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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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    one dimensional s$p$aces may be more than points2013-04-18

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"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.

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    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.2012-11-14