Let $C = A + B$, where $A$, $B$, and $C$ are positive definite matrices. In addition, $C$ is fixed. Let $\lambda (A)$, $\lambda (B)$, and $\lambda (C)$ be smallest eigenvalues of $A$, $B$, and $C$, respectively. Is there any result about the smallest eigenvalues of $C$ in comparison with the sum of smallest eigenvalues of $A$ and $B$? Is it true that : $\lambda (A)$ + $\lambda (B)$ < $\lambda (C)$ ?
Moreover, what is the smallest possible value of $\lambda (A)$ + $\lambda (B)$ given a fixed $C$, and under what condition does this happen?
Many thanks!
Xuan
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Question about $\lambda_{min} (A+B) > \lambda_{min} (A) + \lambda_{min} (B) $ can be seen from Weyl's inequality.
The remaining question is about the smallest attainable value of $\lambda_{min} (A) + \lambda_{min} (B) $ given a fixed $C$?
Best Answer
If $S$ is a symmetric real matrix, then the minimal eigenvalue of $S$, $\lambda_{\min}(S)$ is given by $$\lambda_{\min}(S)=\min_{x\neq 0}\frac{x^tSx}{x^tx}.$$ Indeed, it's not hard to see that when $S$ is diagonal, and in the general case diagonalize $S$ in an orthonormal basis.
By the properties of $\min$, this gives $\lambda_{\min}(A+B)\geq \lambda_{\min}(A)+\lambda_{\min}(B)$. Moreover, if we have equality $\frac{x^tSx}{x^tx}$, it's for an eigenvector for $\lambda_{\min}$, hence we have equality if $A$ and $B$ have a common eigenvector.