# 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 $A$?

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I appreciate any help. I am not sure what the procedure is for forming standard matrices based on basis from transformations – user115241 Dec 12 '13 at 23:53
Did my answer solve your problems, or do you need more help? – Abramo Dec 13 '13 at 12:38
I tried to solve it. Can you correct me if I am wrong. I solved that a1= 24, a2=-9, b1=59, b2=-23. Thank you – user115241 Dec 13 '13 at 17:28
They don't look ok to me. Check your solution – Abramo Dec 13 '13 at 17:33
I'm not understanding how to solve for the a and b values. I tried to do so with substitution and those are the answers that I got. I'm sure this is a basic problem, but I am really weak in Linear Algebra concepts. – user115241 Dec 13 '13 at 17:36

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 $$\begin{pmatrix} a_1 & a_2 \\ b_1 & b_2\end{pmatrix}$$
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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 $M^{\prime}=UMU^{-1}$

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