Inverse eigenvalues theorem: Difference between revisions

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In [[numerical analysis]] and [[linear algebra]], the '''Inverse eigenvalues theorem''' states that, given a matrix A that is [[nonsingular]], with [[eigenvalue]] <math>\lambda>0</math>, <math>\lambda</math> is an eigenvalue of <math>A</math> if and only if <math>\lambda^{-1}</math> is an eigenvalue of <math>A^{-1}</math>.

==Proof of the Inverse Eigenvalues Theorem==

{{math-stub}}

Suppose that <math>\lambda</math> is an [[eigenvalue]] of A. Then there exists a non-zero vector <math>x \in R^n</math> such that <math>Ax = \lambda x</math>. Therefore:

<math>x =I_{n}x = a^{-1}Ax = A^{-1} \lambda x = \lambda A^{-1} x</math>

Since A is [[non-singular]], null(A) = {0} and so <math>\lambda \neq 0</math>. Therefore we may multiply both sides of the above equation by <math>\lambda^{-1}</math> to get that <math>A^{-1}x = \lambda^{-1} x</math>; i.e., <math>\lambda^{-1}</math> is an eigenvalue of <math>A^{-1}</math>. By repeating the previous argument but with A replaced by <math>A^{-1}</math> we see that if <math>\lambda^{-1}</math> is an eigenvalue of <math>A^{-1}</math> then \lambda is an eigenvalue of A.

[[Category:Linear algebra]]