Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 1 down vote accepted

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.

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.