Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In 1971, Thomson and Freede generalized the Lidskii-Wielandt inequalites as follows (version for singular values)

Let $A$, $B$ be $n\times n$ Hermitian matrices. Suppose $\alpha_1\geq \alpha_2 \geq \cdots \geq \alpha_n, \beta_1\geq \beta_2 \geq \cdots \geq \beta_n, \gamma_1\geq \gamma_2 \geq \cdots \geq_n$ are the singular values of $A$,$B$ and $A+B$ respectively. Let $i_1,\cdots,i_m$ and $j_1, \cdots,j_m$ be integers such that $i_m+j_m\leq m+n$ and $1 \leq i_1 < \cdots < i_m \leq n, 1 \leq j_1< \cdots < j_m \leq n$. Then one has $$\sum_{s=1}^m \gamma_{i_s+j_s-s}\leq \sum_{s=1}^m \alpha_{i_s} +\sum_{s=1}^m \beta_{j_s}.$$

Now, my question is : if we define two new vectors $\hat{\gamma}$ and $\hat{\alpha}$ by $\hat{\gamma}_i=\max(\gamma_i,\alpha_i)$ and $\hat{\alpha}_i=\min(\gamma_i,\alpha_i)$ for $i=1,\cdots,n$, does the the following inequality which is similar to the above one still hold? $$\sum_{s=1}^m \hat{\gamma}_{i_s+j_s-s}\leq \sum_{s=1}^m \hat{\alpha}_{i_s}+\sum_{s=1}^m \beta_{j_s}.$$

This is true for the simpler case by considering the Lidskii-Wielandt inequality $$\sum_{s=1}^m (\hat{\gamma}_{i_s} - \hat{\alpha}_{i_s}) = \sum_{s=1}^m \vert\gamma_{i_s}-\alpha_{i_s}\vert \leq \sum_{s=1}^m \beta_s.$$ For the general case, I use the Matlab to generate many random matrices to check and it seems that it is correct. However, I have no idea about how to prove or disprove it. Can anyone help me?

share|improve this question
The inequality that you call Lidskii-Wielandt is incorrect. It contains the inequality $|\gamma_i-\alpha_j|\le\beta_1$. There is no reason why $B$ be non-negative. Take $B$ such that $\beta_1$ is negative and you have a contradiction. –  Denis Serre Dec 28 '10 at 20:28
@Denis: $\alpha_j,\beta_j,\gamma_j$ are singular values, not eigenvalues, so they are all nonnegative. –  Mark Meckes Dec 29 '10 at 3:39
Yes, such the inequality doesn't hold for eigenvalues. You can consider a well-know result, that is $\vert \sigma(X)-\sigma(Y) \vert$ is weakly majorized by $\sigma(X-Y)$, wherer $\sigma(X)$ deontes the vector of singular values of $X$ arranged in the decreasing order. This inequality just means that $\sigma_i(X)$ and $\simga_i(Y)$ can be exchanged in the inequality due to the absolute value. This is just a special case of the Likskii-Wielandt inequality for singular values. My question is about whether the generalized Likskii-Widlandt still have such "exchange" property. –  user11870 Dec 29 '10 at 7:32
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.