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

• mitch_1211
In summary, multiplying an invertible matrix A to a basis of matrices {X1,X2,X3..Xn} creates a new basis {AX1,AX2,AX3..AXn}, as long as A is a linear map and the vectors in the original basis are linearly independent. This is because the number of vectors in the new basis is the same as the number of vectors in the original basis, and A being invertible ensures that the only solution to the equation c1(AX1)+c2(AX2)+c3(AX3)=0 is c1=c2=c3=0.
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...

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.

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).

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

td21 said:
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

What is an invertible matrix?

An invertible matrix is a square matrix that has an inverse, meaning that it can be multiplied by another matrix to produce the identity matrix. This is also known as a non-singular matrix.

What does it mean to create a new basis from X1, X2, X3...Xn?

Creating a new basis from X1, X2, X3...Xn means that the original set of vectors (X1, X2, X3...Xn) can be transformed into a new set of linearly independent vectors that can be used as a basis for the vector space.

How do you prove that an invertible matrix creates a new basis?

To prove that an invertible matrix creates a new basis, you can show that the new set of vectors is linearly independent and spans the same vector space as the original set of vectors. This can be done through various methods such as using the determinant of the matrix or performing row operations.

What are the benefits of using an invertible matrix to create a new basis?

Using an invertible matrix to create a new basis can make certain calculations and transformations easier. It can also provide a better understanding of the underlying structure of the vector space and can be useful in solving systems of linear equations.

Can an invertible matrix always create a new basis from any set of vectors?

No, an invertible matrix can only create a new basis from a set of vectors if the original set of vectors is linearly independent. If the original set of vectors is linearly dependent, then the matrix will not be able to create a new basis.

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