Question

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With the given positive numbers, show that vectors u = (U1, U2) and v = (v1,v2) define an inner product in R2 using the 4 axioms. Set (u, v) = 3u1v1 + 7u2V2

Answer 1

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Answer 2

عمل have to show <u,u) = 34, V, +74 k where U=(4, 42) V=(V),V) detine an inner product on IR? 2 Amap 1 v Recaht that sevectescspace o tome reap sexo : Blues Recall that , map <,> is an inner product on vector space iff <> Satisfies following 4 grejoms for 4, VEV and K 18 any Scaler cis symmetry Axiom - (u,v)a<0,05 (11) Additivity Axiom <4+, W) = 24,V) + < V,w) 10 Homogenecity faciem – <ku, vy s K84,7) positivity queiom <4,4)20 & <4,0) =0 Pff y = Here < > is detinde des as Tuu) = 340, +742V2 U=(4,42) VE(Vis VJ Elke we will show so satisfies all a ascioms 1 symmetry gjo me let U=(4,342) V=(0, ,4) ER² consider Rupru,U)= 9340, +742 V2 = 3 y 4, 7/242 <V, 4) :.<4,0) = Tvu) VYVER? Additivity opriom

let iter 4=(4,4) VECV,,V2,W=( 10, W7.) <4+V, W - 314,4V) Wp +7(42+V.JW2 (** UV-(41V., uztve) :. <utu, w) = 84, W T3V, W, +7442 + 7V₂ Wz (34,6,774 W2)--(31,6,77 v2 wz.) <4, W) + TV,W) Hence 347V, W) = <u, w) +Tuw) 070 ru,u, WER? Homogene city Axiom le let KE AIR Consider aku,w) = 3(K4)U, +7 (K42)^2 =3ky,nit BK 42 Uz =BK (834,0, +342V2) ko <40) Hence <ky,v)=kfu,u) x 4.61K? KEIR V positivity opciom ::consider ::.<4,4) 341427 74242 scene 342 47 42² iwe know grzo taER 341 - 2 <4,4) 20 ЧүЧn € EIR? A UE

Now scy pose 74,4>= 0 342+743 = 0 41=0, 42=0 R & C412044320 o) U=(4,42) = (0,0) 4 = so У чG Hence ER² 4=Ō <4,4)=0 = 0 Aff Here <,> satisfies all areioms inner product on IR? an .:<> is