# Some vector space proofs

by Bertrandkis
Tags: proofs, space, vector
 P: 25 Question 1 Let u, v1,v2 ....... vn be vectors in $$R^{n}$$. 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$$(u.v1)+(u.v2)+.......+(u.vn)=0$$ Let w be a vector in span{v1,v2....vn} therefore $$w=c1v1+c2v2+.......+cnvn$$ $$u.w=u(c1v1+c2v2+.......+cnvn)$$ =>$$c1(u.v1)+c2(u.v2)+.......+cn(u.vn) =0$$ So u is orthogonal to w Question 2 Let $$\{v1,v2....vn \}$$ be a basis for the n-dimensional vector space $$R^{n}$$. Show that if A is a non singular matrix nxn then $$\{Av1,Av2....Avn \}$$ is also a basis for $$R^{n}$$. Let w be a vector in $$R^{n}$$ therefore w can be written a linear combination of vectos in it's basis $$x=c1v1+c2v2+.......+cnvn$$ $$Av1={\lambda}1x1$$,$$Av2={\lambda}2x2$$ ...$$Avn={\lambda}3xn$$ so $$Ax=A(c1v1+c2v2+.......+cnvn)$$ $$Ax={\lambda}1c1v1+{\lambda}2c2v2+.......+{\lambda}ncnvn)$$ therefore $$\{Av1,Av2....Avn \}$$ is also a basis for $$R^{n}$$.
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P: 39,568
 Quote by Bertrandkis Question 1 Let u, v1,v2 ....... vn be vectors in $$R^{n}$$. 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$$(u.v1)+(u.v2)+.......+(u.vn)=0$$ Let w be a vector in span{v1,v2....vn} therefore $$w=c1v1+c2v2+.......+cnvn$$ $$u.w=u(c1v1+c2v2+.......+cnvn)$$ =>$$c1(u.v1)+c2(u.v2)+.......+cn(u.vn) =0$$ So u is orthogonal to w
Yes, that's looks good. And you understand, I assume, that "u orthogonal to v1, v2 ..., vn" means u is orthogonal to each of v1, v2, ..., vn- that's where you get (u.v1)+ (u.v2)+ ...+ (u.vn)= 0+ 0+ ...+ 0= 0.

 Question 2 Let $$\{v1,v2....vn \}$$ be a basis for the n-dimensional vector space $$R^{n}$$. Show that if A is a non singular matrix nxn then $$\{Av1,Av2....Avn \}$$ is also a basis for $$R^{n}$$. Let w be a vector in $$R^{n}$$ therefore w can be written a linear combination of vectos in it's basis $$x=c1v1+c2v2+.......+cnvn$$ $$Av1={\lambda}1x1$$,$$Av2={\lambda}2x2$$ ...$$Avn={\lambda}3xn$$
I don't understand this. Why is $$Av1={\lambda}1x1$$? Are you assuming each of the basis vectors is an eigenvector of A? That is not given in the hypothesis.

 so $$Ax=A(c1v1+c2v2+.......+cnvn)$$ $$Ax={\lambda}1c1v1+{\lambda}2c2v2+.......+{\lambda}ncnvn)$$ therefore $$\{Av1,Av2....Avn \}$$ is also a basis for $$R^{n}$$.
Even if it were true that the original basis consists of eigenvectors of A, what you have done is show that Av1, Av2, ..., Avn span the space. You have not shown that they are independent. Also, you have not used the fact that A is nonsingular.

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-1w?

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