21}\varphi_1 + \dots + K_{2n}\varphi_n), ] [ \dots \dots \dots \dots \dots \dots \dots \dots \dots ] [ f_n - \varphi_n = l(K_{n1}\varphi_1 + \dots + K_{nn}\varphi_n), ]

or shorter

(2) [ f_i - \varphi_i = lK\varphi_i, ] [ f_n - \varphi_n = lK\varphi_n ]

to determine the (n) unknowns (\varphi_1, \varphi_2, \dots, \varphi_n), where the values (f_p) and the coefficients (K_{pq}) are given and (l) is likewise to be regarded as a known parameter. We further relate the properties of the solutions and the context with the problem of the orthogonal transformation of the quadratic form (Kxx) into account.

[Right Page] 5    Kap. I. Lösung des algebraischen Problems.

To solve this algebraic problem, we use the determinants

[ d(l) = \begin{vmatrix} 1 - lK_{11}, & -lK_{12}, \dots, & -lK_{1n} \ -lK_{21}, & 1 - lK_{22}, \dots, & -lK_{2n} \ \dots & \dots & \dots \ -lK_{n1}, & -lK_{n2}, \dots, & 1 - lK_{nn} \end{vmatrix}, ] [ D\left(l, \begin{smallmatrix} x \ y \end{smallmatrix}\right) = \begin{vmatrix} 0 & x_1, & x_2, \dots, & x_n \ y_1, & 1 - lK_{11}, & -lK_{12}, \dots, & -lK_{1n} \ y_2, & -lK_{21}, & 1 - lK_{22}, \dots, & -lK_{2n} \ \dots & \dots & \dots & \dots \ y_n, & -lK_{n1}, & -lK_{n2}, \dots, & 1 - lK_{nn} \end{vmatrix}, ]

whose first is the discriminant of the quadratic form [ [x, x] - lKxx. ]

If we denote by (D\left(l, \begin{smallmatrix} x \ y \end{smallmatrix}\right)) that determinant which arises from (D\left(l, \begin{smallmatrix} x \ y \end{smallmatrix}\right)) by replacing (y) everywhere by (K_y) [ K_y y = K_{y1}y_1 + K_{y2}y_2 + \dots + K_{yn}y_n, ] then it is evidently, as is evident for the variables (x, y) and (l), the equation

(3) [ d(l)[x, y] + D\left(l, \begin{smallmatrix} x \ y \end{smallmatrix}\right) - lD\left(l, \begin{smallmatrix} x \ Ky \end{smallmatrix}\right) = 0. ]

Our problem now consists in finding the (n) unknowns (\varphi_1, \varphi_2, \dots, \varphi_n) from the equations (1) or (2), i.e. a linear form [ [x, \varphi] = x_1\varphi_1 + x_2\varphi_2 + \dots + x_n\varphi_n ] which satisfies identically in (x) the equation [ [f, x] = [\varphi, x] - l[K\varphi, x]. ]

This equation follows immediately from (3), as is immediately evident from the formula:

(4) [ [x, \varphi] = - \frac{D\left(l, \begin{smallmatrix} x \ f \end{smallmatrix}\right)}{d(l)} ] is solved. When, then, the parameter (l) is so constituted that (d(l) \neq 0) occurs, the coefficients of the linear form (4) sought are the values of the unknowns (\varphi_1, \varphi_2, \dots, \varphi_n). This result is independent of the assumption of the symmetry (K_{pq} = K_{qp}).

It is known that the roots of the equation [ d(l) = 0 ] are all real; we denote them by [ l^{(1)}, l^{(2)}, \dots, l^{(n)} ] and assume that they are different from one another.