Question

Problem 4. Let n E N. We consider the vector space R” (a) Prove that for all X, Y CR”, if X IY then Span(X) 1 Span(Y). (b) Let X and Y be linearly independent subsets of R”. Prove that if X IY, then X UY is linearly independent. (C) Prove that every maximally pairwise orthogonal set of vectors in R” has n + 1 elements. Definition: Let V be a vector space and let U and W be subspaces of V. We write V = U OW and call V the direct sum of U and W if V = U + W and U NW = {0}.

Answer 1

of IR. Problem 4: Giren IRA as vector space over IR, 2 let, X, Y Ç IR sit X lY. Then span (x) and spancy) are subppaces also sim (span (X)) Laim (Ron. and dim( spancy) & aim (IR) an sim (span (x) en and dim Espan cet, na dimension of span (x). 120 dim (span (Y)), ) {n.

Let, $41, 42, - - -, &nih be basis of span (x) and {81, B2, -- ßnah be basis of span (Y). Let UE span (X) = a unique scalaro , C ., Che sot us at, at C2 d 2 t - - & Chian let, ve span (Y) = a unique scalers do, da, -, dra Sot Do d, ß, & d2 ß2t ar & draße since, di's ex and Bis GY Given that x 1 Y = &; 1 ; & leich, IEJEN2 = (&;; B;) - 0 i,j Now (tu) = (G&, & Cq d2 t . & Cudni, dißi + d2 ß2+ -- + dr. Pro) - ad, (d., ß, ) & a da (&,, B2) + – – – – a Ch, dng (& mißno) 0 V UG span (x) and UG spany) a span (x) L span (y) na

let, x, y are linearly independent set of vectors in IR. let, 1x1 = p and lyl=9. obviously pen, an Also XIY = for any xEX and Y GY x and y are linearly independent. Because. cxtdy=0 =) C(2,4) + d (4,4) = 0 5) Ciot d (4,4) = 0 5) d=0, Similarly, 0-0, Let, x= {41, 42, - - ., &p) and Y= {ß, ß2 , ßal Then (&i, ß;) = 0 & ij . We have to show that Xuy is linearly independent So It is suffices to show inat i can't be expreaped a linear combingtion of dis. Of not let & B EY say ßo B; (to) i sjel, sot ßt span & - Bo Gant C2d2f - eloxo for scalars C, C2, - - Icp.

(®), B;) - G (Q1,6") + C2 (69, B;) + - - 4 Cp (de Bj) - Bi 20 which is not possible. because, Bi G Y and Y in linearly independent set of vectors in IR. so ß; & Y. Hence, no element of Y can be exforeased as linear combination of vecters on X. Then Xuy is linearly independent set of vectors in IR. . We know that every non null orthogonal set of vectors in IR in linearly independent, again dimension of IR an . so maximally non null or Gogonal set of vectors in B = dimension of IB en Culing Gramschimiot orihogo. nalisation process ? NOW null vector OER sat olx rnt R. This implier that maximally orihogonal set of vectors in R = 1+1 for including o.