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.

Assuming that both these basis are orthonormal, we get
$$||T(x)||^{2} \leq \sum|<v_{n},x>|^{2} = ||x||^{2}$$
where the last inequality is due to Parseval's formula. Hence the operator is bounded.
For the adjoint, calculate $<Tx,y>$ and surmise what $T^{*}y$ should be to get the same inner product. Then prove that what you found is actually a linear operator and is the adjoint.
