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Let $ v_1 = \begin{bmatrix} 1 \\ -1 \\ \end{bmatrix} $ and $ v_2 = \begin{bmatrix} 2 \\ -3 \\ \end{bmatrix} $ Let $ \mathbb R^2 \rightarrow \mathbb R^2 $ be linear transformation satisfying $ T(v_1) = \begin{bmatrix} 7 \\ -8 \\ \end{bmatrix} $ and $ T(v_2)= \begin{bmatrix} 17 \\ -19 \\ \end{bmatrix} .$

Find the image of an arbitrary vector $ \begin{bmatrix} x \\ y \\ \end{bmatrix} .$

Ok. this the point where I don't know where to begin with this type of question. Any input on how to start?

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    $\begingroup$ You don't know "where to begin with... this ...type of questions? You've asked several of this type of questions in the last few days! In fact, I think that if you really make an effort, you should be able to answer this one seeing the answers you got in your very last question before this one. $\endgroup$ – DonAntonio Apr 1 '13 at 22:58
  • $\begingroup$ The truth is, these questions are quite confusing since I'm trying to understand what it is they're asking. When I see explanations, it's easier for me to see how where to solutions come from. I'm having trouble with Linear Transformations. I do not want answers here. $\endgroup$ – Dimitri Apr 1 '13 at 23:01
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Well, first of all you should know that given a linear transformation $T : V \to W$ if you know the values of $T$ on the basis of $V$ then you know $T$ completely by virtue of linearity. The point here is: in an $n$-dimensional vector space, $n$ linearly independent vectors necessarily forms a basis.

In this case $\mathbb{R}^2$ has dimension $2$ so that any set of $2$ linearly independent vector is a basis for $\mathbb{R}^2$. It's easy to see that $v_1$ and $v_2$ are linearly independent because they're not multiple one of the other. Hence the set $\left\{v_1, v_2\right\}$ is a basis of $\mathbb{R}^2$.

Now here comes the point. Given any $(x,y) \in \mathbb{R}^2$, how do we write it in this new basis? Well, it's easy, it should be a linear combination, so that we must have:

$$(x,y) = av_1+bv_2$$

And substituting we get:

$$(x,y) = a(1, -1) + b(2, -3)$$

$$(x,y) = (a+2b, -a-3b)$$

This is a system of linear equations in $a$ and $b$. If you solve, you'll find that $a = 3x+2y$ and $b =-x-y$. So that any vector $(x,y) \in \mathbb{R}^2$ is given in that basis by:

$$(x,y) = (3x+2y)v_1 + (-x-y)v_2$$

And hence

$$T(x,y) = (3x+2y)T(v_1) + (-x-y)T(v_2)$$

Now it's just a question of substituting the values:

$$T(x,y) = (3x+2y)(7, -8) + (-x-y)(17,-19)$$

So that finally, the expression of $T$ on arbitrary $(x,y)\in \mathbb{R}^2$ is:

$$T(x,y) = (4x-3y, -5x+3y)$$

Which is really the expression of $T$. You can easily verify that $T(v_1)$ and $T(v_2)$ are what they are expected to be.

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    $\begingroup$ Very thorough and complete. Sometimes the notation can get confusing, but when you re-write everything properly, it's easier to solve because it looks more "familiar." Thank you. $\endgroup$ – Dimitri Apr 1 '13 at 23:18
  • $\begingroup$ I'm glad it helped, however I must point out to you that in math it's very important that you do some work on your own to get familiar with the concepts, with notation and so on. So I highly sugest that you try other exercises like this with different vector spaces and so on by yourself, just using this example to base yourself. I think you'll be fine after that. $\endgroup$ – user1620696 Apr 1 '13 at 23:29
  • $\begingroup$ Yea, I agree with that. I had asked a similar, yet different example yesterday, but due to this unfamiliarity of the subject, I needed to see some more input on these types of problems. We've only covered one class on this subject and have moved on to other things already and there is not enough time to fully grasp everything. Thanks again though. $\endgroup$ – Dimitri Apr 1 '13 at 23:44
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Hint: Solve

$$\begin{bmatrix} x \\ y \\ \end{bmatrix}=av_1+bv_2,$$

for $a$ and $b$. Then use linearity of $T$, that is

$$T(av_1+bv_2)=aT(v_1)+bT(v_2).$$

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  • $\begingroup$ Thanks! I can see it from this point. $\endgroup$ – Dimitri Apr 1 '13 at 23:05
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Hint: Not a classical and beautiful solution but it might be easier for you... Consider the linear transformation T as a $2\times2 $ matrix: \begin{bmatrix}a & b\\c & d\end{bmatrix} Then do some easy matrix calculations to get your linear transformation.

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$T=\begin{bmatrix}7 & -8\\17 & -19\end{bmatrix}\cdot\begin{bmatrix}3 & 2\\-1 & -1\end{bmatrix}$ represents matrix of T at standard basis and so: $T\begin{bmatrix} x \\ y \\ \end{bmatrix}=\begin{bmatrix}7 & -8\\17 & -19\end{bmatrix}\cdot \begin{bmatrix}3 & 2\\-1 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ \end{bmatrix}$

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Try writing the vector $(x,y)^T$ in terms of $v_1$ and $v_2$ (it will help if you first write the basis vectors $e_1=(1,0)^T$ and $e_2=(0,1)^T$ in terms of $v_1$ and $v_2$).

Then, use the definition of a linear transformation to work out $T((x,y)^T)$.

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