# What have I done?

1. Feb 24, 2010

### dlevanchuk

What have I done???

1. The problem statement, all variables and given/known data
(linear transformation)
Let T: R2 -> R3 be a linear transformation such that T(e1) = u1 and T(e2) = u2, where u1 = [1; 0; -1] and u2 = [2; 1; 0]. Find each of the following:
T([1; 1]) and T([2; -1])

2. Relevant equations

3. The attempt at a solution
Here is the thing. I've been sitting with this problem for a good hour, and cannot figure out. I just started playing around with numbers, toss em around a little bit, and in the end I've got the correct answer... But can somebody explain me, WHY is this the right way of doing it?? lol

First, I looked at u1 and u2 as a matrix A = [1 2; 0 1; -1 0]. Then i did a familiar A*e1 = b1 and A*e2 = b2, where I got b1= [1; 0; -1] and b2 = [2; 1; 0].. (thinking out loud, every matrix multiplication is a transformation, so it must be relating somehow..)

I let b1 and b2 be matrix (i call it transformation matrix) B = [1 2; 0 1; -1 0]

then I decided to multiply this transformation matrix B by a vector that needs to be transformed from R2 to R3 [1; 1] and got
[1 2; 0 1; -1 0] * [1; 1] = [3; 1; -1]

By why in the name of Albert did it work?? lol did I have a correct logic through out the problem??

Last edited: Feb 24, 2010
2. Feb 24, 2010

### HallsofIvy

Staff Emeritus
Re: What have I done???

Yes. If T is a linear transformation on vector space V, and $\{e_1, e_2, ..., e_n\}$ is a basis for V, then the matrix representation of T, relative to this basis (and, if the range space is different, a specific basis for the range) just has $T(e_1)$, $T(e_2)$, etc. as columns.

To see why that is true, remember that we would always have $e_1= 1e_1+ 0e_2+ ...+ 0e_n$ so in matrix multiplication, it would be represented by the column matrix $$\begin{bmatrix}1 \\ 0 \\ 0 \\ ... \\ 0\end{bmatrix}$$. Multiplying any matrix by that just gives the first column of the matrix. Similarly, $e_2$ would be represented by $$\begin{bmatrix}0 \\ 1\\ 0 \\...\\0\end{bmatrix}$$.