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In a book on numerical linear algebra (Deuflhard numerische Mathematik Band 1) there is the following exercise (translated from German):

Give a Formula (using Determinants) for an eigenvector $x \in \mathbb{C}^n$ corresponding to a given simple eigenvalue $\lambda \in \mathbb{C}$ of a Matrix $A \in \mathbb{C}^{n \times n}$.

It really should be a formula and not an algorithm, since the formula is to be used to show that the eigenvector depends on the matrix in a continuous differentiable fashion.
My problem is that any formula I think of yields $0$ instead of a proper eigenvector

Any ideas on how to get such a formula?

Carlos Esparza
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  • When you say that "an eigenvector depends on the Matrix in a continuous differentiable fashion", it is meaningless as such because an eigenvector is defined up to a factor. The less you can do is to normalize your eigenvectors (normed to unity). – Jean Marie Jan 27 '18 at 22:27
  • @JeanMarie what I mean is that there exists a continuously differentiable function which maps every $A \in U \subset \mathbb{C}^{n \times n}$ to one of its eigenvectors (the fact that the eigenvalues "are continuously differentiable" was established before in the book). Obviously this function won't be unique but I don't see any other problem. – Carlos Esparza Jan 27 '18 at 22:41
  • I understand a little more. See similar question with answer (https://math.stackexchange.com/q/807144) – Jean Marie Jan 27 '18 at 22:55
  • see as well the second answer in (https://math.stackexchange.com/q/581057) – Jean Marie Jan 28 '18 at 07:24
  • ... and (https://mathoverflow.net/questions/253584/differentiability-of-eigenvalue-and-eigenvector-on-the-non-simple-case) – Jean Marie Jan 28 '18 at 07:27

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