Spectrum of an operator on a Hilbert space is $\{0\}$ $(x_i)_{i=1} ^n$ and $(y_i)_{i=1} ^n$ are elements of a Hilbert space $H$ such that $\langle x_i,y_j \rangle=0$ for all $i,j.$  Consider the operator $A(a)=\sum_{i=1} ^n \langle a,x_i \rangle y_i$ for $a\in H$ (note that $A(a)\in H$).
How can I show that the spectrum of $A$ is $\{0\}$? I tried to show that $||A||_{op}=0$, but it did not give alot, because I do not know anything about $\langle y_i,y_j \rangle,\langle a,x_i \rangle$.
Can I extend $(x_i)_{i=1} ^n$ and $(y_i)_{i=1} ^n$ to form an orthoormal basis of $H$? If I could do that I would be able to write $\langle a,x_i \rangle$ as a sum of $\langle y_i,y_j \rangle,\langle x_i,x_j \rangle,\langle x_i,y_j \rangle$.
 A: We have $\operatorname{ran}A = \operatorname{span}\{y_i, i = 1,\dots,n\}$, so $A$ has a finite-dimensional range and thus is a compact operator. Since compact operators only have point spectrum outside of the origin, we are left to show that a nonzero number can't be an eigenvalue. Note that the heavy machinery here is the result on the spectrum of compact operators and to an extend seeing that $A$ is compact.
Consider $\lambda \in \mathbb C$ and assume that $A(a)=\lambda a$ for some nonzero $a\in H$. This means in particular that $a$ is part of the range of $A$,  we have $a \in \operatorname{span}\{y_i, i = 1,\dots,n\}$, thus $\langle a,x_i \rangle = 0$. Plugging this into the definition of $A$ yields $A(a)=0$. Since $a\neq 0,$ we get $\lambda = 0$. 
Furthermore, we have $A(y_i)=0$, so $\lambda = 0$ is indeed an eigenvalue. As a bonus we see that  $\operatorname{ran}A \subset \operatorname{ker}A$, hence we have that $A^2 = 0$.
To wrap up: We have shown that the only possible eigenvalue of $A$ is $0$ and that the eigenvalue $0$ has an eigenspace which contains $\operatorname{ran}A=\operatorname{span}\{y_i, i = 1,\dots,n\}$.
Edit: We can also go with a different heavy machinery. As above, we simply show that $A$ is nilpotent (here $A^2 = 0$, since $\operatorname{ran}A \subset \operatorname{ker}A$) and use a result on the spectral radius of a nilpotent operator, which yields $\sigma(A)=\{0\}$
