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Let $A\in\mathbb{R}^{d\times d}$ and $B\in\mathbb{R}^{d\times d}$ be two positive definite matrices. $k$ is a real coefficient. Suppose the largest eigenvalue of $A-kB$ is $\lambda_1$. Is it possible to find a $k$ such that $k+\lambda_1$ is maximized?

Here is my opinion:

When $k$ is positive and large, $A-kB$ may be negtive definite such that $\lambda_1$ is negtive. When $k$ is negtive, $A-kB$ is positive definite such that $\lambda_1$ is positive. So roughly speaking, large $k$ will give small $\lambda_1$, while small $k$ will give large $\lambda_1$. When can their summation be maximized? Thanks.

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    You could start by considering the special case $B=A$. The summation is maximized at $k=\pm \infty$ (the sign depends on whether the largest eigenvalue of A is greater than 1). Does you 'objective' function has some real motivation, or is just ad-hoc?2011-11-23

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First I assume you want $A$ and $B$ to be symmetric. (You stated that $\lambda_1$ is the largest eigenvalue, which I take to imply that your matrices only have real eigenvalues.)

Let $\lambda_m$ be the smallest eigenvalue of $B$, and $\lambda_M$ be the largest. For $k$ with $|k|$ large, if $k$ is positive, we have

$ \lambda_1 \sim -k \lambda_m \implies k+ \lambda_1 \sim |k|(1-\lambda_m)$

and if $k$ is negative we have

$ \lambda_1 \sim -k \lambda_M \implies k+ \lambda_1 \sim |k|(\lambda_M - 1)$

Which means that unless $(1-\lambda_m) \leq 0$ and $(\lambda_M - 1) \leq 0$, the expression $k+\lambda_1$ will not obtain a global maximum at any finite $k$.

The only case where $(1-\lambda_m)\leq 0$ and $(\lambda_M-1) \leq 0$ can hold is if the eigenvalues of $B$ are all $1$s. Using the assumption that $A$ and $B$ are both symmetric, this requires $B$ being the identity matrix, and so $k + \lambda_1$ is independent of $k$, and equals the largest eigenvalue of $A$.

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    There are only two cases. Case (1) $k + \lambda_1$ is constant in $k$. Case (2) in at least one direction $k\to +\infty$ or $k\to-\infty$, the expression $k + \lambda_1$ is **asymptotic** to something that grows linearly in $|k|$. Such a function can never obtain **any** global maximum. It doesn't matter at all what the exact form of $\lambda_1$ is, the above answer proves that $k+\lambda_1$ is either constant or **cannot have an upper bound**. There are no "other (maybe more general) cases".2011-11-24