Using transformations and basis to find standard matrices Let $A =\{(1,3), (2,5)\}$ be a basis of $\mathbb{R}^2$. Let $M =\left[\begin{array}{rr} 1 & -2\\ 3 & 0\end{array}\right]$ be the standard matrix for the linear
transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ with respect to the basis $E =\{(1,0), (0,1)\}$, that is, for a vector (x,y) under the basis E, $T(x,y) =\left[\begin{array}{rr} 1 & -2\\ 3 & 0\end{array}\right]$. What is the standard matrix of this linear transformation with respect to the basis $A$?
 A: The basic fact to keep in mind is the following: The matrix $A$ corresponding to the linear map $f:X\to X$ written in basis $x_1,\dots, x_n$ has the form
$$\begin{pmatrix} \; &\;&\;\\\; &\;&\;\\ f(x_1) & \dots & f(x_n)\\ \; &\;&\;\end{pmatrix}$$
where $f(x_i)$ are the vertical vectors written in the basis $x_1,\dots, x_n$.
So what you could do is to follow these steps:


*

*Compute the images of $x_1=(1,3)$ and $x_2=(2,5)$ under your matrix $M$

*Find $a_1$ and $b_1$ such that $M(x_1) = a_1\cdot x_1 + b_1 \cdot x_2$

*Find $a_2$ and $b_2$ such that $M(x_1) = a_2\cdot x_2 + b_2 \cdot x_2$

*The resulting matrix is
$$\begin{pmatrix} a_1 & a_2 \\ b_1 & b_2\end{pmatrix}$$

A: I figure you can derive the matrix that transforms between the two bases.  Call it $U$.  Further, denote any object expressed in the new basis by a prime.  So we know that for all $x$:
$Mx = U^{-1}M^{\prime}x^{\prime} = U^{-1}M^{\prime}UU^{-1}x^{\prime} = U^{-1}M^{\prime}Ux$
This holds for all $x$, therefore:
$M = U^{-1}M^{\prime}U$ and $M^{\prime}=UMU^{-1}$
