Prove that matrix equation $AX-XA=I$ doesn't have a solution for any $A\in M_n(\mathbb{R})$
If $n=2,$ $$A= \begin{bmatrix} x & y \\ u & v \\ \end{bmatrix} $$ $$X= \begin{bmatrix} \alpha \\ \beta \\ \end{bmatrix} $$
$\Rightarrow AX-XA=O\neq I$
What is the more strict proof?
Note that $X\in\mathcal{M}_n(\mathbb{R}),$ else doing operations like $AX-XA$ have no sense. Use the trace and note that $$\mathrm{Tr}(AX-XA)=\mathrm{Tr}(AX)-\mathrm{Tr}(XA)=\mathrm{Tr}(AX)-\mathrm{Tr}(AX)=0$$ whereas $$\mathrm{Tr}(I)\geq 2.$$
Let $A\in M_n(F)$ where $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n$. If $[A,X]=I$, by mathematical induction we obtain $[A,X^k] = kX^{k-1}$ for each positive integer $k$. It follows that $[A,f(X)]=f'(X)$ for every polynomial $f$. In particular, if we take $f$ as the minimal polynomial of $X$ and $r=\deg(f)$ (so that $1\le r\le n$), then $f'(X)=[A,f(X)]=[A,0]=0$. But this contradicts the minimality of $m$, because the leading coefficient of $f'$ is $r$, which is nonzero over $F$ as we have either $\operatorname{char}(F)=0$ or $\operatorname{char}(F)>n\ge r$. Hence we conclude that $[A,X]=I$ is not solvable.
While this proof is (obviously) not as simple as the classical proof that relies on taking matrix traces, it illustrates the use of the mapping $X\mapsto[A,X]$ as a derivation, which is the key to the classical proof of Jacobson's lemma.
