Eigenvectors and invariant subspaces

Question

If I have a 3 eigenvectors within $\mathbb{R}^3$ and I wanted 2 dimensional invariant subspaces of $\mathbb{R}^3$, could I just take the span of any 2 of of the 3 eigenvectors and they would give me those 2 dimensional invairant subspaces? If so, what is the reason behind this?

Answer 1

When you say you have $3$ eigenvectors, presumably that means that you have some linear transformation, and $3$ linearly independent eigenvectors for that transformation.

In that case, you can use those three eigenvectors as a basis for $\Bbb R^3$. In that basis, your transformation will be diagonal (with the corresponding eigenvalues along the diagonal, if you're representing it as a matrix). And of course, the $xy$-plane, the $xz$-plane and the $yz$-plane (the three possible spans-of-two-eigenvectors) are all invariant subspaces of diagonal operators.

Answer 2

Given a linear endomorphism $f\space\colon E\longrightarrow E$, the subspace of eigenvectors of $f$ with eigenvalue $\lambda$, $E_\lambda$, is invariant to the transformation that $f$ encodes.

This is trivial once you realise that all $f$ is doing to vectors on that subspace is either take them to the zero vector (if $\lambda = 0$, the 0 vector is always invariant to any linear transformation) or scale them by a factor $\lambda$ (so a vector on the span of an eigenvector is mapped to another vector on of it).

The sum of invariant subspaces is trivially invariant, which completes this sketch of the proof of the OP's statement.