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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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    If $A$ and $B$ are not symmetric, then when $B$ has eigenvalues with multiplicity greater than 1, you can get situations where $A - kB$ has complex eigenvalues for all large $k$, so it becomes more difficult to tell what your question is asking for in that case.2011-11-23
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    Thanks. But how to obtain $\lambda_1\sim -k\lambda_m$ when $k$ is positive? Can we prove that? Because of $A$, the relation between $\lambda_1$ and $k, \lambda_m, \lambda_M$ may not be so simple.2011-11-23
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    here $A$ and $B$ are symmetric and positive definite matrices.2011-11-23
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    Write $\lambda_1 = k \mu_1$ where $\mu_1$ is the "largest" eigenvalue of $\frac{1}{k}A - B$. As $|k|\to\infty$ you can apply standard results on the eigenvalue perturbation for linear operators. A (most likely way too powerful) reference is Kato's _Perturbation theory for linear operators_. At the elementary level, observe that the (possibly complex) roots of a polynomial depends continuously (but can fail to be Lipschitz) on the coefficients of the polynomial. You can apply this to get that the eigenvalues of $(1/k)A - B$ is that of $(-B) + o(1)$.2011-11-23
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    Thanks. But when $|k|\rightarrow \infty$, no need to use perturbation, obviously $A-kB\approx -kB$, then $\lambda(A-kB)\approx\lambda(-kB)$. However, this is only valid when $|k|\rightarrow \infty$. The optimal $k$ should not have $|k|\rightarrow \infty$. I think we may need to determine more accurate relationship between $k$ and $\lambda_1$ instead of considering this special case. Is it helpful to use simultaneous diagonalization?2011-11-24
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    The whole point of my answer is, assuming that $B$ is symmetric and not equal to the identity matrix, you **cannot** have an optimal $k$ that is finite, because the mapping $k\to k+\lambda_1$ is unbounded.2011-11-24
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    I understand. In some cases, we can say $\lambda_1$ is a linear function of $k$, and then there is no optimum. But for other (maybe more general) cases, $\lambda_1$ may be a nonlinear function $k$, then there exist optimal solutions.2011-11-24
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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