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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 $C$?

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Are $A,B$ and $C$ symmetric? In this case you can use Rayleigh quotient $\frac{x^tCx}{x^tx}$. –  Davide Giraudo Oct 9 '12 at 20:01
    
A, B, and C are positive definite, therefore, symmetric. Could you clarify on how using Rayleigh quotient to find smallest possible $\lambda (A) + \lambda (B)$ ? –  Ho Xuan Oct 9 '12 at 20:10

2 Answers 2

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.

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Thank you for the answer! I've just realized the first question $\lambda_{min} (A+B) > \lambda_{min} (A) + \lambda_{min} (B) $ can also be seen from Weyl's inequality. However, I am still looking for the answer about the smallest possible value of $\lambda_{min} (A) + \lambda_{min} (B) $ given a fixed $C = A + B$ . –  Ho Xuan Oct 9 '12 at 20:34
    
When $C=I$, we can take $A=\pmatrix{k^{-1}&0\\0&1-k^{—1}}$ and $B=\pmatrix{1-k^{-1}&0\\0&k^{—1}}$ to see that the minimum can be arbitrary small. –  Davide Giraudo Oct 9 '12 at 20:43
    
I think this is only true for certain cases of C. In general, the minimum value is dependent on C and not arbitrarily small. There should be some relationship between $v_{min} (A)$, $v_{min} (B)$, and $v_{min} (C)$ when the minimum value is attained. Here $v_{min}$ is the eigenvector associated with the smallest eigenvalue. Thanks –  Ho Xuan Oct 9 '12 at 21:16
    
So, what are the the cases you are interested in? –  Davide Giraudo Oct 10 '12 at 8:31
    
Hi, thanks for keep answering. I realized that the general case is difficult and there is no obvious connection among eigenvectors corresponding to smallest eigenvalues of $A$, $B$, and $C$. –  Ho Xuan Oct 11 '12 at 15:55

I answer the second part of the question. For $n=1$ the answer is $\lambda_{min}(C)$. For $n > 1$, the answer is, $\lambda_{min}(A) + \lambda_{min}(B)$ can be arbitrarily close to $0$.

Since $C$ is a symmetric positive definite matrix, we consider the singular value decomposition of $C$. $$ C = \sum_{i=1}^n \lambda_i v_i v_i^T,$$ where $0 < \lambda_1 \leq \dots \leq \lambda_n$ are eigenvalues of $C$, and $v_1, \dots, v_n$ is an orthonormal basis. For sufficiently small $\varepsilon > 0$, let \begin{align*} A &= (1-\varepsilon) \lambda_1 v_1 v_1^T + \varepsilon\lambda_2 v_2 v_2^T + \frac{1}{2}\sum_{i=3}^n \lambda_i v_i v_i^T,\\ B &= \varepsilon \lambda_1 v_1 v_1^T + (1-\varepsilon)\lambda_2 v_2 v_2^T + \frac{1}{2}\sum_{i=3}^n \lambda_i v_i v_i^T. \end{align*}

Now $A+B=C$, $A$ and $B$ are positive definite, and $$\lambda_{min}(A) + \lambda_{min}(B) \leq 2\varepsilon,$$ which can be arbitrarily close to $0$.

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