# eigenvalue sign of $M - \lambda_{k} I$

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?

• En inglés, sólo se usa el signo de interrogación derecho. – Arturo Magidin Mar 2 '12 at 5:08
• 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. – Gerry Myerson Mar 2 '12 at 5:13

If $\lambda$ is an eigenvalue of $M$, then $\lambda-\lambda_k$ is an eigenvalue of $M-\lambda_kI$.
• is "If $\lambda$ is an eigenvalue of $M$, then $\lambda-lambda_k$ is an eigenvalue of $M-\lambda_k I$" Because of this? \begin{align} M \mathbf{x}_j &= \lambda_j \mathbf{x}_j \\ M \mathbf{x}_j - \lambda_k I \mathbf{x}_j &= \lambda_j \mathbf{x}_j - \lambda_k I \mathbf{x}_j \\ M \mathbf{x}_j - \lambda_k I \mathbf{x}_j &= \lambda_j \mathbf{x}_j - \lambda_k \mathbf{x}_j \\ (M - \lambda_k I) \mathbf{x}_j = (\lambda_j - \lambda_k) \mathbf{x}_j \end{align} – Pablo Hector Suazo Mar 2 '12 at 14:55
• Yes. ${}{}{}{}$ – Gerry Myerson Mar 3 '12 at 4:18