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Suppose a symmetric matrix $A\in\mathbb{R}^{n\times n}$ is given. Let $J=I-\frac{1}{n}\cdot 1_n1_n^T\in\mathbb{R}^{n\times n}$ be the centering matrix, with $I$ being the identity matrix, and $1_n=[1 \dots 1]^T\in\mathbb{R}^n$.

Now, suppose $B=JAJ$, and $C=AJ$. How do the eigenvalues of $B$ and $C$ compare? I'm interested in the zero and non-zero spectrum behaviour.

With a simple $3\times 3$ example I obtain identical eigenvalues, but different eigenvectors. How could this be formalized mathematically?

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    @draks My second assumption was inappropriate, since $B=C$ then. Also, the assumption on positive semi-definiteness is dropped. Does that change your perception?2012-04-26

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Since $J^2 = J$, $B = J(AJ)$ and $C = (AJ)J$ have the same eigenvalues (see e.g. Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!)).

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    A better reference is http://en.wikipedia.org/wiki/Characteristic_polynomial2012-04-27
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J is just the projector onto the subspace orthogonal to $1_n$. So, consider what A, B and C look like in an (orthonormal) basis in which $1_n$ is the first basis vector. In block form:

$ A = \left[ \begin{array}{cc} a & \vec{b}^T \\ \vec{b} & M \end{array} \right], \quad C = \left[ \begin{array}{cc} 0 & \vec{b}^T \\ 0 & M \end{array} \right], \quad B = \left[ \begin{array}{cc} 0 & 0 \\ 0 & M \end{array} \right] $

As long as M doesn't have any zero eigenvalues then B and C have identical spectra. One sees this as follows. Because A is symmetric, M is symmetric. Let $\vec{v}_1,\ldots,\vec{v}_{n-1}$ be the eigenvectors of M, with corresponding eigenvalues $\lambda_1, \ldots, \lambda_{n-1}$. Then C has the following n-1 right-eigenvectors (in block form):

$ \left[ \begin{array}{c} \lambda_j^{-1} \vec{b} \cdot \vec{v}_j \\ \vec{v}_j \end{array} \right] \quad j=1,\ldots,n-1 $

with corresponding eigenvalues $\lambda_1, \ldots, \lambda_{n-1}$. The final right-eigenvector of C is

$ \left[ \begin{array}{c} 1 \\ \vec{0} \end{array} \right] $ with eigenvalue 0. (Here, $\vec{0}$ indicates the dimension n-1 zero vector.) Similarly, B has the right-eigenvectors

$ \left[ \begin{array}{cc} 0 \\ \vec{v}_j \end{array} \right] \quad j=1,\ldots,n-1 $ with eigenvalues $\lambda_1,\ldots,\lambda_{n-1}$ and

$ \left[ \begin{array}{c} 1 \\ \vec{0} \end{array} \right] $ with eigenvalue 0. I'm not sure what happens when M has one or more zero eigenvalues.