I have a $n$ x $n$ symmetric positive definite matrix $C$, and a symmetric idempotent matrix $M$, given by $I-\frac{1}{n}J_n$ where $J_n$ is a $n$ x $n$ matrix of all ones. Additionally, all the elements along the diagonal of $C$ are ones.

Let's call $\lambda$ the diagonal matrix storing the eigenvalues of $C$, and U the eigenvector matrix of $C$.

I'd like to know if there is a smart way of computing the eigenvalues of the matrix:


Also, is there a proof of the fact that one of the eigenvalues is $0$ (at least this is what I'm expecting to happen)?


1 Answer 1


We assume that $U$ is an orthogonal matrix. Let $K=\lambda^\frac{1}{2}U^TMU\lambda^\frac{1}{2}$.

$U^TMU$ is a symmetric matrix s.t. $spectrum(U^TMU)=spectrum(M)=\{1,\cdots,1,0\}$. Moreover $signature(K)=signature(U^TMU)$. Thus $K$ has $n-1$ positive eigenvalues and one zero eigenvalue. As a consequence, $\det(K)=0$.

Remark. Since $\lambda^{1/2}=U^{-1}C^{1/2}U$, $K=U^{-1}C^{1/2}MC^{1/2}U$ and $spectrum(K)=spectrum(C^{1/2}MC^{1/2})=spectrum(CM)$. Let $\Pi$ be the hyperplane $\sum_i x_i=0$, $P$ be the orthogonal projection on $\Pi$ and let $B=PC_{|\Pi}$. Then $M_{|\Pi}=id_{|\Pi}$ and the non-zero eigenvalues of $K$ are the eigenvalues of $B$.

EDIT. Since $(1,\cdots,1)$ is a basis of $\ker(M)$, $M=id$ on its orthogonal $\Pi$ defined by $\sum_i x_i=0$. There is an orthonormal basis s.t. the matrix of $M$ becomes $diag(I_{n-1},0)$ and the matrix of $C$ becomes $\begin{pmatrix}Q&v\\v^T&s\end{pmatrix}$ where $Q$ is a $n-1\times n-1$ symmetric matrix and $v\in \mathbb{R}^{n-1}$; then the spectrum of $CM=\begin{pmatrix}Q&0\\v^T&0\end{pmatrix}$ is $spectrum(Q)\cup\{0\}$, that is, $spectrum(K)=spectrum(B)\cup\{0\}$.

  • $\begingroup$ Thank you! Could I ask you to clarify the remark? I'm asking you just to expand the notation a little, I'm not familiar with this kind of math stuff. $\endgroup$
    – xanz
    Feb 2, 2016 at 21:12

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