Question

с раиси от к. Show that the function that takes ((X1, X2, X3), (y1, y2, y3)) E to xi yi + x3y3 is not an inner product on R. ((X1, X2, X3), (y1, y2, y3)) E R3 x R3 von SE

Answer 1

c.,.) IR3 R3 be such that < (221, 22,8g).cy, Y292) - Biy,t 243------ Now for <.,.) to be an lines product it . must satistics sutv, w) = <u, w >+ <uw> a cau, nes acu, vs 6 cu o = <u, u © Lurus 20 & cuius 0 l7U7O.. o for u, u, WEIR² & LEIR. - Λέσων, - Let u=cx, 22, 23), o=14, 42, 4₃, w - cz., 22, 23) Butv, w) =<(21, 22, 23)+cy, 42, 42, (Z, Z2 Z3) - cl24+4 , Hat Y2 Zz49) CZ., Z2, Z37 - (@ity, z, + (x3+4) Zz. & cu, wst cu, w> <(2,,x2, 222, <?. , Z₂, Z3) + ºzy, 92, ), (2., 22,29 ) - a xiz, + 2₃ Z3+ y, z, + z 23 - 2 x +4) Zit Cast Y) Zz 2 cute, wd = cu, wt cu, w). Now cau, wy Z <acxi, 22, 23), (Z., Z2, Z32 = caxi, X22, Lxz), (2, 22, 23) - axiz, + a2zzz - 4( 3,z, + 4x2ZX) = 4k u, u)? Now det ut <0,3 o>, too , A u to But cuius = oto= 0 so it does not satisfies the - broberty, Hence it is not an fourth inner poodic.