reduction of a skew-symmetric matrix Birkoff and MacLane state that any real symmetric matrix $A$ has the form
$ A = P^{-1}BP $ where $ B^2 $ is diagonal and they ask for a proof as an exercise.
It seems to me that if $A$ is skew-symmetric then $B$ is also, but for matrices of
order 3, then the square can be diagonal only if it is 0. Where am I going wrong?
 A: Let's go complex.  If $A$ is real and skew-symmetric, then $iA$ is Hermitian.  Hence $iA$ is diagonalizable with real eigenvalues.  Hence $A$ is diagonalizable with purely imaginary eigenvalues.  And since $A$ is real, the non-zero eigenvalues must come in conjugate pairs.
Hence $A$ is similar to the complex matrix $\text{diag}(i\lambda_1,-i\lambda_1,i\lambda_2,-i\lambda_2, \dots)$.  But it is easy to see that $\text{diag}(i\lambda,-i\lambda)$ is similar to $C_\lambda = \begin{bmatrix} 0 & -\lambda \\ \lambda & 0 \end{bmatrix}$.  Hence $A$ is similar to the matrix $B$ which is made up of $(2\times2)$-matrices $C_{\lambda_1}$, $C_{\lambda_2}$, etc, along the diagonal.
And then it is easily seen that $B^2$ is diagonal.
Finally, I don't know if you also want $P$ to be real, but there is a result that says that if two real matrices are similar over the field of complex numbers, then they are also similar over the field of real numbers.
A: First of all, I should have stated that we are  working on quadric forms under the
orthogonal group, so the field is $\mathbb{R}$ and transformations are
orthogonal. So, if $A$ is skew-symmetric then $A = -A'$ and 
$A^2 = -A'A $ is symmetric. It has been prooved that any symmetric matrix
can be diagonalized by an orthogonal transformation so we may suppose
that $ A^2 = P^{-1}DP $ where $D$ is diagonal. If we set $B = PAP^{-1}$
it follows that $A=  P^{-1}BP $ and $B^2 = D $ which is diagonal.
