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Given $\operatorname{tr}(X^TAX) > \operatorname{tr}(X^TBX)$ and $A$ and $B$ are p.s.d then under what conditions will we have $\lambda_{\max}(A)>\lambda_{\max}(B)$ to be guaranteed ? What are required conditions for $\lambda_{\min}(A)>\lambda_{\min}(B)$ ? All matrix entries are real valued and $X$ is a rectangular matrix. Thirdly, what are the conditions for second smallest eigenvalue (algebraic connectivity) of $A$ to be greater than the same for $B$? Also fourthly, what are the conditions for the eigenvalues of $A$ to majorize the eigenvalues of $B$ from above, below and so forth? And by majorization I mean this mathematical property: https://en.wikipedia.org/wiki/Majorization . And finally and most important of all for me..What are the conditions w.r.t $A(X)$ and $B(X)$ for $\operatorname{tr}(X^TA(X)X) > \operatorname{tr}(X^TB(X)X)$ to be true if $A(X)$ and $B(X)$ are functions of X and matrix valued as well?

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If $X$ is fixed there is little that could be said in general. If we have $$\text{tr}(X^TAX)>\text{tr}(X^TBX)$$ for all $X$, then $\lambda_k(A)>\lambda_k(B)$ for all $j$. This follows from $$ \lambda_k(A)=\min_{\dim K=k}\max\{x^TAx:\ x\in K,\ x^Tx=1\}. $$ For a fixed subspace $L$, we have $$\max\{x^TAx:\ x\in L,\ x^Tx=1\}>\max\{x^TBx:\ x\in L,\ x^Tx=1\},$$ since $x^TAx=\text{Tr}(X^TAX)>\text{tr}(X^TBX)=x^TBx$, where $X$ is the matrix with $x$ in the first column and zeroes elsewhere. Then $$ \lambda_k(B)=\min_{\dim K=k}\max\{x^TBx:\ x\in K,\ x^Tx=1\} <\max\{x^TAx:\ x\in L,\ x^Tx=1\}.$$ But now we can do this for any $L$ with $\dim L=k$, and so $\lambda_k(B)<\lambda_k(A).$


Regarding majorization: let $P(X)$ denote the projection onto the diagonal (or "pinching"), i.e. $P(X)$ is the matrix with diagonal $X_{11},\ldots,X_{nn}$ and zeroes elsewhere. Then we have, thanks to the Schur-Horn theorem:

The following statements are equivalent:

  • $\lambda(B)\prec\lambda(A)$

  • There exist unitaries $U,V$ such that $B=VP(UAU^*)V^*$.


Regarding your last question, "functions of $X$" is extremely vague, so I don't think that any conclusion can be drawn.

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