I am supposed to show that for a skew-symmetric matrix $A$ with $det(A) \neq 0$, meaning that is has an even number of columns and rows, there is an invertible matrix $ R $ such that $ R^T A R = M$, where $M$ is a block matrix of the form $\begin{bmatrix} 0 & Id & \\ -Id & 0 & \end{bmatrix}$.

This excercise is so general that I don't know which approach/idea or observation is useful to solve this excercise.

  • $\begingroup$ $Id$ denotes the identity of what? $\endgroup$ – rschwieb May 10 '13 at 16:57

A skew-symmetric matrix $A$ creates an alternating bilinear form $B(x,y):=x^TAy$.

A full discussion of these, including the reexpression of $A$ into a basis so that the matrix has that block form is in Basic Algebra I by Jacobson, starting on page 349.

Added: Ah, here's another set of brief notes which I think has the details.

I didn't spend long enough to confirm it, but it also has to be in Keith Conrad's notes on bilinear forms somewhere...


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