Let $V$ be the space $\mathbb C^2$, with the standard inner product. Let $T$ be the linear operator defined by $T \epsilon_1 = (1, - 2), \,\,\ T\epsilon_2 = (i, - 1)$. If $ \alpha = (x_1, x_2)$, find $T^* \alpha$.
Attempt: Because the base $ \{ \epsilon_1,\epsilon_2 \}$ is orthonormal, I know that the matrix of $T^*$ is the conjugate transpose of the matrix of $T$ (in the same basis). Then
$$[T^*]=\begin{pmatrix}1 & -2\\-i & -1\end{pmatrix}$$.
Therefore
$T^* \epsilon_1 = (1, - i), \,\,\ T^* \epsilon_2 = (-2, - 1)$
and
$T^* \alpha= T^* (x_1,x_2)=x_1T^* \epsilon_1 + x_2T^*\epsilon_2= x_1(1, - i) +x_2(-2, - 1)= (x_1 - 2x_2, -ix_1 -x_2)$.
Is this right?