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Let $A$ and $B$ be two $n \times n$ (symmetric) positive definite matrices, and denote the $k$th smallest eigenvalue of a general $n \times n$ matrix by $\lambda_k(X)$, $k = 1, 2, \ldots, n$ so that $$\lambda_1(X) \leq \lambda_2(X) \leq \cdots \leq \lambda_n(X).$$ I guess the following relation holds: $$\lambda_k(A + B) > \max\{\lambda_k(A), \lambda_k(B)\}, \; k = 1, 2, \ldots, n.$$

This looks intuitive but I have difficulty to prove it, any hints?

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  • $\begingroup$ How do you order eigenvalues? $\endgroup$ – Pierre-Guy Plamondon Jan 22 '16 at 18:18
  • $\begingroup$ It doesn't matter. I am trying to say any eigenvalue of the sum should be greater than the corresponding ones. To be more precise, I will edit. $\endgroup$ – Zhanxiong Jan 22 '16 at 18:20
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For symmetric matrices you have the Courant-Fischer min-max Theorem: $$\lambda_k(A) = \min \{ \max \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=k \}$$ with $$R_A(x) = \frac{(Ax, x)}{(x,x)}.$$ Now, your assertion follows easily, since $R_{A+B}(x) > \max\{R_A(x), R_B(x)\}$.

This theorem is also helpful to prove other nice properties of the eigenvalues of symmetric matrices. For example: \begin{equation*} \lambda_k(A) + \lambda_1(B) \le \lambda_k(A+B) \le \lambda_k(A) + \lambda_n(B) \end{equation*} This shows the continuous dependence of the Eigenvalues on the entries of the matrix, and also your assertion.

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  • $\begingroup$ Does this result also hold for sum of semi definite positive matrices? (changing the > for ≥ ) $\endgroup$ – Manuel Oct 29 '18 at 21:51
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    $\begingroup$ Yes and it's the same proof. $\endgroup$ – gerw Oct 30 '18 at 8:44
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    $\begingroup$ @grew : your last equation is not inline with the question. OP is considering $\lambda_n$ as the lowest eigen value and $\lambda_1$ as highest. But you seem to consider the other way. $\endgroup$ – Rajesh D Mar 19 '19 at 8:58
  • $\begingroup$ @RajeshDachiraju You are right! Now the question has been edited, so that question and answer share the same notation. $\endgroup$ – FormulaWriter Aug 14 '20 at 19:54
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    $\begingroup$ @user777: Yes, this formula only needs the symmetry of $A$ and $B$. $\endgroup$ – gerw May 13 at 19:54

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