Finite-rank operator and choice of basis Suppose $H$ is a Hilbert space and $T(h) = \sum\limits_{i=1}^n \alpha_i \langle h, v_i\rangle u_i$ is a finite-rank operator on $H$, where $\{v_i\}$ and $\{u_i\}$ are orthonormal basis of $H$.
I'm interested in writing this operator in term of just one basis, i.e.
$$T(h) = \sum\limits_{i,j} \lambda_{i,j} \langle h, e_i\rangle e_j,$$
where this summation would be finite.
This suggests me a more general question : given any two orthonormal basis $\{w_i\}$ and $\{z_i\}$, is it possible to find coefficients $\beta_i$ so that
$$T(h) = \sum\limits_{i=1}^k \beta_i \langle h, w_i\rangle z_i$$
and, if yes, do we have $k = n$ ?
 A: In general, the representation of $T(h)$ in terms of $\{w_i\}$ and $\{z_i\}$ will not be finite for arbitrary orthonormal bases. 
This can best be illustrated in a very simple special case: Assume that $n = 1$ in the definition of $T$, and let $v_i = w_i$ for all $i$, so that we only change from $\{u_i\}$ to $\{z_j\}$. When we express $u_1$ in terms of the different basis $\{z_j\}$, the representation
$$u_1 = \sum_{i=1}^\infty d_{1,i}z_i$$ 
will contain infinitely many non-zero terms in most cases (e.g., think of Fourier series). In terms of the changed basis, the operator $T$ now reads:
$$T(h) = \sum_{i=1}^\infty\alpha_1d_{1,i}\langle h, v_1\rangle z_i\equiv \sum_{i=1}^\infty\beta_i\langle h, v_1\rangle z_i,$$
with $\beta_i := \alpha_1d_{1,i}$. Unless $T\equiv 0$, as many $d_{1,i}$ are non-zero as $\beta_i$ are non-zero, hence the representation of $T$ with respect to $\{z_j\}$ is an infinite sum/series (in many cases).
Note that with a similar argument one can show that for $n>1$ the representation of $T$ after the basis change will usually involve a double sum and not a single sum as stated at the end of your question.
