The intuitive explaination of why the eigenvector's number is less than the dimension

I know that the eigenvalue's number is less than the dimension of a matrix, but as the intuition of the eigenvector, each eigenvector keeps the original direction after a linear transform. I think in $\mathbb{R}^n$ there are $n$ vectors which can do this. Why is this not true? Could anyone please give a intuitive explanation of why the eigenvector's number is less than the dimension, or geometric multiplicity is less than algebraic multiplicity?

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A rotation matrix in the plane fixes no vector. So, there is a simple example to supplement your intuition. I'll be interested in what people have to say about the geometry of generalized eigenvectors which I suppose is the true intent of your question. Perhaps begin your post with: "suppose the eigenvalues are all real". – James S. Cook Dec 3 '12 at 14:30
Never heard of the "dimission of a matrix." Googling it turned up nothing. Maybe a translation error? – Thomas Andrews Dec 3 '12 at 14:35
Grammar note: In English, punctuation marks such as , . ? are always followed by a space. I edited to fix it. – Nate Eldredge Dec 3 '12 at 14:58
Well, you certainly couldn't have more eigenvalues than independent directions! – Neal Dec 3 '12 at 15:25

1. The identity $\Bbb R^n\to\Bbb R^n$ fixes every vector, so everyone (except $0$) is an eigenvector with eigenvalue $1$ (there are infinitely many of them), spanning the whole space, that is, dimension $n$.
2. Similarly the reflection about the origo: $x\mapsto -x$ in $\Bbb R^n$: every (nonzero) vector is eigenvector with eigenvalue $-1$.
3. As James S.Cook commented, the rotation in $\Bbb R^2$ doesn't have any (real) eigenvalue. So this case, it is indeed less.
4. Take the 'toppling' funtion in the plane: $(x,y)\mapsto (x+y,y)$. Then you can calculate that it has only $1$ as eigenvalue with a $1$ dimensional eigenspace: the $x$-axis.
And, why is the sum of dimensions of eigenspaces of a transformation $A$ is less or equal than the dimension? It is basically because if $\lambda\ne\mu$, then the eigenspaces $E_\lambda:=\{x\mid Ax=\lambda x\}$ and $E_\mu$ are disjoint: $E_\lambda\cap E_\mu=\{0\}$.