6
$\begingroup$

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 $\mu_2\ge\lambda_1$?

Improve this question
$\endgroup$
6
  • $\begingroup$ 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. $\endgroup$
    – Jochen Glueck
    Commented Jun 13, 2019 at 6:01
  • 3
    $\begingroup$ It is not true that all eigenvectors of a symmetric matrix are orthogonal. See, e.g., the identity matrix. $\endgroup$
    – Gerry Myerson
    Commented Jun 13, 2019 at 6:10
  • 1
    $\begingroup$ 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/… $\endgroup$
    – Christian Remling
    Commented Jun 13, 2019 at 16:26
  • 1
    $\begingroup$ @GerryMyerson: However, the conclusion that $\mu_j\le\lambda_j$ is of course correct, being an immediate consequence of min-max. $\endgroup$
    – Christian Remling
    Commented Jun 13, 2019 at 16:28
  • 1
    $\begingroup$ @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. $\endgroup$
    – Christian Remling
    Commented Jun 13, 2019 at 19:19

1 Answer 1

Reset to default
1
$\begingroup$

I think, yes. The quadratic form $Q(z) :=z^tMz$ agree with that of $\Lambda$ on the hyperplane $\{y^t\Lambda z=0\}$. Therefore you have a space of codimension 1 on which $Q(z)\geqslant \lambda_1 \|z\|^2$, that yields $\mu_2\geqslant \lambda_1$.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .