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I want to show the following inequality regarding schur's majorization theorem $$\overset{n}{\underset{i = 1}{\sum}}a_i\lambda_i\leq \overset{k}{\underset{i = 1}{\sum}}\lambda_i,$$ where $\lambda_1\geq \lambda_2 \geq \dots \geq \lambda_n =0,~ 0\leq a_i\leq 1$ and $\overset{n}{\underset{i = 1}{\sum}}a_i =k$.

Any help will be appreciated.

Extra information: here $a_i (i =1\dots n) $ are diagonal entries of some real symmetric matrix whose eigenvalues are $1$ (with multiplicity $k$) and $0$ (with multiplicity $n-k$).

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1 Answer 1

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$$ \sum_{i=1}^na_i\lambda_i-\sum_{i=1}^k\lambda_i=\sum_{i=1}^k(a_i-1)\lambda_i+\sum_{k+1}^na_i\lambda_i\le\sum_{i=1}^k(a_i-1)\lambda_k+\sum_{k+1}^na_i\lambda_k=\lambda_k\left(\sum_{i=1}^na_i-k\right)=0\;. $$

Intuitively speaking, the sum of the weights is the same in both sums, and in the left-hand sum the weights are shifted more towards the smaller eigenvalues than in the right-hand sum.

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  • $\begingroup$ Thanks @joriki, it completely make sense to me. I also got a reasoning; since each $a_i \leq 1$ and their sum is $k$, so we can take maximum upper bound $1$ for any $k$ $a_i$'s. $\endgroup$
    – user602672
    Commented Dec 13, 2019 at 8:09

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