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I am a little confused with results that I've got. Please look through my solutions set, questions will be listed at the bottom.

Given:

Basis E $\in$ $\Bbb R$2 $$E= \Bigg \{ \begin{bmatrix}1 \\ 2 \\ \end{bmatrix}, \begin{bmatrix}3 \\ 5 \\ \end{bmatrix} \Bigg\}$$

Linear transformation: L: $\Bbb R$2→ $\Bbb R$2 $$L(x,y) = (2x-y, 3x-2y)$$

We need to find transformation matrix, and it is said that the answer is: $$ \begin{bmatrix} -3 & -8 \\ 1 & 3 \\ \end{bmatrix} $$

Solution:

Step 1: I have applied transformation to each column vector from our basis: $$ L \Bigg (\begin{bmatrix}1 \\ 2 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 0\\ -1 \\ \end{bmatrix}$$ $$L \Bigg (\begin{bmatrix} 3 \\ 5 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}1 \\ -1 \\ \end{bmatrix}$$

Step 2:

Let:

$$ A = \begin{bmatrix} 1 & 3 \\ 2 & 5 \\ \end{bmatrix}, A^{-1} = \begin{bmatrix} -5 & 3 \\ 2 & -1 \\ \end{bmatrix}, B = \begin{bmatrix} 0 & 1 \\ -1 & -1 \\ \end{bmatrix} $$

Method 1:

$$ LA = B $$ $$ LAA^{-1}=BA^{-1} $$ $$ LI=BA^{-1} $$ $$ L=BA^{-1} $$ $$ L_1 = \begin{bmatrix} 0 & 1 \\ -1 & -1 \\ \end{bmatrix} \begin{bmatrix} -5 & 3 \\ 2 & -1 \\ \end{bmatrix}= \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ \end{bmatrix} $$

Method 2: if I construct an augmented matrix, from matrix A and a matrix of images of transformation of column-vectors from our basis, as its columns (matrix B), and reduce matrix A to the identity matrix: $$ \left[ \begin{array}{cc|cc} 1&3&0&1\\ 2&5&-1&-1 \end{array} \right]\sim \left[ \begin{array}{cc|cc} 1&0&-3&-8\\ 0&1&1&3 \end{array} \right] $$ $$ L_2 =\begin{bmatrix} -3 & -8 \\ 1 & 3 \\ \end{bmatrix} $$

Method 3: Method 3: I have also tried to apply a method from this comment, where the author uses images as rows (not as columns) , but got neither $L_1$ nor $L_2$

Step 3: I suppose, that transformation should work this way $LA=B$ (when I apply it to a column-vector/matrix of column vectors, I should receive my images from Step 1), so: $$L_1A=B$$ $$ \begin{bmatrix} 2 & -1 \\ 3 & -2 \\ \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 5 \\ \end{bmatrix}= \begin{bmatrix} 0 & 1 \\ -1 & -1 \\ \end{bmatrix}$$

which seems fair enough for me, but I cannot say that about $L_2$ solution: $$L_2A=B$$ $$ \begin{bmatrix} -3 & -8 \\ 1 & 3 \\ \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2 & 5 \\ \end{bmatrix}= \begin{bmatrix} -19 & -48 \\ 7 & 18 \\ \end{bmatrix}$$

So my questions are:

  1. Which solution is correct (maybe none of these), and where am I mistaken
  2. Is it correct to refer to column-vectors as "images"
  3. Do I understand transformation matrix in a right way when I say that $LA=B$

I understand that I lack understanding, so I'll be glad to receive any help. Thanks.

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Your method 1 should write as $$AT=B$$ where $T$ is the transformation matrix. In this expression, you are treating columns of $A$ as basis vectors. This is equivalent to say $$A\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}0\\-1\end{pmatrix}, \quad A\begin{pmatrix}y_1\\y_2\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix}.$$ This is to find how the columns of $B$ can be expressed in terms of the basis.

Method 2 works fine, since this is exactly what method 1 does, but using Gauss-Jordan elimination instead of finding inverse and multipling.

Method 3 is not correct. You confused the image with the step afterwards. You already found the image, which is $B$. You are now trying to find an expression of $B$ in terms of the basis $A$. In the answer you referred, they used row instead of column, but the result is the transpose too. Also that problem has standard basis, so it is different from yours.

To answer your question 3, $LA=B$ is a correct expression, but not what you intended. It is just what you already did in step 1.

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