I Proving a result about invertibility without determinants

1. Feb 27, 2017

Mr Davis 97

Let A and B be nxn matrices over an arbitary field such that AB = -BA. Prove that if n is odd then A or B is not invertible.

This is rather easy when we use determinants. However, I am curious, how hard would it be to prove this without the use of determinants? What would be involved in such a proof, and how much more work would it be when compared to just using determinants?

2. Feb 27, 2017

Staff: Mentor

I have no idea except to describe the determinant by geometric means which wouldn't really be another approach. However, "arbitrary field" is wrong. One has to require a certain condition on the field.

3. Feb 27, 2017

mathwonk

assume A invertible and prove B is not.

i think it implies they (B and -B) are similar, hence have the same jordan form, and that blocks with diagonal constant c are balanced by equal size diagonal blocks with constant -c. Hence odd size suggests some c = -c = 0. (Pass to algebraic closure to use jordan form.) Of course as fresh 42 warned, it is prudent to require 1 ≠ -1 in the field used.

so this is using ideas actually more powerful than determinants but also more conceptual.

normal forms may be obtained by diagonalizing the characteristic matrix with no specific mention of determinants, but they are there as the constant term of the characteristic polynomial (defined as the product of the diagonal entries of the diagonalized characteristic matrix.)

Last edited: Mar 1, 2017