Is the signature of this matrix zero?

[lavinia]

Given are two square matrices of the same dimension, M and N.

M is symmetric. N is non singular.

From M and N form the symmetric matrix,

M N
N* 0

Where N* is the transpose of N.

Is the signature of this matrix necessarily zero? Counterexample?

 

[AlephZero]

My "engineer's intuition" says it is true, if you think about finding the extrema of x*Mx subject to the constraints Nx = 0, using Lagrange multipliers. The situations where this runs into trouble are when the constraints are not all independent, but that is excluded because N is non-singular.

Basically, each eigenvalue of M is turned into a pair of eigenvalues of the augmented matrix, one positive and one negative.

But my brain isn't working well enough to turn that idea into a proof right now - sorry!

 

[Vargo]

I think a topological argument works here.

If you continuously deform such a matrix (within the constraints you mentioned), then the signature cannot change without an eigenvalue crossing 0. But that cannot happen without the determinant being zero. But such a matrix cannot have zero determinant because its rows are independent. In other words, the signature cannot change under a continuous deformation that maintains the given constraints on M and N.

Every symmetric matrix can be continuously deformed to 0 (by scaling it down to 0). And GL(n) only has two connected components, so each non-singular matrix can be deformed to I or I' (which is I with one entry flipped to -1). So it suffices to check M=0, N=I, and M=0,N=I'.

 

[I like Serena]

Pick for instance M=(0) and N=(1).

Determinant=-1
Trace=0
Eigenvalues are +1 and -1.

 

[lavinia]

I think a topological argument works here.

If you continuously deform such a matrix (within the constraints you mentioned), then the signature cannot change without an eigenvalue crossing 0. But that cannot happen without the determinant being zero. But such a matrix cannot have zero determinant because its rows are independent. In other words, the signature cannot change under a continuous deformation that maintains the given constraints on M and N.

Every symmetric matrix can be continuously deformed to 0 (by scaling it down to 0). And GL(n) only has two connected components, so each non-singular matrix can be deformed to I or I' (which is I with one entry flipped to -1). So it suffices to check M=0, N=I, and M=0,N=I'.

Very cool argument. It took me a while to get it.

I am going to try to generalize this to the matrix

M T 0
T* M N
0 N 0

where the conclusion now - I think - would be that the signature is this matrix is the signature of M. Can I just shrink T to zero?

 

[lavinia]

My "engineer's intuition" says it is true, if you think about finding the extrema of x*Mx subject to the constraints Nx = 0, using Lagrange multipliers. The situations where this runs into trouble are when the constraints are not all independent, but that is excluded because N is non-singular.

Basically, each eigenvalue of M is turned into a pair of eigenvalues of the augmented matrix, one positive and one negative.

But my brain isn't working well enough to turn that idea into a proof right now - sorry!

This is the approach I have tried - more or less. I get a system of quadratic equations.

 

[lavinia]

I am going to try to generalize this to the matrix

M T 0
T* M N
0 N 0

where the conclusion now - I think - would be that the signature is this matrix is the signature of M. Can I just shrink T to zero?

If T is shrunk towards zero the the matrix becomes singular only if M is singular so your proof goes through.

 

[Vargo]

I am going to try to generalize this to the matrix

M T 0
T* M N
0 N 0

where the conclusion now - I think - would be that the signature is this matrix is the signature of M. Can I just shrink T to zero?

Lets see... The determinant of this matrix will equal (-1)^n(\det M) (\det N)^2, where n is the block size. If M is non-singular, the determinant is never zero, so no signature changes could happen by shrinking T down to zero.

If M is singular, the conclusion still sounds plausible, but I'm not sure whether the same proof could be adapted.

 

[lavinia]

If M is diagonal then T can be shrunk to zero in the plane of non-zero eigen values.

Choose a basis of eigen vectors for the copy of M that is in the upper left corner of the matrix.