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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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What do you expect? Did you try an example? –  draks ... Apr 26 '12 at 15:16
    
@draks My second assumption was inappropriate, since $B=C$ then. Also, the assumption on positive semi-definiteness is dropped. Does that change your perception? –  user506901 Apr 26 '12 at 20:22
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2 Answers 2

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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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Thanks. I'm only not sure about the statement "the non-zero points of the spectrum of AB coincide with those of the spectrum of BA" from the reference you gave. So, regardless of the zero eigenvalues existing or not, the spectrum of $B$ and $C$ coincide? –  user506901 Apr 27 '12 at 9:29
    
A better reference is en.wikipedia.org/wiki/Characteristic_polynomial –  Robert Israel Apr 27 '12 at 16:13
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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.

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