# Finding a matrix for a linear transformation

1. May 22, 2016

### patricio2626

'1. The problem statement, all variables and given/known data

Find the matrix A' for T: R2-->R2, where T(x1, x2) = (2x1 - 2x2, -x1 + 3x2), relative to the basis B' {(1, 0), (1, 1)}.

2. Relevant equations

B' = {(1, 0), (1, 0)} so B'-1 = {(1, -1), (0, 1)}.

3. The attempt at a solution

I'm confused at what exactly a transform matrix relative to a given basis is to mean. Does this mean that some vector vB', when multiplied by A', will equal T(v)B'? I have the solution from the book: B-1AB', but if I had to put in English what it looks to me that this does: B'-1 converts vstd to vB', then A is the transform matrix which should take vstd and output T(v)std, then B' will convert vB' to vstd. It therefore look as if this will take vectors in standard bases and output them in standard bases after transforming? That can't be right because the answer given is not the same as the transform. What is the question asking, exactly?

2. May 22, 2016

### tommyxu3

You have to find $Te_1,Te_2$ under the expression of $e_1,e_2,$ where $\{e_1,e_2\}$ is the given basis.

3. May 22, 2016

### patricio2626

Sure, T(e1), T(e2) is easy: {(2, -1), (0, 2)}

4. May 22, 2016

### tommyxu3

You have to express them $\textbf{under}$ $e_1,e_2.$
That is, you have to get $Te_1=ae_1+be_2,Te_2=ce_1+de_2,$ likewise.

5. May 22, 2016

### tommyxu3

6. May 23, 2016

### patricio2626

I think that if I can get an answer to this then I will be able to figure out what is going on in the answer and explanation in the book:
I'm confused at what exactly a transform matrix relative to a given basis is to mean. Does this mean that some vector vB', when multiplied by A', will equal T(v)B'?

7. May 23, 2016

### tommyxu3

Take your problem for example. For the given ordered basis $e_1,e_2,$ If I get $Te_1=ae_1+be_2,Te_2=ce_1+de_2,$ and let the matrix $A$ be $$\begin{pmatrix} a & c \\ b & d \end{pmatrix}.$$
Then for any vector $v=xe_1+ye_2,$ if you want to get $Tv,$ besides calculating it directly, you can use the matrix like
$$M \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ax+cy \\ bx+dy \end{pmatrix} ,$$
which you can notice would be the coefficients of $Tv,$ for $(ax+cy)e_1+(bx+dy)e_2=x(ae_1+be_2)+y(ce_1+de_2)=xT(e_1)+yT(e_2)=T(xe_1+ye_2)=Tv,$ satisfying the rules of linear maps.