
#1
Jan1508, 09:45 AM

P: 25

Question 1
Let u, v1,v2 ....... vn be vectors in [tex]R^{n}[/tex]. Show that if u is orthogonal to v1,v2 .....vn then u is orthogonal to every vector in span{v1,v2....vn} My attempt if u is orthogonal to v1,v2 .....vn then[tex] (u.v1)+(u.v2)+.......+(u.vn)=0[/tex] Let w be a vector in span{v1,v2....vn} therefore [tex] w=c1v1+c2v2+.......+cnvn [/tex] [tex] u.w=u(c1v1+c2v2+.......+cnvn)[/tex] =>[tex] c1(u.v1)+c2(u.v2)+.......+cn(u.vn) =0 [/tex] So u is orthogonal to w Question 2 Let [tex] \{v1,v2....vn \}[/tex] be a basis for the ndimensional vector space [tex]R^{n}[/tex]. Show that if A is a non singular matrix nxn then [tex] \{Av1,Av2....Avn \} [/tex] is also a basis for [tex]R^{n}[/tex]. Let w be a vector in [tex]R^{n}[/tex] therefore w can be written a linear combination of vectos in it's basis [tex] x=c1v1+c2v2+.......+cnvn [/tex] [tex] Av1={\lambda}1x1[/tex],[tex] Av2={\lambda}2x2[/tex] ...[tex] Avn={\lambda}3xn[/tex] so [tex]Ax=A(c1v1+c2v2+.......+cnvn) [/tex] [tex]Ax={\lambda}1c1v1+{\lambda}2c2v2+.......+{\lambda}ncnvn) [/tex] therefore [tex] \{Av1,Av2....Avn \} [/tex] is also a basis for [tex]R^{n}[/tex]. 



#2
Jan1508, 10:00 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

Better, I think, would be to use "proof by contradiction". Suppose Av1, Av2, ..., Avn were NOT independent. What would that tell you about v1, v2, ..., vn (remember that since A in nonsingular, it has an inverse matrix). Suppose Av1, Av2, ..., Avn does NOT span the space. That is, suppose there were some w such that a1Av1+ a2Av2+ ...+ anAvn was NOT equal to w for any choice of a1, a2, ..., an. What does that tell you about v1, v2, ..., vn and A^{1}w? 


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