2
$\begingroup$

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 $\Lambda$?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.