# Can we classify the general linear maps such that for a fixed matrix $A \in M_n(\mathbb R)$ the conjuagation action fixes the first column?

Let $$A \in GL_n(\mathbb R)$$ be fixed. Let us consider the conjugation action by $$G \in GL_n(\mathbb R)$$, i.e., $$G^{-1}AG$$. I would like to see a way to identify the matrices such that the action fixes the first column of $$A$$. That is, what can we say about the set \begin{align*} \mathcal E = \{G \in GL_n(\mathbb R): G^{-1}AGe_1 = a_1\}, \end{align*} where $$e_1$$ is the standard basis vector and $$a_1$$ is the first column of $$A$$. In particular, is the set connected? Let us exclude the trivial case: $$A$$ is a multiple of $$I$$ or $$-I$$.

If we define a linear map $$\phi: M_n(\mathbb R) \to \mathbb R^n$$ by $$X \mapsto (XA-AX)e_1$$, then $$\mathcal E = \ker(\phi) \cap GL_n(\mathbb R)$$.

## 1 Answer

One can not say that $$\mathcal E$$ is always connected.

For $$n=3,$$(and similarly $$n>3$$) let $$A$$ be a matrix with $$a_{i1}=1,\quad \forall i \in \{1,2,\ldots,n\}$$

Then $$\mathcal E$$ contains the identity matrix whose determinant is $$1$$ and it also contains the following matrix with determinant $$-1$$:

$$\begin{pmatrix} 1&0&0\\0&0&1\\0&1&0 \end{pmatrix}$$

According to the previous version of your question about the (semi)group structure of $$\mathcal E$$, observe that if you can prove that $$\mathcal E$$ is a semigroup then it is a group, too. Because the inverse of a every matrix $$G$$, can be written in the form of a polynomial in $$G$$.

I think that $$\mathcal E$$ is not a Lie group because it is contained in a sub vector space $$F$$, the kernel you mentioned, whose dimension is equal to the dimension of $$G$$. Moreover $$Exp(F)$$ is contained in $$\mathcal E$$ But in general case, $$F$$ is not a Lie algebra with the commutator Lie bracket. Please see the comment by Corbennick

• I was in mind probably $\mathcal E$ would have two (or more) components corresponding to the positive determinant and negative determinant. Would you comment on how many components would $\psi(\mathcal E)$ have, where $\psi: B \mapsto B^{-1}AB$ provided $A$ has a real eigenvalue? In this case, the positive and negative determinant would be mapped into the same image. – user1101010 Oct 18 '18 at 16:29
• @user9527 In fact one can generalize your interesting question in the following form: is there a linear subvector space of $M_n(\mathbb{R}$ which intersect $GL_n(\mathbb{R}$ in more than 2 components?Is there an example of this situation with codimension is equal $n$?(In your question that kernel has codimension n) – Ali Taghavi Oct 18 '18 at 16:54
• In $2\times 2$ matrices the matrices of the form $[[x,y],[y,x]]$ intersect $GL_2$ in the subset of the plane with equation $xy\neq 0$, which has 4 components. – YCor Oct 18 '18 at 17:04
• @user9527 Yes. In size $n$, in the subspace of upper triangular matrices, there are $2^n$ components of invertible matrices. – YCor Oct 18 '18 at 17:06
• So, $G\to F$ is an immersion between manifolds of the same dimension, so has an open image; $F$ is the the lie algebra of $G$. – YCor Oct 19 '18 at 12:19