About calculating adjoint operator Let inner product space V be defined over F or C and linear operators T on V, evaluate $T^{*}$ at the given vector in V.
$V=R^2, T(a,b)=(2a+b,a-3b), x=(3,5)$
I know $T^{*}$ is the conjugate transpose. But how am I supposed to approach this, need some general idea.
 A: Using the definition of adjoints, you know that,
$$\langle T^*(3, 5), (a, b)\rangle = \langle (3, 5), T(a, b)\rangle$$
for all $(a, b) \in \Bbb{R}^2$. In $\Bbb{R}^2$, this inner product is the dot product. Let $(x, y) = T^*(3, 5)$. We then have
$$(x, y) \cdot (a, b) = (3, 5) \cdot T(a, b) = (3, 5) \cdot (2a + b, a - 3b),$$
for all $(a, b) \in \Bbb{R}^2$. We must solve this for $x$ and $y$.
First, try $(a, b) = (1, 0)$. Then,
$$x = (x, y) \cdot (1, 0) = (3, 5) \cdot (2, 1) = 11.$$
Next, try $(a, b) = (0, 1)$. Then,
$$y = (x, y) \cdot (0, 1) = (3, 5) \cdot (1, -3) = -12.$$
Thus, $T^*(3, 5) = (x, y) = (11, -2)$.
A: I arrive to another solution: As we know the representative matrix of the adjoint operator is the adjoint of the representative matrix of the operator(in a orthonormal basis) : 
B the standard basis
If T(a,b)=(2a+b,a-3b)
$[T]_B=\begin{pmatrix}
2& 1\\
1 & -3
\end{pmatrix} $
And so the representative matrix of the adjoint is :
$\begin{pmatrix}
2 & 1\\
1& -3
\end{pmatrix}$
therefore T*=T
T*(3,5)=T(3,5)=(11,-12)
