Prove that $AB-BA = I$ has no solution in $M_{n\times n}(\mathbb R)$ without using matrix trace

Question

The title is self-explanatory. Prove that $AB-BA = I$ has no solution in $M_{n\times n}(\mathbb R)$ without using matrix trace. A,B are both from $M_{n\times n}(\mathbb R)$ and $AB$ is matrix multiplication. I am aware that these matrices' traces coincide and hence $tr(AB-BA) = 0$, while $tr(I) = n$ but are there other ways to solve this?

Answer 1

We prove by induction that we have $A^nB-BA^n=nA^{n-1}$ for $n\geq 1$. (For $n=0$, we get $B-B=0$)For $n=1$ this is clear. Suppose that we have the result for $0\leq k\leq n$. By multipling by $A$ we get $A^{n+1}B-ABA^n=nA^n$ and $A^nBA-BA^{n+1}=nA^n$. Hence adding the two, we get $$A^{n+1}B-BA^{n+1}+A(A^{n-1}B-BA^{n-1})A=2nA^n$$ and we finish easily.

Now Let $M(x)$ the minimal polynomial of $A$, of degree $d\geq 1$. We get immediately that $$M(A)B-BM(A)=M^{\prime}(A)=0$$ But this is a contradiction with the definition of the minimal polynomial.

Answer 2

Here is another way, albeit it involves traces at some level.

Let $\phi:M_{n\times n}(\mathbb R) \to M_{n\times n}(\mathbb R)$ be given by $\phi(A) = AB-BA$. Using the inner product induced by the Frobenius norm, we have $\phi^T(X)= X B^T -B^T X$, and clearly $I \in \ker \phi^T$. Since $\ker \phi^T = ({\cal R} \phi)^\bot$, we cannot have $I \in {\cal R} \phi$.