# Minimum and maximum eigenvalue

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 $$\mu_2\ge\lambda_1$$?

• It might be worthwhile to mention that the answer is positive if $y$ is an eigenvector of $\Lambda$, say for the eigenvalue $\lambda_k$, since we then the eigenvalues of $M$ are $0, \lambda_1, \dots, \lambda_{k-1}, \lambda_{k+1}, \dots, \lambda_n$. Hence, $\mu_2 = \lambda_2$ if $k=1$ and $\mu_2 = \lambda_1$ else. – Jochen Glueck Jun 13 at 6:01
• It is not true that all eigenvectors of a symmetric matrix are orthogonal. See, e.g., the identity matrix. – Gerry Myerson Jun 13 at 6:10
• You do have interlacing eigenvalues under rank one perturbations, but on the cyclic subspace generated by $Ay$ (and obviously unchanged spectrum on the orthogonal complement), so your conjecture as stated isn't true. See my answer here for some background: mathoverflow.net/questions/193527/… – Christian Remling Jun 13 at 16:26
• @GerryMyerson: However, the conclusion that $\mu_j\le\lambda_j$ is of course correct, being an immediate consequence of min-max. – Christian Remling Jun 13 at 16:28
• @Ivan: This is only those ev's of the part of the operator in the reducing subspace spanned by $Ay$ (I make the assumption that $v$ is cyclic in the linked answer). Those in the orthogonal complement won't change. – Christian Remling Jun 13 at 19:19