# Change of basis?

• Mar 11th 2010, 12:25 PM
chickeneaterguy
Change of basis?
Let B = {(1, -4), (-2, 9) and notice that B is a basis for R2. Therefore each (x,y) in R2 can be written as a linear combination of the vectors in B. If (x,y) = a(1,-4) + b(-2,9) then the coordinates of (x,y) relative to the basis B are said to be (a,b). (i.e. (x,y) = (a,b)B . Thus, associated with each (x,y) is another point in R2, (a,b). The function (sometimes called a transformation) that sends (x,y) to (a,b) is called a coordinate transformation. This transformation can be represented by a matrix. In other words, there is a matrix A such that the new coordinates (a,b) can be found by multiplying (x,y) on the left by A. Find A.

I have an answer but I'm not sure if I'm doing this right.
• Mar 11th 2010, 12:35 PM
icemanfan
You have that

$\displaystyle \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix}$.

Now you need to find A such that

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = A \begin{bmatrix}x \\ y \end{bmatrix}$.

But

$\displaystyle A \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = A \begin{bmatrix} x \\ y \end{bmatrix}$.

Therefore, A is the inverse of

$\displaystyle \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}$.
• Mar 11th 2010, 12:41 PM
chickeneaterguy
I got
$\displaystyle \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4 \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}$
• Mar 11th 2010, 12:53 PM
icemanfan
How did you arrive at that answer? I calculate

$\displaystyle A = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix}$,

from which you get

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x + 4y \\ 2x + y \end{bmatrix}$.
• Mar 11th 2010, 01:24 PM
chickeneaterguy
Quote:

Originally Posted by icemanfan
How did you arrive at that answer? I calculate

$\displaystyle A = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix}$,

from which you get

$\displaystyle \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 9 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x + 4y \\ 2x + y \end{bmatrix}$.

Note, $\displaystyle M^-$ indicates M inverse since I couldn't get M^(-1) to appear correctly.

$\displaystyle MA=B \ \ \ \ A=M^-B$

Where $\displaystyle M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}$

$\displaystyle B = \begin{bmatrix} x \\ y \end{bmatrix}$

$\displaystyle M^- = \frac{1}{ad-bc}\*\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

$\displaystyle M^- = 1\*\begin{bmatrix} 9 & 2 \\ 4 & 1 \end{bmatrix}$

$\displaystyle A = \begin{bmatrix} 9 & 2 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4 \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}$
• Mar 11th 2010, 07:30 PM
chickeneaterguy
So...what's right?
• Mar 12th 2010, 04:30 AM
HallsofIvy
Quote:

Originally Posted by chickeneaterguy
Note, $\displaystyle M^-$ indicates M inverse since I couldn't get M^(-1) to appear correctly.

$\displaystyle MA=B \ \ \ \ A=M^-B$

Where $\displaystyle M = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ -4 & 9 \end{bmatrix}$

$\displaystyle B = \begin{bmatrix} x \\ y \end{bmatrix}$

$\displaystyle M^- = \frac{1}{ad-bc}\*\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

$\displaystyle M^- = 1\*\begin{bmatrix} 9 & 2 \\ 4 & 1 \end{bmatrix}$

$\displaystyle A = \begin{bmatrix} 9 & 2 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 9x+2y \\ 4x+y \end{bmatrix} = x\begin{bmatrix} 9 \\ 4 \end{bmatrix} + y\begin{bmatrix} 2 \\ 1 \end{bmatrix}$

The question asked for a matrix. A, here, is vector, not a matrix.