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Let $M$ be a symmetric $n\times n$ tri-diagonal matrix, with positive values in its main diagonal. and let $\mathbf{1} \in R^n$ be the vector of all 1, such $M \mathbf{1} = 0$

Suppose $M$ has eigenvalues $0=\lambda_1 < \lambda_2 \leq \cdots \leq \lambda_n$, and let $\mathbf{v}_k$ be an eigenvector of $\lambda_{k}$, the corresponding $k$-th eigenvalue of $M$ with multiplicity 1.

Why $(M - \lambda_{k} I)$ has $k - 1$ negative eigenvalues?

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    Actually, Richard Guy uses the double question mark notation, which he says he borrowed from "the Hungarians", though he uses it for a conjectural or hypothetical statement.2012-03-02

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If $\lambda$ is an eigenvalue of $M$, then $\lambda-\lambda_k$ is an eigenvalue of $M-\lambda_kI$.

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    Yes. ${}{}{}{}$2012-03-03