Take the 2-minute tour ×
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.

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.

share|improve this question
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

1 Answer 1

up vote 3 down vote accepted

"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.

share|improve this answer
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

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.