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grei
Posts:
104
Registered:
11/27/12


Re: Eigenvector problem
Posted:
Nov 12, 2013 10:44 AM


I don't understand your question. To have "eigenvalues" and "eigenvectors" you have to have a linear operator on a vector space. What vector space and operator are you talking about? Is this equation the characteristic equation for a linear operator on a three dimensional vector space?
In any case, x^3+ 8x^2 16x= x(x^2 8x+ 16)= x(x 4)(x 4)= 0 when x= 0 or x= 4. There is no "third root, which could represent any real number", the three roots are 0, 4, and 4 with "4" being a "double root".
"Could this eigenvalue or root represent an infinite number of eigenvectors?" ??? EVERY eigenvalue has a corresponding "eigenspace". That is, the set of eigenvectors corresponding to a given eigenvalue form a subspace, necessarily containing an infinite number of eigevectors.
Depending on the precise linear transformation (we cannot tell from the characteristic equation) the eigenspace corresponding to eigenvalue 4 may have dimension 1 or dimension 2. But in either case there are an infinite number of eigenvectors corresponding to eigenvalue 2, just as there are an infinite number of eigenvectors corresponding to eigenvalue 0. (Its eigenspace has dimension 1.)


Date

Subject

Author

11/9/13


Kyle

11/12/13


grei


