Proving Invertible Matrix A Creates New Basis from {X1,X2,X3..Xn}

[mitch_1211]

does multiplying the invertible matrix A to the basis {X1,X2,X3..Xn} create a new basis; {AX1,AX2,Ax3..AXn}? where Xn are matrices

I can prove that for eg if {v1,v2,v3} is a basis then {u1,u2,u3} is a basis where u1=v1 u2=v1+v2, u3=v1+v2+v3

I setup the equation c1(u1)+c2(u2)+c3(u3)=0 and determine if the only solutions are c1=c2=c3=0 or if there are others. This then gives linear independence or dependence and because there are the right number of vectors (3 u's and 3v's) spanning will automatically follow if vectors are linearly independent.

I'm not sure how to approach the matrix basis case...

 

[HallsofIvy]

If A is an invertible n by n matrix and {vn} is a basis for n dimensional vector space then, yes, {Avn} is also a basis.

 

[Bacle]

This is only true if the transformation A is a linear map, i.e., a vector space isomorphism; otherwise, one may have either that Axi=Axj for A:V-->V , i.e.,
from a vector space to itself, or if you have A:V-->W, with dimV<dimW, then , even if Axi=/Axj , the set {Ax1,...,Axn} will not be a basis for W, tho it will be a basis for
A(V).

 

[td21]

c1(Au1)+c2(Au2)+c3(Au3)=0
A(c1(u1)+c2(u2)+c3(u3))=0
c1(u1)+c2(u2)+c3(u3)=0
c1=0,c2=0,c3=0

 

[mitch_1211]

c1(Au1)+c2(Au2)+c3(Au3)=0
A(c1(u1)+c2(u2)+c3(u3))=0
c1(u1)+c2(u2)+c3(u3)=0
c1=0,c2=0,c3=0

Thank you everyone for all those ideas and thank you td21 for making me realize what a simple task this is, A is invertible => therefor A is non-zero, u forms a basis so is also non-zero so the only solution would be c1=c2=c3=0