Re: Change of co-ordinates

presumably, you mean $\displaystyle B = \{1,x,x^2\}$, that is, $\displaystyle p_0(x) = 1, p_1(x) = x, p_2(x) = x^2$. suppose we have a vector in $\displaystyle P_2(\mathbb{R})$ ginve in C-coordinates, that is:

$\displaystyle v = c_1h_1 + c_2h_2 + c_3h_3$ or $\displaystyle v = [c_1,c_2,c_3]_C$. in other words,

$\displaystyle h_1 = [1,0,0]_C,h_2 = [0,1,0]_C,h_3 = [0,0,1]_C$.

now if we had such a matrix A. then:

$\displaystyle A([h_i]_C)$ would be its i-th column. but that would just be the i-th basis vector of C in B-coordinates. so, for example,

$\displaystyle A([h_1]_C) = [h_1]_B = [3,2,7]_B$, since:

$\displaystyle h_1(x) = 3 + 2x + 7x^2 = 3p_0(x) + 2p_1(x) + 7p_2(x)$

convince yourself that a linear combination of the basis vectors in C is taken to the same linear combination of the images of the basis vectors in B-coordinates, because a matrix is a linear mapping.

Re: Change of co-ordinates

Another approach: by a well known theorem if $\displaystyle B=\{u_1,\ldots,u_n\},\;B=\{u'_1,\ldots,u'_n\}$ are basis of a vector space $\displaystyle V$ and $\displaystyle \begin{Bmatrix} u'_1=a_{11}u_1+\ldots +a_{1n}u_n\\...\\ u'_n=a_{n1}u_1+\ldots +a_{nn}u_n\end{matrix}$ then, $\displaystyle [x]_B=P\;[x]_{B'}$ where $\displaystyle P=\begin{bmatrix} a_{11} & \ldots & a_{n1}\\ \vdots&&\vdots \\ a_{1n} &\ldots & a_{nn}\end{bmatrix}$

In our case, we immediately get $\displaystyle [x]_B=\begin{bmatrix}{3}&{7}&{\;\;1}\\{2}&{5}&{\;\;1} \\{7}&{4}&{-9}\end{bmatrix}[x]_C$

Re: Change of co-ordinates

Quote:

Originally Posted by

**Deveno** presumably, you mean $\displaystyle B = \{1,x,x^2\}$, that is, $\displaystyle p_0(x) = 1, p_1(x) = x, p_2(x) = x^2$. suppose we have a vector in $\displaystyle P_2(\mathbb{R})$ ginve in C-coordinates, that is:

$\displaystyle v = c_1h_1 + c_2h_2 + c_3h_3$ or $\displaystyle v = [c_1,c_2,c_3]_C$. in other words,

$\displaystyle h_1 = [1,0,0]_C,h_2 = [0,1,0]_C,h_3 = [0,0,1]_C$.

now if we had such a matrix A. then:

$\displaystyle A([h_i]_C)$ would be its i-th column. but that would just be the i-th basis vector of C in B-coordinates. so, for example,

$\displaystyle A([h_1]_C) = [h_1]_B = [3,2,7]_B$, since:

$\displaystyle h_1(x) = 3 + 2x + 7x^2 = 3p_0(x) + 2p_1(x) + 7p_2(x)$

convince yourself that a linear combination of the basis vectors in C is taken to the same linear combination of the images of the basis vectors in B-coordinates, because a matrix is a linear mapping.

Yes $\displaystyle A([h_1]_C) = [h_1]_B = [3,2,7]_B$, $\displaystyle A([h_2]_C) = [h_2]_B = [7,5,4]_B$ and $\displaystyle A([h_3]_C) = [h_3]_B = [1,1,-9]_B$

so can we say that M=A (inverse), where

M=3 7 1

2 5 1

7 4 -9

and just find the inverse of M to find our matrix A?

Re: Change of co-ordinates

no, the matrix:

$\displaystyle \begin{bmatrix}3&7&1\\2&5&1\\7&4&-9\end{bmatrix}$

is the matrix A, it changes C-coordinates into B-coordinates.

if you have a non-standard basis, and you want to change coordinates from the non-standard basis to the standard basis,

you write down the matrix whose columns are the non-standard basis vectors coordinates in the standard basis,

that is the "change of coordinate" matrix. if you want to change from B-coordinates to C-coordinates, THEN you would take the inverse of A.