Leading eigenvalues - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T17:14:19Z http://mathoverflow.net/feeds/question/43815 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43815/leading-eigenvalues Leading eigenvalues filiz 2010-10-27T15:42:20Z 2010-10-27T16:12:05Z <p>If I know about the leading eigenvalues and the eigenfunctions of two operators, is there any result about the leading eigenvalue of the sum of the two operators?</p> http://mathoverflow.net/questions/43815/leading-eigenvalues/43817#43817 Answer by ohai for Leading eigenvalues ohai 2010-10-27T15:49:57Z 2010-10-27T16:00:31Z <p>I would guess that the magnitude of the leading eigenvalue of the sum is at most the sum of the magnitudes of the leading eigenvalues of the two operators, because the size of the leading eigenvalue is like a norm and the norms have the triangle inequality.</p> <p>Added: I guess this assumes the operators are self adjoint.</p> http://mathoverflow.net/questions/43815/leading-eigenvalues/43818#43818 Answer by Homology for Leading eigenvalues Homology 2010-10-27T15:53:39Z 2010-10-27T15:53:39Z <p>You need to add some assumption, otherwise $\begin{pmatrix} 1 &amp; n \\ 0 &amp; 1 \end{pmatrix}$ and $\begin{pmatrix} 0 &amp; 0 \\ 1 &amp; 0 \end{pmatrix}$ add to a matrix with eigenvalues $1 \pm \sqrt{n}$. Maybe your operators are self-adjoint?</p> http://mathoverflow.net/questions/43815/leading-eigenvalues/43823#43823 Answer by David Speyer for Leading eigenvalues David Speyer 2010-10-27T16:12:05Z 2010-10-27T16:12:05Z <p>If $A$ is self adjoint, and $\lambda$ its leading eigenvalue, then <code>$\lambda = \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle$</code>. If $A$ and $B$ are self-adjoint, we have the obvious consequence <code>$$\mathrm{sup}_{\langle u,u \rangle =1} \langle u, (A+B) u \rangle = \mathrm{sup}_{\langle u,u \rangle =1} \left( \langle u, A u \rangle + \langle u, B u \rangle \right) \leq \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle + \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Bu \rangle$$</code> so the leading eigenvalue of the sum is bounded by the sum of the leading eigenvalues.</p>