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All matrices I consider are complex hermitian $N\times N$ matrices. None of them are Diagonal matrices. Consider a positive semi definite matrix $A$ and a matrix $B$. I am interested in the sum $C=A-\gamma B$ which $\gamma$ is a non-negative scalar.


What is the smallest gamma such that $C$ is negative semi-definite when $B$ is 1) positive definite 2) positive semi-definite?

If $\gamma=0$, $C$ is positive semidefinite, as $\gamma$ increases, the "positive semi-definiteness" of $C$ decreases. At some point, it becomes, indefinite, and after a while, it will become a negative semidefinite matrix. I want to know for what gamma this happens.

Solution of Case 1

It is quite straightforward to prove that the solution of case 1 is $\gamma=\lambda_{max}(B^{-1}A)$ which follows from expressing $\gamma$ as the solution to a generalized eigenvalue problem of $A$ and $B$ and using the fact $B$ is invertible.

(please see this MSE question for a distantly related problem, but from which solution can't be deduced for this one).

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1 Answer 1

up vote 1 down vote accepted

For diagonal matrices, the above question can't be true!!. (thanks to the author of this answer, this edit was added by the questioneer).

Let $A=\pmatrix{1 & 0 \\ 0 & 0 }$, $B=\pmatrix{0 & 0 \\ 0 & 1 }$. Then $C=\pmatrix{1 & 0 \\ 0 & -\gamma }$, which can never be negative semi-definite.

Addendum (after question was modified):

Let $A=\pmatrix{1 & 1 \\ 1 & 1 }$, $B=\pmatrix{1 & -1 \\ -1 & 1 }$. Then $C=\pmatrix{1-\gamma & 1+\gamma \\ 1+\gamma & 1-\gamma }$. Then we note that $C \binom{ 1 }{ 1 } = 2 \binom{ 1 }{ 1 }$, and hence $C$ always has an eigenvalue at 2 and can never be negative semi-definite.

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oops! I missed that, edited the question, for non-diagonal matrices. Thanks a lot!! –  dineshdileep Nov 4 '12 at 4:20
A slight modification yields non-diagonal matrices. –  copper.hat Nov 4 '12 at 8:00

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