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Consider two $n \times n$ similar matrices $A$ and $B$. With Jordan decomposition $$ V_a^{-1} A V_a = J\ \mbox{ and } \ V_b^{-1} B V_b = J$$ I find the similarity transformation $P = V_b V_a^{-1}$ and thus $$ A = P^{-1} B P.$$

The similarity transformation matrix $P$ is not unique as it can be multiplied by a scalar, and $P_2 = cP$ is also a similarity transformation for any scalar $c\ne 0$.

I wonder what a more general way to parameterize $P$ is (e.g., by introducing some weighting/scaling in Jordan decomposition), and how I can describe all possible similar transformation matrices?

A similar question was asked here but without the Jordan decomposition setup and, unfortunately, without ensuring the invertibility of $P$.

Arastas
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  • You are effectively asking for a description of the centralizer of a matrix with given JNF. Check for duplicates, it's very likely that the question was asked before. – Moishe Kohan Apr 23 '25 at 12:50
  • @MoisheKohan sorry, I do not see why it’s about the centraliser (of which matrix)? Centraliser of a matrix is the set of all matrices commuting with the given one, right? – Arastas Apr 23 '25 at 13:42
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    Right. Your set will be of the form $gC(A)$, where $g$ is a fixed element of your subset and $C$ denotes the centralizer. – Moishe Kohan Apr 23 '25 at 13:44
  • See for instance this question for a discussion of centralizers in terms of JNF. (It's complicated...) – Moishe Kohan Apr 23 '25 at 13:51
  • @MoisheKohan thanks! If I got it right, you say that if $PA=BP$ and $AX =XA$, then $(PX)A=B(PX)$. How can we choose only those $X$ that $PX$ is invertible? – Arastas Apr 23 '25 at 14:02
  • You just consider the centralizer group, restricting to invertible elements of the centralizer (which is an open and dense subset of the centralizer). – Moishe Kohan Apr 23 '25 at 14:04
  • @MoisheKohan yes, thanks. Do we know if all possible similarity transformations matrices are necessarily of this form? – Arastas Apr 23 '25 at 14:07
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    Yes, they are. I will add a proper answer later on, I was assuming that you know all this already... – Moishe Kohan Apr 23 '25 at 14:08

1 Answers1

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Let $M_n(F)$ denote the algebra of $n\times n$ matrices over a field $F$ (in the case of interest, $F=\mathbb C$). For a matrix $A\in M_n(F)$ the commutant $C(A)$ of $A$ is $$ C(A)=\{X\in M_n(F): XA=AX\}. $$ Then $C(A)$ is a subalgebra of $M_n(F)$. In $C(A)$ we also have the subgroup $Z(A)$ consisting of invertible matrices, called the centralizer of $A$. (This terminology is frequently conflated, I am trying to separate the two notions.) If $F=\mathbb C$, then for every $A$, the centralizer $Z(A)$ is open and dense in $C(A)$, moreover, it is a connected subset of $C(A)$ (since it has complex codimension 1). With a bit more thought, one sees that $Z(A)$ is a Lie subgroup of $GL_n(\mathbb C)$ whose Lie algebra is $C(A)$ and $Z(A)=\exp(C(A))$. Since $C(A)$ is a linear subspace of $M_n(\mathbb C)$, this gives us a "parameterization" of $Z(A)$. Describing $C(A)$ explicitly in terms of the JNF of $A$ is a bit complicated, it was discussed several times on this site, see for instance this post. (The easiest case is when $A$ is diagonalizable, then $C(A)$ is a direct sum of matrix subalgebras corresponding to the eigenspaces of $A$.)

Next, some general formalism. Let $X$ be a set and $G\times X\to X$ be a (left) group action on $X$. For $x\in X$ the stabilizer $G_x$ of $x$ in $G$ is the subgroup of $G$ consisting of elements $g$ such that $gx=x$. For $x, y\in X$, the transporter subset $G(x,y)$ is the subset $$ \{g\in G: gx=y\}. $$ It is clear that if $g\in G(x,y)$ and $h\in G_x$, then $gh\in G(x,y)$. Conversely, if $g, g'\in G(x,y)$, then $g^{-1}g'\in G_x$, hence, $g'\in g G_x$. Thus, one can describe the transporter subset $G(x,y)$ as the coset $gG_x$ for some fixed element $g\in G(x,y)$ (provided that $G(x,y)$ is nonempty of course).

In your setting, $G=GL_n(F)$, $X=M_n(F)$ (here, again, $F$ is an arbitrary field) and the action of $G$ on $X$ is by conjugation: $$ (g,x)\mapsto gxg^{-1}. $$
The stabilizer $G_x$ then is nothing but the centralizer $Z(x)$ defined above. By combining these observations, we see that for any two conjugate matrices $x, y\in M_n(F)$ (i.e. satisfying $y=g_0xg_0^{-1}$ for a fixed $g_0\in G$) we have $$ G(x,y)=\{P\in GL_n(F): PxP^{-1}=y\}, $$ which, in turn, can be described in terms of the centralizer as $$ g_0Z(x). $$ Or, in terms of the commutant, it can be described as $$ g_0\exp(C(x)). $$ (This is a nonlinear analogue of describing the solution set of a system of linear equations $Lx=b$ in terms of a particular solution $x_0$ and the kernel of $L$.)

Moishe Kohan
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