I don't know if this is the right place to post this question, but I find it interesting and have not gotten an answer elsewhere. If it violates any rules, I will gladly delete it.

Let $\Lambda$ be a real, positive definite, symmetric $n\times n$ matrix with ordered eigenvalues $0<\lambda_1\le\dots\le\lambda_n$. For any unit vector $y$, we can construct another matrix in the following fashion: $$M = \Lambda - \frac{(\Lambda y)(\Lambda y)^t}{y^t\Lambda y}$$ $M$ is symmetric and positive semi-definite with a zero eigenvector $y$. Let its eigenvalues be labeled $0=\mu_1\le\dots\le\mu_n$.

Now, since $M$ is symmetric, all other eigenvectors will be perpendicular to $y$. Take any such $x$, then $$x^tMx = x^t\Lambda x - \frac{(y^t\Lambda x)^2}{y^t\Lambda y}\le \lambda_n x^tx$$ and we conclude that the other eigenvalues cannot exceed the largest one of $\Lambda$, i.e. $\mu_n\le\lambda_n$. My question: is it also true that $\mu_2\ge\lambda_1$?