# On number of negative eigenvalues of a matrix

1. Aug 11, 2010

### Sina

1. The problem statement, all variables and given/known data
When trying to solve a question about parameter independence of certain aspects of the Jacobian of a real valued function on a manifold I came to the point where I have to show the following:

Let A be a matrix, J be the Jacobian of an orthogonal transformation (I suppose we can assume non-reflection) then define B to be B= JTAJ, where A and B are real symmetric. I have to show that both B and A have same number of negative eigenvalues.

3. The attempt at a solution
This problem I think translates into following:

Both A and B are real symmetric so suppose their diagonal forms are respectively D1 and D2 (ie matrices which have eigenvalues for the diagonal entries). Then there is a unitary transformation U such that D2 = UTD1U (an be shown by direct calculation). I have to now show that the number of negative diagonal entries on each D is the same. It seems logical at first because U is an orthogonal matrix but direct calculation does not yield the answer. Moreover when I try to put it into a geometric setting (i.e rotation of vectors) it seems wrong. Am I doing something wrong? edit: I think this might be a wrong approach though because I haven't used the fact that there is a Jacobian as a part of U

Thanks.

Last edited: Aug 11, 2010
2. Aug 11, 2010

### lanedance

if you could show $$J^T = J^{-1}$$ it would follow pretty quickly

3. Aug 12, 2010

### Sina

I dont think it is the case, it is just the Jacobian of any orthogonal coordinate transformation (not the transformation itsself)

4. Aug 12, 2010

### lanedance

for a 2D rotation
$$u = x cos(\theta)+ y sin(\theta)$$
$$v = x sin(\theta)- y cos(\theta)$$

the jacobian is
$$J(x,y) = \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y }\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix}$$

which is looking very similar to the transformation itself...maybe within a T... if you agree, could you generalise that?

Last edited: Aug 12, 2010