Assume $A$ is a complex-valued square matrix, i.e. $A\in \mathbb{C}^{n\times n}$, and $A$ has a full set of eigenvectors denoted as $V=[v_1, v_2, \cdots, v_n]$. Then we known the following facts \begin{align} Av_i &= \lambda_i v_i,\cr [w_1, w_2, \cdots, w_n] &= [v_1, v_2, \cdots, v_n]^{-*}, \cr A^*w_i &= \bar\lambda_i w_i,\cr w_i^*v_j &= 1, ~\text{if}~ i=j;~ \text{otherwise}~~ w_i^*v_j = 0, \end{align} where the superscript $*$ denotes the conjugate transpose of a matrix.
My question is given a vector $x\in \mathbb{C}^n$, how to prove the reality below \begin{align} x = \sum_{i=1}^{n} \langle w_i, x \rangle v_i, \end{align} where $\langle \cdot, \cdot \rangle$ denotes the inner product between two complex-valued vectors?
Since $V$ is a basis, we may expand $x$ as $$ x = \sum_{i = 1}^n a_i v_i $$ Then \begin{align} w_j^* x &= w_j^* \left(\sum_{i = 1}^n a_i v_i \right) \\&= \sum_{i =1}^n w_j^* (a_i v_i) \\&= \sum_{i = 1}^n a_i (w_j^* v_i) \\&= a_j (w_j^* v_j) + \sum_{i = 1, i \neq j}^n a_i (w_j^* v_i) \\&= a_j \cdot (1) \sum_{i = 1, i \neq j}^n a_i \cdot (0) \\&= a_j + \sum_{i = 1, i \neq j}^n 0 \\&= a_j \end{align} Hence $w_j^* x = a_j$. This holds for all $j \in \{1, \ldots, n\}$. Therefore, we may substitute into our expansion of $x$ in $V$ to obtain $$ x = \sum_{i = 1}^n (w_i^* x) v_i $$ Since $w_i^* x = \langle w_i , x \rangle$, we finally find $$ x = \sum_{i = 1}^n \langle w_i , x \rangle v_i $$ as desired.
It is worth noting that we never used the fact that the $w_i$ or the $v_i$ are eigenvectors. All we really needed was that $V$ is a basis and that $w_i^* v_j = 1$ if $i = j$ and is $0$ otherwise.