Let $A$ be diagonalizable. Let $x_{1},\cdots,x_{n}$ $\in C^{n}$ be $n$ linearly independent right eigenvectors, i.e., $Ax_{i} = \lambda_{i}x_{i}$; and $y_{1},\cdots,y_{n}$ $\in C^{n}$ be $n$ linearly independent left eigenvectors, i.e.,$y^{T}_{i}A = \lambda_{i}y^{T}_{i}$. Show that there is a choice of left and right eigenvectors of $A$ such that any vector $v\in C^{n}$ can be expressed as
$v = \sum\limits_{i=1}^{n} (y^{T}_{i}v)x_{i}$
If we write $X = [x_{1}, \cdots, x_{n}] \in C^{n\times n}$ and $Y = [y_{1}, \cdots, y_{n}] \in C^{n\times n}$. What is the relation between X and Y ?
I try to solve this but I fail, I don;t know where to start, by the way, could I assume $x_{1},\cdots,x_{n}$ and $y_{1},\cdots,y_{n}$ are two basis of $C^{n}$
Yes, $x_{1},\cdots,x_{n}$ and $y_{1},\cdots,y_{n}$ are two bases of $C^{n}$ . You can show it with Vandermonde matrix or Lagrange interpolation.