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Can one find some square matrix $A$ and a square positive semidefinite matrix $B$, such that the largest eigenvalue of $C=A+B$ is smaller than the largest eigenvalue of $A$?

danielrch
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  • even better: can one find such $A$, $B$ so that all the eigenvalues of $C$ are smaller than the corresponding eigenvalues of $A$ when put in the order from the smallest eigenvalue to the largest eigenvalue? – danielrch Aug 24 '14 at 16:59
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    No, because $tr(C) = tr(A) + tr(B) \geq tr(A)$ and the trace is the sum of the eigenvalues. (In response to your comment) – Mike F Aug 25 '14 at 04:44

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Yes, here is a random example: $$ A = \pmatrix{-2&1\\ -1&1}, \ B = \pmatrix{ 1&0\\ 0&0}, \ C = A+B = \pmatrix{-1&1\\ -1&1}. $$ As $\det(A)$ is negative, $A$ has a positive eigenvalue and a negative eigenvalue. However, the characteristic polynomial of $C$ is $x^2-\operatorname{tr}(C)x+\det(C)=x^2$. Hence both eigenvalues of $C$ are zero (and $C$ is similar to a Jordan block of size 2). So, the largest eigenvalue of $C$ is strictly smaller than the largest eigenvalue of $A$, regardless of whether "largest" means "maximum" (i.e. most positive) or "largest sized".

user1551
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