Cramer's rule for eigenvectors I know that the above doesn't exist, but do bear with me. I need estimates/formulas for entries of certain eigenvectors and Cramer's rule keeps popping up in my mind. So, what can play an anlogous role in this case?
 A: If $u$ is a unit eigenvector with eigenvalue $\lambda$ of a Hermitian matrix
$$ A_n = \begin{pmatrix} a & X^* \\\\ X & A_{n-1} \end{pmatrix}$$
with $a$ a real number, $X$ an $n-1$-dimensional row vector, and $A_{n-1}$ an $n-1 \times n-1$ Hermitian matrix, then (provided that $\lambda$ is not an eigenvalue of $A_{n-1}$) the magnitude of the first coefficient $u_1$ of $u$ is given by the formula 
$$ |u_1|^2 = \frac{1}{1 + \| (A_{n-1} - \lambda)^{-1} X \|^2};$$
see e.g. Lemma 41 of this paper of mine with Van Vu (and we learned of this formula from this previous paper of Erdos, Schlein, and Yau).  Similarly for other components of $u$.  The proof proceeds by expanding out the bottom $n-1$ components of the eigenvalue equation $A_n u = \lambda u$ and using the resulting equation and the formula $\|u\|^2=1$ to solve for $u_1$.  
The expression $\| (A_{n-1} - \lambda)^{-1} X \|$ can be expanded further by a number of formulae (e.g. Cramer's rule); in our applications, it turns out that the spectral theorem applied to $A_{n-1}$ is useful.  The above identity is particularly useful for establishing delocalisation of eigenvectors for random matrices (i.e. that the energy of the coefficients of a unit vector are spread out almost uniformly).  It may seem a bit strange that the top left coefficient $a$ plays no explicit role in this formula, but it is implicitly present due to its influence on the eigenvalue $\lambda$.
The phase $u_1$ of the unit eigenvector does not have a clean formula, because eigenvectors are only determined up to phase rotations unless one somehow selects an artificial normalisation.
A: http://en.wikipedia.org/wiki/Jordan_normal_form 
http://www.wolframalpha.com/input/?i=jordan+normal+form+calculator
Let $J$ be the Jordan normal  form of a matrix $A$ (that is , $A=PJP^{-1}$ ) . Then the  $V_{An}$ eigenvectors of $A$ can be written as $PV_{Jn}$ , where $V_{Jn}$ are the eigenvectors of $J$  . Although calculating $J$ and $P$ can be rather complicated , calculating the eigenvectors of $J$ is trivial .
Also recomend reading this 
http://www.cs.berkeley.edu/~wkahan/MathH110/DownDets.pdf 
At least to me it seems like the best tutorial on linear algebra out there .(edit:I did not intend for this to sound  condescending or anything , I recommend that paper to anyone who does linear algebra)
