Leading eigenvalues

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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?

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I asked my question in a very general way. In fact, what I have is:   My operator is the transfer operator P on L1 functions defined on compact X. It is the pre-dual of the operator U:L∞ \rightlarrow L∞ defined by U(ϕ)=ϕ \circ f for a fixed map f on X. I have PU=Id and UP is the projection.   Now my specific question is, if P1(h)=h and P2(g)=g for g,h∈L1 and if 1 is the leading simple isolated eigenvalue for both P1 and P2, then does (P1+P2)/2 have 1 as a leading eigenvalue and what about the corresponding eigenfunction?  – filiz Oct 27 2010 at 20:48
ask this as a separate question? – ohai Oct 27 2010 at 20:56

3 Answers

You need to add some assumption, otherwise $\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ add to a matrix with eigenvalues $1 \pm \sqrt{n}$. Maybe your operators are self-adjoint?

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If $A$ is self adjoint, and $\lambda$ its leading eigenvalue, then `$\lambda = \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle$`. If $A$ and $B$ are self-adjoint, we have the obvious consequence `$$\mathrm{sup}_{\langle u,u \rangle =1} \langle u, (A+B) u \rangle = \mathrm{sup}_{\langle u,u \rangle =1} \left( \langle u, A u \rangle +  \langle u, B u \rangle \right) \leq \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Au \rangle + \mathrm{sup}_{\langle u,u \rangle =1} \langle u, Bu \rangle$$` so the leading eigenvalue of the sum is bounded by the sum of the leading eigenvalues.

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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.

Added: I guess this assumes the operators are self adjoint.

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"size of the leading eigenvalue is like a norm" is only true if you make additional assumptions. See Homology's example.  – Willie Wong Oct 27 2010 at 15:55
@ohai. The modulus of the leading eigenvalue is called spectral radius. It is not a norm at all, since it vanishes on non-zero nilpotent matrices. It does not satisfy a triangle inequality either. Take the sum of the matrices $$\begin{pmatrix} 0 & 1 \\\\ 0 & 0 \end{pmatrix},\qquad\begin{pmatrix} 0 & 0 \\\\ 1 & 0 \end{pmatrix}.$$ – Denis Serre Oct 27 2010 at 16:23
However, the spectral radius equals the operator norm when the matrix is Hermitian. – Denis Serre Oct 27 2010 at 16:24
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