# Thread: Transformation in Euclidean vector space.

1. ## Transformation in Euclidean vector space.

Consider the Euclidean vector space $R^2$ with the basis $B = {(2 \quad 0),(1 \quad 1)}$
and the linear transformation $ img.top {vertical-align:15%;} $T:x \rightarrow Ax$" alt=" $T:x \rightarrow Ax$" />, where

A = [1 2 ; 0 1]

Find the $B$-matrix of $T$, that is find $[T]_B$

I thought that I was just going to find the transform matrix from B to A, but that doesn't seem right?

2. ## Re: Transformation in Euclidean vector space.

what you need to do is a two-step process:

step one: find the "change of basis matrix" P. to be explicit, the P we are going to use is the one that changes B-coordinates to standard coordinates.

since [1,0]B = 1(2,0) + 0(1,1) = (2,0), we have:

$P\begin{bmatrix}1\\0 \end{bmatrix} = \begin{bmatrix}2\\0 \end{bmatrix}$

which makes it clear P is of the form:

$P = \begin{bmatrix}2&\ast \\ 0&\ast \end{bmatrix}$

similarly, since [0,1]B = (1,1), we have that:

$P = \begin{bmatrix}2&1 \\ 0&1 \end{bmatrix}$

so the overall procedure is this:

B-coordinate input ---> change to standard coordinates --->apply standard T (matrix A) ---> change back to B-coordinates.

to change back to B--coordinates, we need to use P-1, which is:

$P^{-1} = \frac{1}{(2^*1 - 1^*0)}\begin{bmatrix}1&-1\\0&2 \end{bmatrix} = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2}\\0&1 \end{bmatrix}$

step 2: find the desired matrix for T, which is:

$[T]_B = P^{-1}AP = \begin{bmatrix} \frac{1}{2}&-\frac{1}{2}\\0&1 \end{bmatrix} \begin{bmatrix}1&2\\0&1 \end{bmatrix} \begin{bmatrix}2&1 \\ 0&1 \end{bmatrix}$

$= \begin{bmatrix}1&1 \\ 0&1 \end{bmatrix}$.

let's verify that this actually works:

we know that T(x,y) = (x+2y,y).

writing (x,y) in B-coordinates, we get: (x,y) = x(1,0) + y(0,1) = x[1/2,0]B + y[-1/2,1]B = [(x-y)/2,y]B.

applying our [T]B to [(x-y)/2,y]B (to get [T(x,y)]B), we obtain:

[T]B([(x-y)/2,y]B) = [(x-y)/2+y,y]B = [(x+y)/2,y]B.

changing this back to standard coordinates, we have:

[(x+y)/2,y]B = ((x+y)/2)[1,0]B + y[0,1]B = ((x+y)/2)(2,0) + y(1,1) = (x+y,0) + (y,y) = (x+2y,y) = T(x,y).