If the eigenvalues are distinct then the eigenspaces are all one dimensional

Theorem: Diagonalizable matrices share the same eigenvector matrix $S$ iff $AB=BA$

Proof: $(\Leftarrow)$ Suppose $AB=BA$.

$$ABx=BAx=B\lambda x=\lambda Bx$$ Thus $x$ and $Bx$ are both eigenvectors of $A$, sharing the same $\lambda$. Assume that the eigenvalues of $A$ are distinct (it means the eigenspaces are all one dimensional) then $Bx$ must be a multiple of $x$. In other words $x$ is an eigenvector of $B$ as well as $A$.

Above, I can't follow the meaning of "if the eigenvalues are distinct then the eigenspaces are all one dimensional".

For example, let $$A=\begin{pmatrix}4 & -5 \\ 2 & -3\end{pmatrix}$$ It has two distinct eigenvalues, $-1$ and $2$. But its eignevectors are $x_1=\begin{pmatrix}1 &1\end{pmatrix}^\mathrm{T}$ and $x_2=\begin{pmatrix}5 &2 \end{pmatrix}^\mathrm{T}$, they are not one dimensional.

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In your example the eigenspace for - 1 is spanned by $(1,1)$. This means that it has a basis with only one vector. It has nothing to do with the number of components of your vectors. – kalvotom Nov 11 '12 at 6:43
vectors have no dimension, vector spaces have – Marc van Leeuwen Nov 11 '12 at 7:20
one dimensional spaces may be more than points – user73164 Apr 18 '13 at 4:01

"one dimensional" refers to the dimension of the space of eigenvectors for a particular eigenvalue. All the eigenvectors corresponding to the eigenvalue -1 are multiples of $x_1$. In other words, they are spanned by one vector, so the space of eigenvectors has dimension one.
Do not confuse this with the fact that $x_1 = (1,1)$ has 2 coordinates. That just means that $x_1$ lives in a 2 dimensional space $\mathbb{R}^2$ (or whatever your field is). The space of eigenvectors is a subspace of that 2 dimensional space, and that subspace is 1 dimensional.
So that means the eigenspace consists of only one vector-$x_1$ or $x_2$, right? Thanks! – pulier Nov 11 '12 at 6:56
No. The eigenspace consists of the set of vectors spanned by $x_1$ (or $x_2$), i.e., the multiples of $x_1$ (or $x_2$). Remember that the eigenspace for a particular eigenvalue is always a subspace. – Ted Nov 11 '12 at 6:58
Set of vectors? Then, for example, 2$x_1$+3$x_1$ can span the eigenspace of $\lambda_1$? – pulier Nov 11 '12 at 7:07
I don't quite understand what you're asking. The span of $x_1$ is the set of all multiples of $x_1$. If an eigenspace is one dimensional, it will be spanned by any of the nonzero vectors in the eigenspace. – Ted Nov 14 '12 at 8:34