These questions refer to this post: https://terrytao.wordpress.com/2008/10/28/when-are-eigenvalues-stable/#more-904
The general setting is considering a matrix that depends continuously on a parameter A(t). The post answers the question: what can we say about the derivative of the eigenvalues and eigenvectors of this matrix with respect to t?
My first question pertains to deriving the adjoint eigenvector equation. Specifically, let $A$ be an n x n matrix, and let $v_{1}, v_2, ..., v_n$ be the eigenvectors of $A$ forming a basis of $\mathbf{R}^n$. Let $w_{1}, w_2, ..., w_n$ be the corresponding dual basis (so $w_{j}^{\ast}v_k = \delta_{jk}$).
Using the reproducing formula for any vector $u \in \mathbf{R}^n$ $$ u = \sum_{1}^{n} (w_{j}^{\ast}u)v_j \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (1)$$
and the eigenvector equations
$$ Av_k = \lambda_k v_k \;\;\;\;\; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2)$$
what are the steps to derive the adjoint eigenvector equations (3)?
$$w_{k}^{\ast}A = \lambda_k w_{k}^{\ast} \;\;\;\;\;\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (3)$$
My second question also pertains to applying the reproducing formula (1). After taking the derivative of (2) with respect to t and making various substitutions, Tao arrives at the formula
$$ w_{j}^{\ast}\dot{A}v_k + (\lambda_j - \lambda_k)w_{j}^{\ast}\dot{v}_k = 0 \;\;\;\;\;\; (4)$$
Then, Tao applies the reproducing formula (1) to this expression (4) to derive the formula
$$ \dot{v}_k = \sum_{j \neq k} \frac{w_{j}^{\ast}\dot{A}v_k}{\lambda_k - \lambda_j}v_j + c_k v_k \; \; \; \;\;\;\; (5)$$
where c_k is a scalar that "reflects the freedom to multiply $v_k$ by a scalar."
My question is: what steps are necessary to arrive at this formula (5) using (1) and (4)?