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Let $v_1 = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$ and $v_2=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ and let $$A= \begin{bmatrix} 3 & 2 \\ -2 & 1 \end{bmatrix}$$ be a matrix for $T\colon \Bbb R^2\to \Bbb R^2$ relative to the basis $B = \{v_1, v_2\}$.

From this I found that:

$T(v_1) = \begin{bmatrix} 4 \\ -1 \end{bmatrix}$ and $T(v_2) = \begin{bmatrix} 5 \\ -3 \end{bmatrix}$

How would I find a formula for $T\begin{pmatrix} \begin{bmatrix}x_1 \\ x_2\end{bmatrix} \end{pmatrix}$ The answer in my book is $\begin{bmatrix} -x_1-6x_2 \\ 2x_1+5x_2 \end{bmatrix}$

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How is the matrix $A$ related to the linear transformation $T$? Is $A$ the matrix of $T$ in the standard basis or in the basis $\{v_1, v_2\}$? You have to make this clear in order to make sense of the question. – Sammy Black Nov 16 '13 at 18:38
@SammyBlack Sorry i just updated it and it is in the basis {$v_1,v_2$} – Name Jonathan Nov 16 '13 at 18:42
From what I gather, and I could be wrong, you're doing two mistakes. One is thinking that $T(v_i)=Av_i$, the other is miscalculating $Av_i$. – Git Gud Nov 16 '13 at 18:53
@GitGud Well this is a possibility but a small one as my answers for $v_1$ and $v_2$ are the same as the books answers. – Name Jonathan Nov 16 '13 at 18:57
$v_1,v_2$ are linearly independent (they are not a scalar multiple of each other) and as there are two of them they form a basis for any 2 component vector, so you can decompose the vector $(x_1,x_2)$ into a linear sum of $v_1,v_2$ and you know how $T$ acts on those. – Graham Hesketh Nov 16 '13 at 19:19
up vote 1 down vote accepted

The real thing to clarify here is how the transformation is being expressed. The answer given is correct if the goal is to rewrite the transformation in terms of the standard basis $w_1=[1,0]^T,w_2=[0,1]^T$.

The change of basis matrix (the one that converts $w$-basis coordinates to $v$-basis coordinates is $C=\begin{bmatrix}1&1\\-1&-2\end{bmatrix}$, which has inverse $C^{-1}=\begin{bmatrix}2&1\\-1&-1\end{bmatrix}$.

Then the new matrix of $T$ with respect to the $w$-basis is

$$ A'=C^{-1}AC=\begin{bmatrix}-1&-6\\2&5\end{bmatrix} $$

Thus when the expression $T(x_1w_1+x_2w_2)$ is written in terms of the $w$-basis coordinates, it becomes $A'\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}-x_1-6x_2\\2x_1+5x_2\end{bmatrix}$

This quantity on the left hand side must be what is meant by "$T(\begin{bmatrix}x_1\\x_2\end{bmatrix})$". Since people often get lost when they mix the notation of $T$ with coordinate vectors, I prefer to avoid writing them together as this question statement seems to have done.

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So, we should probably take the inverse of $\begin{pmatrix} 2&1 \\ -1&-1 \end{pmatrix}^{-1} = \begin{pmatrix} 1&1 \\ -1&-2 \end{pmatrix}$. This is called the change of basis. So, $\begin{pmatrix} 4&5 \\ -1&-3 \end{pmatrix}\cdot \begin{pmatrix} x_{1}\\-x_{1} \end{pmatrix} + \begin{pmatrix} 4&5 \\ -1&-3 \end{pmatrix}\cdot \begin{pmatrix} x_{2}\\-2x_{2} \end{pmatrix} = \begin{pmatrix} -x_{1}-6x_{2}\\2x_{1}+5x_{2} \end{pmatrix}$.

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