# Change of Coordinate Matrix

1. Apr 7, 2009

### jeff1evesque

In geometry the change of variable,

$$x = (2 / sqrt(5))x' - (1 / sqrt(5))y'$$ (#1)
$$y = (1 / sqrt(5))x' + (2 / sqrt(5))y'$$ (#2)
can be used to transform the equation $$2x^2 - 4xy + 5y^2 = 1$$ into the simpler equation $$(x')^2 + 6(y')^2 = 1$$, in which form it is easily seen to be the equation of an ellipse.

$$B and B'$$ are the standard ordered basis and new rotated basis respectively

My question:
Why is BB' have such a representation with B and B'? Why wouldn't it be B'B?

2. Apr 8, 2009

### Cantab Morgan

Can you try to state your question differently? I was following what you wrote up to that point, but I'm not familiar with the notation you're using at the end.

3. Apr 9, 2009

### jeff1evesque

So in this particular example,
B' = { 1/sqrt(5)(2, 1), 1/sqrt(5)(-1, 2) }
Question: what about the basis B, what is it's values?

Geometrically the change of variable is (x,y) --> (x',y') is a change in the way that the position of a point P in the plane.

The change of variable is actually a change from $$[P]_B = (x, y)$$, the coordinate vector of the point P relative to the ordered basis B = {e1, e2}, to $$[P]_B' = (x', y')$$, the coordinate vector of P relative to the new rotated basis B'.

Notice also that the matrix
| 2 -1 |​
Q = 1/sqrt(5) | 1 2 |

equals $$^B_B'$$, where I denotes the identity transformation on $$R^2$$. Thus $$[v]_B = Q[v]_B_'$$ for all v in $$R^2$$. A similar result is true in general.

Two questions for the last two sentences from above:
1.) What exactly is [v]_B? What are the values for this vector, how is it obtained?
2.) What do they mean by a similar result is true?

thanks,

JL

4. Apr 9, 2009

### Cantab Morgan

Sorry, I still don't know what that notation means.

But here's what I see is happening. The ellipse $$2x^2 -4xy + 5y^2 = 1$$ represents a quadratic form, meaning a particular inner product on R^2. Imagine that there is a symmetric matrix S with positive eigenvalues...

$$S = \left( \begin{array}{cc} 2 & -2 \\ -2 & 5 \end{array} \right)$$

Then we're looking at the set of all points $$v = \left( \begin{array}{c} x \\ y \end{array} \right)$$ satisfying $$Sv \cdot v = 1$$. These form an ellipse. By finding the eigenvectors of that matrix S, and normalizing them, we assemble your rotation matrix Q. Q rotates the plane. Consider its transpose and see that $$v' = Q^Tv$$, and we get the ellipse $$x'^2 +6y'^2 = 1$$. Note that these coefficients 1 and 6 are the eigenvalues of S, and they represent the maximum and minimum values of the ellipse's distance from the origin.

So, what is v? This question I understand. What you are calling $$v_B$$. Well, it can really be any point in the plane that you transform by $$Q^T$$, but in our case we can limit it to those points in the plane on the ellipse. That is, all points for which $$Sv \cdot v = 1$$.

I don't understand your question about the values of the basis B. You wrote down the basis B.

Like you, I also don't know what they mean by "a similar result is true".