invariant 1-plane

Can someone explain this:

Let be a linear transformation.

Prove that f has an invariant 1-plane.

Quote:

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Originally Posted by dwsmith

Can someone explain this:

Let be a linear transformation.

Prove that f has an invariant 1-plane.

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This is because a cubic equation with real coefficients always has at least one real root. It follows that f has at least one real eigenvalue and hence an eigenvector. The 1-dimensional subspace spanned by that eigenvector will be invariant under f.

Quote:

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Originally Posted by Opalg

This is because a cubic equation with real coefficients always has at least one real root. It follows that f has at least one real eigenvalue and hence an eigenvector. The 1-dimensional subspace spanned by that eigenvector will be invariant under f.

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How do I go about proving that now. I understand what meant now at least.

I am guess this will have to do with the cubic characteristic polynomial. However, how can the cubic be guaranteed to not have 3 real solutions and hence have one invariant 2-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.

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?