Practicing Success
If A and B are square matrices such that $A^* = A$ and $B^* = B$, where $A^*$ denotes the conjugate transpose of A, then $(AB-BA)^* =$ |
Null matrix $AB-BA$ $BA - AB$ none of these |
$BA - AB$ |
We have, $(AB-BA)^* = (\overline{AB-BA})^T= (\overline{A}\, \overline{B}- \overline{B}\,\overline{A})^T=(\overline{A}\, \overline{B})^T - (\overline{B}\,\overline{A})^T$ $⇒(AB-BA)^* = (\overline{B})^T (\overline{A})^T (\overline{A})^T (\overline{B})^T - = -B^* A^* A^* B^*$ $⇒(AB-BA)^* = BA-AB$ |