Show that the operator $T$ is bounded and find the adjoint I need help with this problem:

Let $\{u_{n}\}$ and $\{v_{n}\}$, $(n \in \mathbb{N})$ be two different Hilbert bases in a Hilbert space $H$. Define a linear operator $T$ so that, for $x \in H$:
  $$T(x) = \sum_{n=1}^{\infty} \lambda(n)u_{n}⟨v_{n} \mid x⟩$$ where $\lambda(n)$ is a fixed sequence of complex numbers with $|\lambda(n)| < 1, \forall n$ . Show that $T$ is bounded, and find its adjoint operator.

So to figure out that it is bounded I guess I could compute the norm somehow?
And to find the adjoint I use the definition that for $x,y \in H$ then:
$$\langle T(x) \mid y\rangle = \langle x \mid T^{*}(y) \rangle$$
where $T^*$ is the adjoint of $T$.
I am quite new to functional analysis and would be grateful is someone have a solution for this.

We have for $x \in H$ $\def\norm#1{\left\|#1\right\|}\def\sp#1{\left<#1\right>}$
\begin{align*}
  \norm{Tx}^2 &= \norm{\sum_{n=1}^\infty \lambda_n u_n\sp{v_n, x}}^2\\
    &= \sp{\sum_{n=1}^\infty\lambda_n\sp{v_n, x}u_n, \sum_{m=1}^\infty \lambda_m \sp{v_m, x}u_m}\\
    &= \sum_{n=1}^\infty\sum_{m=1}^\infty \overline{\lambda_n \sp{v_n, x}}\lambda_m \sp{v_m, x} \underbrace{\sp{u_n, u_m}}_{\delta_{nm}=}\\
    &= \sum_{n=1}^\infty |\lambda_n\sp{v_n, x}|^2\\
    &\le \sup_n |\lambda_n|\sum_{n=1}^\infty|\sp{v_n, x}|^2\\
    &\le \norm x^2
\end{align*}
So $T$ is bounded with $\norm T \le 1$, we moreover see that $\norm T = \sup_n |\lambda_n|$.
To compute the adjoint, let $x,y\in H$, we have
\begin{align*}
  \sp{T^*x, y} &= \sp{x, Ty}\\
     &= \sp{x, \sum_{n=1}^\infty \lambda_n \sp{v_n, y}u_n}\\
     &= \sum_{n=1}^\infty \lambda_n \sp{v_n, y}\sp{x, u_n}\\
     &= \sp{\sum_{n=1}^\infty \overline{\lambda_n\sp{x,u_n}}v_n, y}\\
     &= \sp{\sum_{n=1}^\infty \overline{\lambda_n}\sp{u_n, x}v_n, y}
\end{align*}
So for each $x \in H$:
$$ T^* x = \sum_{n=1}^\infty \overline{\lambda_n}\sp{u_n, x}v_n. $$
