By Woerle K.
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Additional resources for 200 Aufgaben aus der Trigonometrie mit Loesungen
Suppose that we have three linear spaces G, , G 2 , and H ; the space Mult(G, , G,; H ) consists of functions f(g, ,g,), g, E G,, r = 1,2, with values in H, such that if either argument is fixed, f depends linearly on the other. ” Bilinear functions are often considered as “products”; the bilinear property requires that the “product”,f(g, ,g,) have the distributive property expected of a product. Thus, the scalar product of two vectors in real euclidean n-space is a bilinear function. The product of a linear operator, acting from a space X to a space Y, and a vector in X , constitutes a bilinear function from Hom(X, Y ) and X , to Y.
8) On applying the left to g , we obtain n C brthr r=l = 0, and since the h, are linearly independent, we have b,, = 0, as was to be proved. We determine the coefficients b,, so that the two sides agree in their action on the basis elements g,. 6), as required. The operators A , CCbrSArsmust then coincide on the whole of G. 3. 9) dim G = dim G*. 2. since dim K = 1. The details of the argument deserve an indication for the present specialization. If dim G = m, 16 1 LINEAR SPACES and g , , ... ,g , are a basis of G, we can form a basis of G * , say h , , ...
To show that equality holds here, we prove that if n 2 max(r,, ... , r,,), then A"+'g = 0 implies that Ang = 0. 7), a representation m g =Tg,, and here m A"+'g = g , G,, ~ A"+lg,, s = 1, ... 7), we then have, if A"+'g A"+'g, = 0 , If ii 2 r F ,s = s = 1, = ... ,m. 1 . I 1) 1, ... , 111, it then follows that Ang, = 0, and so that A": = s = 1, ... , m, 0. Thus r I max(r, , . . ,Y,,), and the proof is complete. 2 The Case of Equal Ascent and Descent It is possible for only one of the ascent and descent to be finite.