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I am reading the paper "Continuous methods for extreme and interior eigenvalue problems" by G.H. Golub and L.-Z. Liao. The papers says for the following problem, (Lemma 2.1 (i), $0>\lambda_i-c\ge \lambda_1-c$, $i=1,...,n$ and $c$, $\lambda_i$s are constants)

$$\begin{equation} \begin{aligned} & \min\limits_{\alpha}\sum^n_i\alpha_i^2(\lambda_i-c) \\ s.t.~~~& \sum_{i=1}^n\alpha_i^2\le 1 \\ \end{aligned} \end{equation}$$, any local minimiser is also a global minimiser. But I cannot figure out why. Would someone please help?

user112758
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Negate your coefficients, so you have a diagonal matrix $H$ with entry $h_{ii} = c - \lambda_i > 0.$ You have the quadratic form $$X^T H X,$$ where $X$ is the column vector of the $\alpha_i.$

The gradient of this, as a column vector, is $$ 2 H X. $$ This really is the zero vector at the origin, but that value is $0,$ not a maximum.

There is an orthonormal basis of eigenvectors of $H,$ in fact they are the coordinate axes. The gradient of the quadratic form defining the sphere is $2X.$ So, in Lagrange multipliers, we are asking when $HX$ is parallel to $X,$ which are the eigenvectors. The largest value of $X^THX$ occurs at the largest eigenvalue. However, they permit there to be multiplicity larger than one. That is, the maximum could occur at two diametrically opposed points. However, with multiplicity two, the max could occur along a circle. With multiplicity three, a two sphere given by intersecting the original sphere with a 3-space.

Will Jagy
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