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?
 A: Less or equal. And, not exactly the 'number' of the eigenvalues, but the sum of the dimensions of the eigenspaces.
The best is to understand it by simple examples.


*

*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 $n$.

*Similarly the reflection about the origo: $x\mapsto -x$ in $\Bbb R^n$: every (nonzero) vector is eigenvector with eigenvalue $-1$.

*As James S.Cook commented, the rotation in $\Bbb R^2$ doesn't have any (real) eigenvalue. So this case, it is indeed less.

*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: $E_\lambda\cap E_\mu=\{0\}$.
