Diagonalization of alternating matrix

Question

I am reading the book "Algebra", written by Serge Lang and having difficulty in an explaining from that book on page 588.

The problem is the following. Let $G \in M_n(\mathbb{R})$ be an alternating matrix ($G^t = - G)$. Then there exists a nonsingular matrix $C$ such that $C^t G C$ is the matrix:

\begin{pmatrix} 0& I_r & 0\\ -I_r & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}

in which, $I_r$ is the identity matrix of dimension $r$. The detail proof is given on page 587.

Normally, I see that it should be $C^{-1}GC$ (instead of $C^tGC)$.

Is it true that $C^{-1} = C^t$ if $G$ is alternating? If that, please give me some hint to prove that result.

Thank you in advance.

Answer 1

The eigenvalues of $K:=\pmatrix{0&I&0\\ -I&0&0\\ 0&0&0}$ are $\pm i$ and $0$, but $G$ in general can have other eigenvalues. So, clearly, the author doesn't mean that $C$ is orthogonal, otherwise $G$ would be always similar to $K$.

As a user has pointed out in the comment, here $G$ is the matrix representation not of a linear transformation, but of a bilinear form. When the basis is changed, the old and new matrix representations of a bilinear form in general are related by matrix congruence, not by similarity.

Answer 2

Assume $G\neq \bf{0}.$
$G$ is a real skew-symmetric matrix, so its non-zero eigenvalues are coming in pairs $\pm i\lambda.$ The rank of $G$ is even, say $rank(A)=2m.$
The matrix $$K_r=\begin{pmatrix} 0& I_r & 0\\ -I_r & 0 & 0\\ 0 & 0 & 0 \end{pmatrix}$$ is skew-symmetric with rank $2r.$
Skew-symmetric matrices are congruent iff they have the same rank.
Thus $G,K_m$ are congruent: there exists a nonsingular matrix $C$ such that $$C^tGC=K_m.$$ The matrix $C$ can be unitary $(C^t=C^{-1})$ only if the eigenvalues of $G$ are those of $K_m,$ (incl. their multiplicity).

Answer 3

in case of orthogonal matrix A'=inverse (A), since alternating matrices are orthogonal one use A' and A^(-1) interchangeably in case of Alternate Matrices