Change of co-ordinates

**breitling** · January 4th 2012, 01:10 AM

I have the following
B:=[p_0,p_1,p_2] denote the natural ordered basis for P_2(R)
C:=[h_1,h_2,h_3] where h_1(x)=7x^2+2x+3; h_2(x)=4x^2+5x+7 and
h_3(x)=-9x^2+x+1

Construct the change of coordinate matrix A which converts C-coordinates to B-coordinates.

Now my first thought was the rows in C are given by the columns in B, however when going through the calculations to construct f_c, in a later calculation. Do i have to find an expression for p_0, p_1 and p_2 using the equations?

many thanks in advance.

**Deveno** · January 4th 2012, 02:25 AM

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

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

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

now if we had such a matrix A. then:

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

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

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

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.

**FernandoRevilla** · January 4th 2012, 03:20 AM

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

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

**breitling** · January 4th 2012, 06:07 AM

Originally Posted by Deveno

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

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

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

now if we had such a matrix A. then:

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

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

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

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

**Deveno** · January 4th 2012, 12:17 PM

no, the matrix:

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

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.

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