Minimum and maximum eigenvalue

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 $$\Lambda$$?

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

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