For a Hermitian operator $A$ on a Hilbert space $H$ with eigenvalues $\{a_i\}$ and eigenvectors $v_i$ forming a complete basis, one can write $A$ as: $$ A = \sum_i a_i v_i v_i^\dagger, $$ where $v^\dagger$ is the Hermitian conjugate of $v$.
Suppose there is another complete basis $\{w_i\}$. Can I also use the spectral decomposition with this $w$-basis, instead of the eigenbasis $\{v_i\}$? $$ A = \sum_i a_i w_i w_i^\dagger $$
No, you cannot. The point of an eigenbasis is that it diagonalises the operator; it works on each basis vector separately and independently of the others. That diagonalisation is exactly what you see in the sum.
Take, for instance, the two-dimensional case with $A$ represented by $\left[\begin{smallmatrix}1&0\\0&2\end{smallmatrix}\right]$ in some basis $u_i$. Then that basis is an eigenbais, and you get $$ A = 1\cdot u_1u_1^\dagger + 2\cdot u_2u_2^\dagger $$ If you take a new basis, say $v_1 = u_1+u_2$ and $v_2 = u_1-u_2$, then in that basis we have that $A$ is represented by $\left[\begin{smallmatrix}1.5&-0.5\\-0.5&1.5\end{smallmatrix}\right]$. Of course, the matrix is still symmetric / Hermitian, because change of basis cannot change that fact. But if you want to decompose it in something like the above sum it's not enough with $v_1v_1^\dagger$ and $v_2v_2^\dagger$. You need $v_1v_2^\dagger$ and $v_2v_1^\dagger$ terms as well, as there are now non-zero off-diagonal elements: $$ A = 1.5\cdot v_1v_1^\dagger + 1.5\cdot v_2v_2^\dagger - 0.5\cdot v_1v_2^\dagger - 0.5\cdot v_2v_1^\dagger $$