linear algebra - To prove there are no Matrices $A$ and $B$ such that $AB-BA=kI$

Prove that there are no Matrices $A$ and $B$ such that $AB-BA=kI$ where $k \ne 0$

Now since the products $AB$ and $BA$ are both defined and a subtraction exists between them so obviously both are square matrices of same order.

Actually i have proved this by considering generic $2\times 2$ matrices.

Letting $$A=\begin{bmatrix}
a & b\\
c& d
\end{bmatrix}$$

Letting $$B=\begin{bmatrix}

p & q\\
r& s
\end{bmatrix}$$

Now $$AB-BA=\begin{bmatrix}
br-qc & q(a-d)+b(s-p)\\
c(p-s)+r(d-a)& cq-br
\end{bmatrix}=\begin{bmatrix}
k & o\\
0& k

\end{bmatrix}$$

$\implies$

$$br-qc=k$$ and

$$br-qc=-k$$

which is not valid unless $k =0$

is there a formal proof?