# Diagonalized matrix

1. Jun 24, 2008

### off-diagonal

1. The problem statement, all variables and given/known data
Show that we can always write this line-element

ds^2=Adx^2+Bdy^2+Cdxdy ; which A B and C are any real function of x and y

to this form ds^2 = du^2 + dv^2

2. Relevant equations

3. The attempt at a solution
I try to solve it in two way
First , I write total derivative of u as du = (du/dx)dx + (du/dy)dy and then dv = (dv/dx)dx + (dv/dy)dy

and I square both of du and dv then I can write du^2 + dv^2 = ()dx^2 + ()dy^2 + ()dxdy
but I'm not sure this is correct.

Another way to do,I have to find a matrix transform that can diagonalize my metric tensor from the first line-element. I found that, I have to find eigenvalue of these metric and then eigenfunction after that ,I compose its eigenfunction into a new metric (called U) then by follow this equation U^-1 A U = D when U^-1 is U inverse
A is the original metric
D is diagonal metric

but I have a little confused because my metric has component as an any function that make me a difficult to find metric U

2. Jun 24, 2008

Your first method was on the right track, though it may be easier to see if you write u = f(x,y) and v=g(x,y), and then you follow through with differentiation getting du=(df/dx)dx+(df/dy)dy and dv=(dg/dx)dx+(dg/dy)dy, and then realizing that the derivatives of f and g are functions of x and y.

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Last edited by a moderator: Apr 23, 2017 at 1:46 PM
3. Jun 24, 2008

### Dick

You might want to compare this with the problem of finding a standard form conic section for A*x^2+B*y^2+C*xy wherein you want eliminate the cross term xy by transforming to new coordinates. BTW, I don't think you can transform any such thing to du^2+dv^2, unless you are allowed to use an imaginary coordinate transformation. The signature of the metric should be invariant under real transformations. And yes, this is the same roughly the same thing as diagonalizing a matrix, except that you can change the eigenvalues by rescaling a coordinate.

Last edited: Jun 24, 2008
4. Jun 25, 2008

### off-diagonal

Thank a lots both of you so Can I following the first method? or I have to fiind and imaginary coordinate transform?

5. Jun 25, 2008

### HallsofIvy

Staff Emeritus
ds2= Adx2+ Bdxdy+ Cdy2 is equivalent to the matrix multiplication
$$ds^2= \left[\begin{array}{cc}dx & dy \end{array}\right]\left[\begin{array}{cc}A & \frac{B}{2} \\ \frac{B}{2} & C\end{array}\right]\left[\begin{array}{c}dx \\ dy\end{array}\right]$$

Find the eigenvalues and eigenvectors of that 2 by 2 matrix. Since it is symmetric it has two real eigenvalues. Taking x' and y' coordinate axes in the directions of the eigenvectors, the matrix becomes diagonal with the eigenvalues on the diagonal.