Prove Lie algebra $L=\{x\in \mathfrak{gl}_n:xA+Ax^{T}=0\}$ is semisimple

Question

Let $L=\{x\in \mathfrak{gl}_n(\mathbb{C}):xA+Ax^{T}=0\}$ with $A=\begin{pmatrix} 13 & 7 & 1 & 2 & -1 & 1 & 5\\ 9 & 6 & 3 & 4 & 1 & 1 & 4 \\ 5 & 3 & 2 & 3 & 2 & 10 & 1 \\ 4 & 2 & 1 & 2 & 1 & 4 & 1 \\ 1 & 1 & 0 & 1 & 1 & 5 & 0 \\ 3 & 1 & −10 & −4 & −5 & 1 & 2 \\ 5 & 2 & 1 & 1 & 0 & 0 & 2 \\ \end{pmatrix}$.

1. Show that $L$ is semispimple
2. Find how its decomposes into simple algebras.

It reminds me of the way we define the classical Lie algebras so maybe $A$ has some property I fail to see?

Edit: If $L_1=\{x\in\mathfrak{gl}_n(\mathbb{C})\}: x(A+A^T)+(A+A^T)x^T=0\}$ then $L\subseteq L_1$ where $A+A^T$ is symmetric and thus we can use the signature argument of @KentaS. But in general this is just a subalgebra so I am not sure if I get anything of essence. Maybe if I have $A$ diagonizable I can then find the signature and proceed that way?

Some motivation: Notice that the definition of the classical Lie algebras is similar but for signature matrices so this question is essentially a special case of "how do you study Lie algebras defined in a similar manner to the classical ones for arbitary matrices?"

Answer 1

TL;DR. The Lie algebra in question is isomorphic to the Lie algebra $\mathfrak{so}_3\oplus\mathfrak{sp}_4$, and, thus; it is semisimple. The property of $A$ which makes it possible is the fact that $A$ is invertible and $rk B+rk C=rk A$ where $B,C$ are the symmetric and skew-symmetric parts, respectively. For any matrix with these properties, the corresponding Lie algebra will be semisimple.

First, I want to present another perspective on the way the Lie algebra in question is defined.

The group $G=GL_7(\mathbb{C})$ acts on the vector space $V$ of $7\times 7$-matrices via $g\cdot M = gMg^T$. This is not the classical conjugation action that we are familiar with, however, it is still a Lie group representation. This representation also induces a Lie algebra action of $\mathfrak{gl}_7(\mathbb{C})$ on $V$ via $M\mapsto xM+Mx^T$. I omit the details. I'd also like to point out that this representation is not irreducible, in fact, it splits as $$ V = S^2(n)\oplus \Lambda^2(n) $$ where $S^2(n), \Lambda^2(n)$ are the spaces of symmetric and skew-symmetric $n\times n$-matrices, respectively.

The Lie algebra $L=\{x\in\mathfrak{gl}_7\mid xA+Ax^T=0\}$ that you have defined is then the Lie algebra of the stabilizer $C_G(A)=\{g\in G\mid gAg^T=A\}$ of $A$. This immediately explains why $L$ is always a Lie algebra. This observation also implies that we can replace $A$ with any $gAg^T$ since the stabilizers of two vectors in the same orbit are isomorphic. This was pointed out in the comments. Moreover, if $A=B+C$ where $B\in S^2(n), C\in \Lambda^2(n)$, then we have $$ C_G(A) = C_G(B)\cap C_G(C). $$

I claim that $A$ is in the same orbit with the matrix $$ M = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -1 & 0 & 0 \end{bmatrix} $$ The intuition behind this normal form is the following. The identity matrix $I_n$ is the standard normal form of a positive definite symmetric matrix under the transpose action. Similarly, the matrix $$ J_n = \begin{bmatrix} 0 & I_n\\ -I_n & 0 \end{bmatrix} $$ is the standard normal form of a skew-symmetric matrix under the transpose action. The above matrix consists of the normal forms of its symmetric and skew-symmetric parts. Assuming this claim, the rest of the proof goes easily. For a matrix $x\in\mathfrak{gl}_7$, write $$ x = \begin{bmatrix} x_1 & x_2 \\ x_3 & x_4 \end{bmatrix} $$ where $x_1$ is $3\times 3$ and $x_4$ is $4\times 4$. Then, we have $$ x M+Mx^T = 0\quad\iff\quad x_1+x_1^T = 0,\; x_4J_2+J_2x_4^T=0,\; x_2 = 0,\; x_3=0. $$ Note that $\{x_1\mid x_1+x_1^T=0\}$ is $\mathfrak{so}_3$ and $\{x_4\mid x_4 J_2+J_2 x_4^T=0\}$ is $\mathfrak{sp}_4$. Thus, the Lie algebra of the stabilizer of $M$ is $\mathfrak{so}_3\oplus\mathfrak{sp}_4$ and since $M$ and $A$ are in the same orbit, the Lie algebra $L$ of the stabilizer of $A$ is isomorphic to $\mathfrak{so}_3\oplus\mathfrak{sp}_4$.

