# Thread: Problem with simple Linear Map from R3 to P1

1. ## Problem with simple Linear Map from R3 to P1

I'm having a great deal of trouble seeing how the following map was done. It isn't even a problem but simply an example (#1.3) which can be seen on the page the following link leads to:
Linear Algebra/Representing Linear Maps with Matrices - Wikibooks, open books for an open world

the spaces are given to be
where the bases are

and the action of h on B is given by

and the final mapping is:

which only makes some sense to me. I really don't know how to set the equations in D to map to these values and don't remember ever going over it in this book (specifically, moving from a space larger than the target space which is the case here, R3 --> P1). What's really confusing is the two negative maps for -x. I can see how maybe one is arrived at by this procedure:
1 + x
-1 + x

(1p --> 2p)

0 + 2x =
(where we then add our value from R3)

0 + 2x = - x

x = -1/2

??????
I don't believe this is correct but I really have no idea how this was done. I also notice that doing the dot product from the final maps to R3 gives us our initial values, for example:
-1/2*(1 + x) + -1/2*(-1 + x) =
(-1/2 - 1/2x) + (1/2 -1/2x) =
- 1/2x -1/2x =
-x

2. The matricial equation of $h$ is $Y=AX$ with

$A=\begin{bmatrix}{-1/2}&{\;\;1}&{\;\;2}\\{-1/2}&{-1}&{-2}\end{bmatrix}$

where $X=(x_1,x_2,x_3)^t$ are the coordinates of $v\in\mathbb{R}^3$ with respect to $B$ and $Y=(y_1,y_2)^t$ are the coordinates of $h(v)\in\mathcal{P}_1$ with respect to $D$

3. h is a linear map from R^3 to P1. any linear map is completely determined by its action on a basis. rather than describe h abstractly in terms of its image on an arbitrary domain element (which is the usual way we describe a function), we can assign NUMERICAL values to a matrix which "gives us the same information".

these numbers (the entries in our matrix representation of h) aren't intrinsic to the mapping h, they depend on a choice of basis for the domain and co-domain.

now, we are given that h((0,0,1)) = -x, h((0,2,0)) = 2, and h((2,0,0)) = 4. but to construct the matrix for h relative to the bases B and D, we need to express these images of the basis element of B in terms of D = {1+x, -1+x}, in other words, their D-coordinates.

let's look at the first one: h((0,0,1)) = -x. how can we express -x as a(1+x) + b(-1+x)?

well, a(1+x) + b(-1+x) = a + ax -b + bx = (a-b) + (a+b)x. if we want this to equal -x, we should like:

a-b = 0, so a = b, and a+b = -1, so 2a = -1, so a = -1/2, and thus b = -1/2 as well. so -x = [-1/2,-1/2]D.

in a similar vein, the D-coordinates of 2 are [1,-1]D, and the D-coordinates of 4 are [2,-2]D.

now, (0,0,1), our first element of B, in B-coordinates is [1,0,0]B (kind of confusing, right? think of it this way:

call the elements of B, b1,b2,b3. since (0,0,1) = b1, what we really mean by [1,0,0]B is 1b1 + 0b2 + 0b3 = b1).

the matrix of h (let's call it M) should satisfy the following:

M[1,0,0]Bᵀ = [-1/2,-1/2]Dᵀ or in matrix form:

$\begin{bmatrix} {m_{11}}&{m_ {12}}&{m_{13}}\\{m_{21}}&{m_{22}}&m_{{23}} \end{bmatrix}\begin{bmatrix} {1}\\{0}\\{0}\end{bmatrix} = \begin{bmatrix} {-1/2}\\{-1/2}\end{bmatrix}$

but by actually doing the multiplication on the left, we get that [1,0,0]B picks out the first column of M. similarly, M[0,1,0]Bᵀ picks out the 2nd column of M (which reflects h's action on b2 = (0,2,0)), and M[0,0,1]Bᵀ gives us the 3rd column of M.

therefore, M has the images of the B basis elements (in D-coordinates) as its columns, so

$M=\begin{bmatrix}{-1/2}&{\;\;1}&{\;\;2}\\{-1/2}&{-1}&{-2}\end{bmatrix}$

4. Originally Posted by Deveno
let's look at the first one: h((0,0,1)) = -x. how can we express -x as a(1+x) + b(-1+x)?

well, a(1+x) + b(-1+x) = a + ax -b + bx = (a-b) + (a+b)x. if we want this to equal -x, we should like:

a-b = 0, so a = b, and a+b = -1, so 2a = -1, so a = -1/2, and thus b = -1/2 as well. so -x = [-1/2,-1/2]D.
Although a seemingly simple calculation, I couldn't find this anywhere. Thank you so much, Deveno!