## 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?

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Maybe your question will ring a bell for someone else, but this strikes me as being a bit vague. What kind of estimates/formulas do you need? – Deane Yang May 7 2012 at 8:14
If you know an eigenvalue $\lambda$, then the associated eigenvector is the kernel of $A-\lambda I$ and you can use Cramer-like formulas to express this kernel (tell me if you are interested in more detail, but it's not so hard). If you don't know the eigenvalue, then I don't think you can do much. Determinants are rational functions and eigenvalues/eigenvectors are not, so I am afraid everything that you can do will essentially be computing the characteristic polynomial and finding its roots. – Federico Poloni May 7 2012 at 10:19
Yes, I am interested..:) – Felix Goldberg May 7 2012 at 21:30

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 $n-1 \times n-1$ Heritian 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.

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Thanks, this looks very interesting and I'll be sure to delve into it! – Felix Goldberg May 7 2012 at 21:31

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 .