Assume symmetricmatrix $B\in\mathbb{R}^{n\times n}$ is given, and a transformation $$A=JBJ,$$ where $J=I - \frac{1}{n}1_n1_n^T$ and $I$ denoting the identity matrix, hence centering its rows and columns to zero (sum of row entries equals zero, for all rows; same holds for columns). It is known that at there is one (at least) eigenvalue zero ($0$); I wonder if it can be shown that the corresponding eigenvector is $1_n=[1~\dots~1]^T\in\mathbb{R}^n$ (ie. vector of all $1$'s)?

I'm trying to establish relation with the geometric implication in case $1_n$ is an eigenvector, similarly as is done with arrows in example from http://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors

The thing that bothers me is that in case of identical eigenvalues related to different eigenvectors. An explanation on this would be appreciated. It has to do with the method I use for spectral decomposition, and it actually inspired the question: sometimes, on different invocation of the method for dominant eigenvalue extraction I get different eigenvectors, but the eigenvalue is always the same (what does that tell about the matrix?). Since there might be more zero eigenvalues, I wonder if it's possible to somehow restrict the method to get the eigenvectors different than the one with all equal elements.

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    $\begingroup$ If every row of $A$ adds up to $k$, then $A(1,1,\ldots,1)^T=(k,k,\ldots,k)$, so $(1,1,\ldots,1)^T$ is an eigenvector of $A$ with eigenvalue $k$. $\endgroup$ – Arturo Magidin Mar 15 '12 at 16:53

Certainly. Since $1_n^\top1_n=n$, we have $J1_n=(I-\frac1n1_n1_n^\top)1_n=I1_n-\frac1n1_n1_n^\top1_n=1_n-\frac nn1_n=0$.

  • $\begingroup$ Could you consider the update I made to the question? $\endgroup$ – user506901 Mar 16 '12 at 8:01
  • $\begingroup$ @user506901: I don't think I understand the question. For eigenvalue $0$, the arrow is just squashed to the point $0$, but I suspect that's not what you meant. Also I don't know how "identical eigenvalues related to different eigenvectors" relates to the rest of the question. $\endgroup$ – joriki Mar 16 '12 at 8:05
  • $\begingroup$ You're right about the relation; still it's what inspired the question. See the edit I made. $\endgroup$ – user506901 Mar 16 '12 at 8:18

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