Let $A\in M_{n\times n}(\mathbb{R})$ be a matrix. Let $\mathcal{B}$ be a basis of $\mathbb{R}^n$ and $X:=\mathcal{M}_{\mathcal{B}}(A)$. If $\mathcal{S}$ is the basis for which $\mathcal{M}_{\mathcal{S}}(X)$ is the Jordan Canonical Form of $X$, which is the basis $\mathcal{T}$ for which $\mathcal{M}_{\mathcal{T}}(A)$ is the JCF of $A$?
My Proof. Let $\mathcal{E}$ be the canonical basis of $\mathbb{R}^n$. Let $J:=\mathcal{M}_{\mathcal{S}}(X)$. We know that $$A=\mathcal{M}_{\mathcal{E},\mathcal{B}}(Id)\cdot X\cdot\mathcal{M}_{\mathcal{E},\mathcal{B}}(Id)^{-1},$$ $$X=\mathcal{M}_{\mathcal{S},\mathcal{E}}(Id)\cdot J\cdot\mathcal{M}_{\mathcal{S},\mathcal{E}}(Id)^{-1}.$$ Therefore, $A=\mathcal{M}_{\mathcal{E},\mathcal{B}}(Id)\mathcal{M}_{\mathcal{S},\mathcal{E}}(Id)\cdot J\cdot(\mathcal{M}_{\mathcal{E},\mathcal{B}}(Id)\mathcal{M}_{\mathcal{S},\mathcal{E}}(Id))^{-1}=\mathcal{M}_{\mathcal{S},\mathcal{B}}(Id)\cdot J\cdot\mathcal{M}_{\mathcal{S},\mathcal{B}}(Id)^{-1}$. We conclude that $\mathcal{T}$ is the basis $\mathcal{S}$ written in basis $\mathcal{B}$.
Can you tell me if my proof is correct? Thanks in advance.