Can someone explain this:

Let $\displaystyle f:\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation.

Prove that f has an invariant 1-plane.

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- Nov 29th 2011, 02:34 PMdwsmithinvariant 1-plane
Can someone explain this:

Let $\displaystyle f:\mathbb{R}^3\to\mathbb{R}^3$ be a linear transformation.

Prove that f has an invariant 1-plane. - Nov 30th 2011, 12:30 AMOpalgRe: invariant 1-plane
- Nov 30th 2011, 06:51 AMdwsmithRe: invariant 1-plane
- Nov 30th 2011, 10:49 AMDevenoRe: invariant 1-plane
it might have 3 real solutions. but say it had an invariant 2-dimensional eigenspace. then certainly that eigenspace has a one-dimensional invariant subspace (just pick one of the eigenvectors). or more, simply, if v is an eigenvector of f corresponding to any real eigenvalue (and there must be at least one) λ, then:

f(cv) = c(f(v)) (since f is linear)

= c(λv) = (cλ)v. so if U = {av: a in R}, then f(U) ⊆ U, hence U (which is a line going through the origin in the direction of v) is invariant under f. - Nov 30th 2011, 01:50 PMdwsmithRe: invariant 1-plane
For showing the invariant 2-plane, it can't be done the same way can it? If so, it is just done twice with two separate vectors corresponding to the complex conjugate?