# What is the canonical form of real symmetric 2n\times 2n matrix under unitary congruence?

If $M$ is a $2n\times 2n$ real symmetric matrix, I would like to ask what could be its canonical form under unitary congruence. We view a unitary $n\times n$ matrix $U$ as a real $2n\times 2n$ matrix, and try to see if we can diagonalize'' $M$ by $$U^T M U.$$

In addition, what is the canonical form if we only require that $U$ is symplectic?

-

## 1 Answer

The difficulty is that the space $S_{2n}(\mathbb{R})$ of $2n$-by-$2n$ symmetric matrices with real entries is not irreducible under the action of $\mathrm{U}(n)$. In fact, there is a $\mathrm{U}(n)$-irreducible decomposition $$S_{2n}(\mathbb{R}) = \mathbb{R}\cdot I_{2n}\oplus A_n\oplus B_n$$ where $A_n$ is the space of matrices of the form $$A = \begin{pmatrix}s&a\cr-a&s\end{pmatrix}$$ where $s$ is traceless symmetric $n$-by-$n$ and $a$ is antisymmetric $n$-by-$n$ and where $B_n$ is the space of matrices of the form $$B = \begin{pmatrix}p&q\cr q&-p\end{pmatrix}$$ where $p$ and $q$ are $n$-by-$n$ symmetric matrices. (In this realization, $\mathrm{U}(n)$ is the set of matrices of the form $$U = \begin{pmatrix}x&y\cr -y&x\end{pmatrix}$$ where $x$ and $y$ are $n$-by-$n$ and $U = x+iy$ is unitary, i.e., $UU^\ast=I_n$.)

When you write $M = \lambda I_{2n} + A + B$ with $A\in A_n$ and $B\in B_n$, then you can use $U$ to diagonalize $A$ or you can use it to diagonalize $B$, but, usually, you cannot diagonalize both when $n>1$.

As for only requiring that $U$ be symplectic, now you are asking for a classification of the adjoint orbits of $\mathrm{Sp}(n,\mathbb{R})$ acting on its Lie algebra. Of course, you can conjugate the semi-simple elements into a maximal torus (of some signature), but a full classification of, say, the nilpotent elements, is complicated.

-
Thanks Prof. Bryant. I guess I can use symplectic matrices for my purpose. Then what is the criterion (or a sufficient condition) that a symmetric matrix corresponds to a semi-simple element in the Lie algebra of $Sp(n, {\mathbb R})$? –  UVIR Mar 9 '14 at 20:46
Oh, if you premultiply the symmetric matrix $S$ by $J$ (the skew-symmetric matrix that defines the symplectic structure), then $JS$ lies in the Lie algebra of $\mathrm{Sp}(n,\mathbb{R})$, and you then check to see whether that matrix $JS$ is semi-simple as a linear transformation on $\mathbb{R}^{2n}$. If it is, then it can be conjugated by an element of $\mathrm{Sp}(n,\mathbb{R})$ into one of the maximal tori. (Because this Lie algebra has rank $n$, there will be $n$ independent eigenvalues that show up as roots of the characteristic polynomial of $JS$.) –  Robert Bryant Mar 10 '14 at 1:32