3 ms·
One does not need determinant to define eigenvalues. For example: If T is a linear operator on vector space V, a scalar a is an eigenvalue if there is a v in
by richrichie 2y ago
One does not need determinant to define eigenvalues. For example:
If T is a linear operator on vector space V, a scalar a is an eigenvalue if there is a v in V s.t. Tv = av.
This is the approach the book takes.
- 3PS 2y agoI agree, but the definition alone isn't sufficient to actually calculate eigenvalues. Hence the standard approach which says that for matrix A, vector v, and eigenvalue λ, we have Av = λv => Av - λv = 0 => (A - λI)v = 0 => det(A - λI) = 0 Which then yields the characteristic polynomial. Skipping the determinant means you need a different approach.
- richrichie 2y agoYou can prove many fundamental results of Linear Algebra with the definition that does not directly use determinants. In fact, one would define trace and determinant as sum and product of eigenvalues. Definition of characteristic polynomial would then follow. If "computation" is what you are after then Av = λv is about solving a system of equations and you can try elimination, etc.