# Linear Algebra Working with Linear Transformations

Let $v_1=[-3;-1]$ and $v_2= [-2;-1]$

Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation satisfying:

$T(v_1)=[15;-6]$ and $T(v_2)=[11;-3]$

Find the image of an arbitrary vector $[x;y]$

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Welcome to math.SE. Is this a homework question? Also, it is more helpful if you point out what did you try? and where did you get stuck? –  user2468 Apr 5 '12 at 3:06

One method would be to find the image of the standard unit vectors first. Then using linearity, you can find the image of an arbitrary vector.

In a bit more detail:

To find $T(0,1)$, first write $(1,0)$ as a linear combination of $v_1$ and $v_2$. Here you have to solve the equation $$(1,0)=\alpha v_1+\beta v_2.$$ The solution is $$(1,0) =v_2-v_1.$$ Now using the fact that $T$ is linear $$T(1,0)=T( v_2-v_1 )=T(v_2)-T(v_1)=(11,-3)-(15,-6)= (-4,3).$$

Now, do the same procedure to figure out what $T(0,1)$ is.

Then you can say $$T(x,y)=T\bigl( (x,0)+(0,y)\bigr) =x\,T(1,0)+y\,T(0,1).$$

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Note sure if (homework) yet. So hint:

Let $$T = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

We can re-interpret the given $T(v_1)$ and $T(v_2)$ as:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 \\ -1 \end{pmatrix} = \begin{pmatrix} 15 \\ -6 \end{pmatrix} , \\ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 11 \\ -3 \end{pmatrix}$$ Or more succinctly as, $$\begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 & -2 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 15 & 11 \\ -6 & -3 \end{pmatrix} \tag{1}$$ Can you take it from here?

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The matrix whose columns are T(v1) and T(v2) is the representation matrix of T in the basis B=(v1, v2) of R2 (as domain) and E=((1,0), (0,1)) of R2 as codomain. If you multiply this matrix by the coordinates vector of (x,y) in the basis B you will get T(x,y) (since E is the canonic basis).

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