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I'm having trouble understanding the steps in working with bases. Suppose T is a linear transformation from $R^2$ to $R^2$ defined by $T(x,y)$ = $(2x-y, x+y)$ and we want to find the matrix of T with respect to the bases $$B_1 = [(1,1),(2,1)] \ \ \text{ and }B_2 = [(-1,2), (1,1)].$$

I know the matrix $A[u_1]B_1 = [T(u_1)B_2$ but i'm not sure how to apply that with the information given.

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    $\begingroup$ In general do you know how to represent a linear transformation using a matrix (given a basis)? $\endgroup$ Jun 8, 2020 at 4:18
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    $\begingroup$ It might be useful to note that matrix $B_{1}$ when written as follows $\begin{bmatrix}1 & 2 \\ 1 & 1 \end{bmatrix}$ (i.e. basis vectors in the columns) indicates that if you right multiply it by $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ you obtain $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and likewise, if you right multiply it by $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$ you obtain $\begin{bmatrix}2 \\ 1\end{bmatrix}$. I hope this helps. $\endgroup$
    – ad2004
    Jun 8, 2020 at 23:17
  • $\begingroup$ I know that B and T can form matrices, but I think the formula mentioned is the only one I know, so I was confused as to the value of $u$ $\endgroup$
    – DuncanK3
    Jun 10, 2020 at 3:39

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Let us denote the standard basis $$ e_1 := \begin{pmatrix} 1 \\ 0 \end{pmatrix}\quad\textrm{and}\quad e_2 := \begin{pmatrix} 0 \\ 1 \end{pmatrix}. $$ In this basis $$ T = \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix}. $$ The basis transformation matrices from $(e_1,e_2)$ to $B_1$ or $B_2$ are given by $$ B_1 = \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\quad\textrm{and}\quad B_2 = \begin{pmatrix} -1 & 2 \\ 1 & 1 \end{pmatrix}. $$ The transformation matrix in each basis will be given by $$ TB_1^{-1} \quad\textrm{and}\quad TB_2^{-1}. $$

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  • $\begingroup$ Ah so there is no $[u]$ required. But if $T$ had a vector, say $u = (2,3)$ then it would have to be multiplied to that formula? $\endgroup$
    – DuncanK3
    Jun 10, 2020 at 3:36

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