Linear Algebra - Isomorphism, Matrix of Linear Transformation Let T:Rn→Rn be a linear transformation. Let A be the standard matrix for T . Let β be an ordered basis for Rn and B is a matrix whose columns are vectors in β. Prove[T]β=B−1AB.

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Linear Algebra - Isomorphism, Matrix of Linear Transformation  Let T:Rn→Rn be a linear transformation. Let A be the standard matrix for T . Let β be an ordered basis for Rn and B is a matrix whose columns are vectors in β. Prove[T]β=B−1AB.

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Solution: Let A be the standard matrix for T where T:Rn→Rn is a linear transformation. Let β be an ordered basis for Rn and B is a matrix whose columns are vectors in β. Define T(x)=Ax. Let the vectors of βbev1,v2,…vn. Then, a1v1+a2v2+…+anVn=Ax [v1v2...vn][a1a2⋮an]=Ax Further we have B[a1a2⋮an]=Ax(given:B=[v1v2...vn]) [a1a2⋮an]=B−1Ax(Right multiply by B−1) Suppose the vector x is any vector from , then x represents any column of B. Thus, we have[a1a2⋮an]=B−1AB Conclusion The column matrix [a1a2⋮an] is coordinate vector with respect to B. that is [T]β=[a1a2⋮an] Hence, it is proved that [T]β=B−1AB.

Linear Algebra - Isomorphism, Matrix of Linear Transformation Let T : R^{

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