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I am having difficulty solve this problem in my homework:

(In my notation, $[x;y]$ represents a matrix of 2 rows, 1 column)

Let $\mathbf{x}=[x_1;x_2]$, $v_1$=[−3;5] and $v_2=[7;−2]$ and let $T\colon\mathbb{R}^2\to\mathbb{R}^2$ be a linear transformation that maps $\mathbf{x}$ into $x_1v_1+x_2v_2$. Find a matrix $A$ such that $T(\mathbf{x})$ is $A\mathbf{x}$ for each $\mathbf{x}$.

I am pretty clueless. So I assume that I start off with the following:

$x_1v_1 + x_2v_2 = x_1[−3;5] + x_2[7;−2]$

But I do not know what to do from here, or if this is even the correct start!

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PS - Sorry, I don't know how to make the formulas look prettier. – Derek Feb 20 '12 at 7:28

First: matrices with "two rows, one column" are called vectors (or column vectors).

Second: What does your function $T$ do to $\mathbf{x}=[1;0]$? What does it do to $\mathbf{x}=[0;1]$?

Or if you prefer:

Third: $A$ will be a $2\times 2$ matrix; write $$A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right).$$ When you multiply $A$ by $\mathbf{x}$, you get $$A\mathbf{x} = \left(\begin{array}{cc} a&b\\ c&d\end{array}\right)\left(\begin{array}{c}x_1\\x_2\end{array}\right) = \left(\begin{array}{cc} ax_1 + bx_2\\ cx_1 + dx_2 \end{array}\right).$$ In order for this to be the same as $$T(\mathbf{x}) = x_1\left(\begin{array}{r}-3\\5\end{array}\right) + x_2\left(\begin{array}{rr}7\\-2\end{array}\right) = \left(\begin{array}{c}-3x_1 + 7x_2\\ 5x_1 -2x_2 \end{array}\right),$$ what are the values of $a$, $b$, $c$, and $d$?

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In the step "When you multiply A by x, you get..", shouldn't there be a multiply sign after the matrix A instead of the '=' sign? – Inquest Feb 20 '12 at 7:37
@Nunoxic: Yes. Thank you. – Arturo Magidin Feb 20 '12 at 7:40
Arturo, thank you. Is the "in order for this to be the same as" step, is that based upon the transformation rule that T(x+y) = T(x) + T(y)? – Derek Feb 20 '12 at 7:49
It is based on basic properties of vectors (Scalar multiplication and addition) – Inquest Feb 20 '12 at 7:54
@Derek : You're looking for the values of $a$,$b$,$c$ and $d$ that will set $\begin{pmatrix} ax_1 + bx_2 \\ cx_1 + dy_2 \end{pmatrix}$ equal to $\begin{pmatrix} -3x_1 + 7x_2 \\ 5x_1 - 2x_2 \end{pmatrix}$ for every vector $[x_1;x_2]$. The obvious choice is in front of you. – Patrick Da Silva Feb 20 '12 at 13:37

If I understand you correctly, I would say that

$$A = \left(\begin{array}{rr}-3&7\\5&-2\end{array}\right) \ \textrm{and} \ x'=Ax.$$ You can see this if you use $$x' = \left(\begin{array}{cc}x_1\\x_2\end{array}\right).$$ Then $$x_1'= -3\cdot x_1 + 7\cdot x_2 = x_1 \cdot v_{11} + x_2\cdot v_{21}$$ and $$x_2'= 5\cdot x_1-2\cdot x_2 = x_1\cdot v_{12} + x_2\cdot v_{22}$$ (here $v_{12}$ is the second element of the first $v_1$).

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Also good answer – grayQuant Feb 5 '14 at 20:55
Thank you very much for your comment! I am happy that my answer was also of help. – partial81 Feb 5 '14 at 22:08

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