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Question

I am trying to show that if A is a nxn Hermitian matrix (A* = A where A* is the adjoint aka conjugate transpose of A) then A is congruent to A^3 (the third power of A). 

Congruent here means there is invertible S such that S*AS = A^3. 

I also need a counter-example to show the same conclusion does not hold if A^3 is replaced with A^2.

This should have to do with the inertia of a matrix and Sylvester's law.

Anonymous · Accepted Answer

Part 1:

This is trivial if A is non-singular: Just let S = A

So you need to separate out the "singular part" of A and the obvious way to do that (especially given the hint about inertia and Sylvester's law) is to use the eigenvectors of A as basis vectors and transform A into a diagonal matrix D.

In the transformed basis system, you can take S = D but with the zero diagonal terms replaced by 1 (or any other nonzero number) to make  S non-singular. Transform S back into the original basis system.

Part 2:

Take A to be the 1x1 matrix [ -1 ].

Let S = [ s ]. Since S is Hermitian, s is real.

But if S*AS = A^2, then s^2 = -1 which has no real solution.

The connection with the inertia of the matrix is that cubing the matrix preserves the signs of all the eigenvalues, but squaring the matrix does not.

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