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 $y$.
The question: what can be said about the other eigenvalues? For example, 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 $\Lambda$. Is the same true for the smallest eigenvalue, i.e. the smallest non-zero eigenvalue of $M$ is at least as large as the smallest eigenvalue of $\Lambda$?
Best Answer
In short the answer is yes. Actually you can prove that there exists an ordering of the eigenvalues of the two matrices. The proof is easy; just use the min-max theorem, which states that for non-zero vectors $x$:
$$\lambda_k=\min_{U}\max_{x\in U, \dim U=k}\frac{x^t\Lambda x}{{||x||}^2}$$
but since we know that:
$$\frac{x^t\Lambda x}{{||x||}^2}\geq \frac{x^tM x}{{||x||}^2}$$
The result that the k-th larger eigenvalue of $\Lambda$ exceeds that of $M$ readily follows:
$$\lambda_k\geq \mu_k$$
where $\Lambda x_k=\lambda_kx_k$ , $M y_k=\mu_ky_k$, and the eigenvalues are ordered.