Choose a basis for W, [tex]\{M_1, M_2, ..., M_n}. Taking the column matrices with exactly one "1" and the rest of the entries "0", so that , will work nicely. For linear transformation L, write as a linear combination of the the basis. That coeffientes in that linear combination will give you the first column of A. Continue with , etc. to find the other columns.

This is a general method. If L is a linear transformation from vector space U to vector space V, choose bases for U and V. Apply L to each basis vector of U in turn and write the resulting vector (in V) as a linear combination of the chosen basis for V. The coefficients will give the columns of the matrix representation of L.

You've lost me here. You say "let a be an ordered basis" but then say "for each a in V". Did you mean "let b be an ordered basis"? Does "[a]b" mean "the coefficients when a is written as a linear combination in that basis"? (I was try to think of "b" as a vector!) "W" is still the column matrices, right? Remember the basis I suggested for W above? Do you see that U maps each vector of b into the corresponding member of that basis. And any linear transformation that maps basis vector to basis vectors is an isomorphism.Now suppose V is an n-dimensional vector space over the field F, and let a

be an ordered basis for V. For each a in V, define Ua = [a]b. Prove that U is an

isomorphism of V onto W.

[quote]Basically, what I said before. A is the matrix representing T in this basis. Its columns are the coefficients in the linear combination of each in the basis .If T is a linear operator on V, then UTU^(-1) is a linear

operator on W. Accordingly, UTU^(-1) is left multiplication by some n X n matrix A.

What is A?

Any help on these two problems would be appreciated.