It is not true in general that any matrix has such a normal form, however, the matrix in question has some interesting properties. Using a computer algebra, you may verify the following claims:

1. The symmetric and skew-symmetric parts $B,C$ of $A$ satisfy $$ rk(B) = 3,\quad rk(C) = 4, \quad rk(A)=rk(B+C)=7. $$

2. $B$ is positive semi-definite, i.e., its $3$ nonzero eigenvalues are all strictly positive.

Note that for any two matrices $rk(B+C)\leq rk(B)+rk(C)$ is satisfied. The equality holds if and only if the column spaces of $B,C$ have no intersection, i.e., the image of $B+C$ is the direct sum of images of $B,C$. If we choose a basis of $\mathbb{C}^7$ consisting of eigenvectors of $B$ and $C$, then we deduce that there exists $g\in GL_7$ such that $$ gBg^{-1} = \begin{bmatrix} B' & 0\\ 0 & 0 \end{bmatrix},\quad gCg^{-1} = \begin{bmatrix} 0 & 0\\ 0 & C' \end{bmatrix} $$ where $B',C'$ are diagonal matrices with diagonal entries consisting of eigenvalues of $B,C$, respectively. For our purposes, it is not important what the eigenvalues exactly are, but their signs matter. For $B'$, we know by the fact that $B$ is PSD, the diagonal entries of $B'$ are strictly positive, say $\lambda_1,\lambda_2,\lambda_3$. For $C'$, the eigenvalues of skew-symmetric matrices come in pairs and they have to be purely imaginary, so we may assume that $C'$ is of the form $$ \begin{bmatrix} \lambda_4 i & 0 & 0 & 0\\ 0 & \lambda_5 i & 0 & 0\\ 0 & 0 & -\lambda_4 i & 0\\ 0 & 0 & 0 & -\lambda_5 i \end{bmatrix}$$ where $\lambda_4,\lambda_5$ are strictly positive real numbers.

Consider the diagonal matrix $h$ with diagonal entries $(1/\sqrt{\lambda_1},1/\sqrt{\lambda_2},1/\sqrt{\lambda_3},1/\sqrt{\lambda_4},1/\sqrt{\lambda_5},1/\sqrt{\lambda_4},1/\sqrt{\lambda_5})$. Then, we have $$ h (g A g^{-1}) h^T = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & i & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & i & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & -i & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & -i \end{bmatrix} $$ The lower $4\times 4$ block of the matrix is actually similar to $$ \begin{bmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{bmatrix} $$ by comparing the eigenvalues. Hence, there exists yet another $k\in GL_7$ such that $$ k( h (g A g^{-1}) h^T )k^{-1} = M $$ where $M$ is the normal form above. We are almost done, but not yet. Note that I claimed $A,M$ are in the same orbit with respect to the transpose action, i.e., there exists $g$ such that $gAg^T = M$. However, the transformations I applied also include conjugations. To overcome that, we need the following theorem from Ganthmacher's Theory of Matrices, Volume 2, Page 41, Theorem 6.

Theorem: Let $A,M$ be two $n\times n$ matrices. Let $A=B+C, M=N+L$ be the decomposition into symmetric and skew-symmetric parts respectively. Then, there exists $g\in GL_n$ with $gAg^T=M$ if and only if there exist $h,k\in GL_n$ with $h B k = N, h C k = L$.

Note that the transformations that I applied send the symmetric part $B$ of $A$ to the symmetric part of $M$ and similarly send the skew-symmetric part $C$ of $A$ to the skew-symmetric part of $M$. Hence, the second condition of the theorem is satisfied. Using the theorem, there exists $g\in GL_7$ with $gAg^{T}=M$.