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Hi there could anybody help me with this question maybe just showing me the first part than guiding me through the rest myself that would be so so appreciated.

Let U be a vector space with ordered basis P = [e1,e2] and let V be a vector space with ordered basis Q = [ f1, f2]. Let T : U → V be a linear map with matrix $$ A = \pmatrix{2&3\\3&5} $$ with respect to the ordered bases P and Q. Suppose Q' =[f1',f2'] is another basis of V and suppose that T has matrix $$ B = \pmatrix{4&6\\1&7} $$ with respect to the ordered bases P and Q'. Express the elements of Q as linear combinations of elements of Q'.

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Recall that the columns of $A,B$ describe the associated transformation. For example, the second column of $A$ tells us that $$ T(e_2) = 3f_1 + 5f_2 $$ So, we therefore know that $$ \Big(T(e_1) =\Big) 2f_1 + 3f_2 = 4f_1' + f_2'\\ \Big(T(e_2) =\Big) 3f_1 + 5f_2 = 6f_1' + 7f_2' $$ Rewrite these equations in terms of the coordinate vectors relative to the basis $Q'$. We have $$ 2[f_1]_{Q'} + 3[f_2]_{Q'} = \overbrace{\pmatrix{4\\1}}^{[4f_1' + f_2']_{Q'}}\\ 3[f_1]_{Q'} + 5[f_2]_{Q'} = \pmatrix{6\\7} $$ Rewriting this with matrices, we have $$ \pmatrix{2&3\\3&5} \pmatrix{[f_1]_{Q'} & [f_2]_{Q'}} = \pmatrix{4&6\\1&7} $$ So, the coefficients in our desired linear combination are the columns of $$ \pmatrix{[f_1]_{Q'} & [f_2]_{Q'}} = \pmatrix{2&3\\3&5}^{-1}\pmatrix{4&6\\1&7} $$

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  • $\begingroup$ You're amazing. Thank you $\endgroup$ – Elena Mar 15 '17 at 17:49
  • $\begingroup$ No problem. I've rewritten some of the middle to make it clearer, by the way. $\endgroup$ – Omnomnomnom Mar 15 '17 at 19:27

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