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 $\Lambda$?

  • $\begingroup$ How about if $\Lambda=I$ (with a zero row and column concatenated at the end) and $y=e_1$? One eigenvalue will vanish in $M$. $\endgroup$ – broncoAbierto Jun 12 at 18:49
  • $\begingroup$ $\Lambda$ wouldn't be positive definite. $\endgroup$ – Ivan Jun 12 at 18:50
  • $\begingroup$ Oh, sorry, I thought zero was an eigenvalue. $\endgroup$ – broncoAbierto Jun 12 at 18:52

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.

  • $\begingroup$ This doesn't answer the question about the lowest non-zero eigenvalue of $M$, $\mu_2$, and $\lambda_1$. $\endgroup$ – Ivan Jun 12 at 21:47

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