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Consider the basis $\{\vec{p},\vec{q}\}$ where $\vec{p}=(1,1)$ and $\vec{q}=(-1,0)$. Let $T:\mathbb{R}^2\to\mathbb{R}^2$ be the linear operator such that $T(\vec{p})=(1,-2)$ and $T(\vec{q})=(4,1)$. Then find a formula for $T(x_1,x_2)$ and use this formula to find $T(5,-7)$.

I figured this might make the transformation matrix $A$ be $ A = \left[ \begin{array}{rr} 1 & 4 \\ -2 & 1 \\ \end{array} \right]$ such that $T_A(\vec{v}) = A\vec{v}$, but this doesn't seem to pan out.

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  • $\begingroup$ That transformation works only if you're using $\{\vec{p}, \vec{q}\}$ as a basis. But usually, for $(x_1, x_2)$, we're using $(1, 0)$ and $(0, 1)$ as a basis. $\endgroup$ – pjs36 Dec 17 '15 at 16:17
  • $\begingroup$ @pjs36 What is the proper approach? $\endgroup$ – alxmke Dec 17 '15 at 16:19
  • $\begingroup$ The components of $T(\vec p)=(1,-2)$ and $T(\vec q)=(4,1)$ are in the basis $\{\vec p, \vec q\}$ or in the standard basis? $\endgroup$ – Emilio Novati Dec 17 '15 at 16:34
  • $\begingroup$ @EmilioNovati Sorry, I don't quite follow. It was given that the basis is $\{\vec{p}, \vec{q}\}$ and then says what happens to them under the linear transformation. I'm not sure how to relate this to determining the linear transformation. $\endgroup$ – alxmke Dec 17 '15 at 16:38
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If we interpret the question as if all vectors are expressed in the standard basis, than we have to find a matrix : $$ A_s=\begin{bmatrix} a&b\\ c&d \end{bmatrix} $$ such that $$ \begin{bmatrix} a&b\\ c&d \end{bmatrix} \begin{bmatrix} 1\\ 1 \end{bmatrix}= \begin{bmatrix} 1\\ -2 \end{bmatrix} \quad \land \quad \begin{bmatrix} a&b\\ c&d \end{bmatrix} \begin{bmatrix} -1\\ 0 \end{bmatrix}= \begin{bmatrix} 4\\ 1 \end{bmatrix} $$ this gives the system: $$ \begin{cases} a+b=1\\c+d=-2\\-a=4\\-c=1 \end{cases} $$ with solutions $a=-4,b=5,c=-1,d=-1$ and the matrix $A$ is: $$ A_s=\begin{bmatrix} -4&5\\ -1&-1 \end{bmatrix} $$

IN this interpretation the fact that $\{\vec p,\vec q\}$ is a basis is true but not relevant.

If the components of $T(\vec p)$ and $T(\vec q)$ in OP are expressed in the basis $t=\{\vec p,\vec q\}$, than ,since in this basis we have $\vec p=[1,0]^T$ and $\vec q=[0,1]^T$, than it is represented , in this basis, by the matrix: $$ A_t=\begin{bmatrix} 1&4\\ -2&1 \end{bmatrix} $$ but note that this transformation is different from the other one.

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