Let $A$ be an $n\times n$ symmetric matrix and $S$ be an $n\times n$ positive definite matrix. Suppose the eigenvalues of $A$ and $SAS$ are arranged in non decreasing order; that is $$ \lambda_1 (A) \leq \cdots \leq \lambda_n (A)\; \mbox{and}\; \lambda_1 (SAS) \leq \cdots \leq \lambda_n (SAS). $$ Assume that $\lambda_1 (A) + \cdots + \lambda_k (A) > 0$. Does it hold or not that $$ \lambda_{i_1} (SAS) + \cdots + \lambda_{i_k} (SAS) = \theta_{i_1\cdots i_k} (\lambda_{i_1} (A) + \cdots + \lambda_{i_k} (A)) $$ for some positive $\theta_{i_1\cdots i_k} $?
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3$\begingroup$ Why would you suspect this to be true? The $n$-sum would be a trace, and that of $SAS$ is basically a weighted sum. So take $A = \mathrm{diag}(-1,2)$ and $S = \mathrm{diag}(10,1)$ then $SAS = \mathrm{diag}(-100,2)$. You have that $A$ has positive trace but $SAS$ has negative trace. $\endgroup$Willie Wong– Willie Wong2025-10-20 18:25:27 +00:00Commented Oct 20 at 18:25
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$\newcommand\la\lambda$The answer is no. E.g., take $n=3$, $$A=\left( \begin{array}{ccc} -1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \\ \end{array} \right),\quad S=XX^\top,\quad X=\left( \begin{array}{ccc} -10 & 10 & -10 \\ 7 & 3 & -10 \\ -3 & -10 & 10 \\ \end{array} \right),$$ $k=2$, $i_1=1$, and $i_2=2$, so that $$\la_{i_1}(A)+\dots+\la_{i_k}(A) =\la_1(A)+\dots+\la_k(A)=-1+2=1>0,$$ whereas $$\la_{i_1}(SAS)+\dots+\la_{i_k}(SAS)\approx -44980+283<0.$$ So, there is no positive $\theta_{i_1,\dots,i_k}$ such that $$\la_{i_1}(SAS)+\dots+\la_{i_k}(SAS)=\theta_{i_1,\dots,i_k} (\la_{i_1}(A)+\dots+\la_{i_k}(A)).$$
