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## Is the signature of this matrix zero?

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?
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 Recognitions: Science Advisor 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!
 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'.

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Homework Help

## Is the signature of this matrix zero?

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

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

Recognitions:
 Quote by 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'.
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?

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

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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.