Prove that $AB-BA=I_2$ cannot hold whatever the real $2\times 2$ matries $A, B$

Question

How can I prove that $AB-BA=I_2$ cannot hold whatever the real $2\times 2$ matries $A, B$.

Suppose $A=\pmatrix{ a_{11} & a_{12}\\ a_{21} & a_{22}}$ and $B=\pmatrix{b_{11} & b_{12} \\ b_{21}& b_{22}}$.

Then $AB=(c_{ij})$, where $c_{ij}=\displaystyle\sum_{r=1}^2a_{ir}b_{rj}, i,j=1,2$ and $BA=(d_{ij})$, where $d_{ij}=\displaystyle\sum_{s=1}^2b_{it}a_{tj}, i,j=1,2.$

Now $AB-BA=I_2\implies c_{ij}-d_{ij}=\displaystyle\sum_{r=1}^2a_{ir}b_{rj}-\displaystyle\sum_{s=1}^2b_{it}a_{tj}$.
$\implies a_{12}b_{21}-b_{12}a_{21}=1, a_{21}b_{12}-a_{12}b_{21}=1,$
$b_{12}(a_{11}-a_{22})+a_{12}(b_{22}-b_{11})=0,$
$a_{21}(b_{11}-b_{22})+b_{21}(a_{22}-a_{11})=0$.

From the first two, $a_{12}b_{21}=b_{12}a_{21}$ is ambiguous. Hence the result follows.

I am looking alternative method to show the result. Is there any other way to show this result?

Answer 1

We always have $\operatorname{tr} (AB-BA) = 0$ but $\operatorname{tr} I_n = n$.

Answer 2

Taking the trace on both sides , we have $$trace(AB-BA)=trace(I_2)$$

But $$trace(A+B)=trace(A)+trace(B)$$. Hence $$trace(AB)-trace(BA)=2$$. But $trace(AB)$ is same as $trace(BA)$. hence you get $2=0$ which is absurd.

So the conclusion follows.