Can the sum of the dimension of eigenspaces of a linear operator $A : \mathbb V\to \mathbb V$ exceed dim $V$?

Question

I don't think so... a proof would be nice. I don't know anything about Jordan forms though, so please keep that in mind.

Answer 1

No, the eigenvectors are vectors from your vector space.

You can not have more linearly independent vectors in your space than the dimension of the space.

Answer 2

Never. The dimension of an eigenspace is at most the algebraic multiplicity of the corresponding eigenvalue, and the sum of the algebraic multiplicities of the eigenvalues is the degree of the characteristic polynomial of the linear operator, i.e. the dimension of $V$.

Answer 3

Impossible. At least in finite dimensional case, dimension of eigenspace is less or equal of the exponent of its polynomial factorization. And by fundamental theorem of algebra, it is impossible

Answer 4

This is a very detailed proof:

You need to know the following facts:

1. If $ V$ is a vector space with $ dim\;V=n$ and $ M \subseteq V$ is linearly independent, then $M$ is finite and $|M|\leq n $.
2. If $ W_1, W_2, \cdots, W_k$ are vector spaces such that their sum is a direct sum (i.e $W_i \cap \left(W_1 + W_2 + \cdots + W_{i-1} \right) =\{ \vec{0}\}$ for every $1\leq i \leq k $) then $ dim \left( W_1+ W_2+ \cdots+ W_k\right)=dim(W_1)+\cdots+dim(W_k)$
3. The sum of the eigen spaces is always a direct sum.

Now, since $W:= W_1+\cdots+W_k $ is a sub vector space of $ V$ (here the$ W_i$ are the eigen spaces), by the first fact a basis of $ W$ must have less than $ dim\;V$ vectors, and since a basis of $ W$ has size $ dim(W_1)+\cdots+dim(W_k)$ (by the facts 2 and 3) your question follows