$AB-BA=I$ having no solutions The following question is from Artin's Algebra. 

If $A$ and $B$ are two square matrices with real entries, show that $AB-BA=I$ has no solutions. 

I have no idea on how to tackle this question. I tried block multiplication, but it didn't appear to work.

Eigenvalues of $AB \text{ and }BA$ are equal.Therefore, 0 must be  the eigenvalue of $AB-BA$. The product of all eigenvalues is the determinant of the operator. Hence, $$|AB-BA|=|I| \implies 0=1, \text{ which is a contradiction }$$
