2
$\begingroup$

Equation 7 of this paper (Ramazan Türkmen, Zübeyde Ulukök, Inequalities for Singular Values of Positive Semidefinite Block Matrices, International Mathematical Forum, Vol. 6, 2011, no. 31, 1535 - 1545) says that it is well known that, given two complex-valued matrices $A$ and $B$, it holds

$$\sigma_j(A+B) \geq \sigma_j(A)+\lambda_n(B)$$

where $\sigma_j$ is the $j$-th singular value and $\lambda_n$ is the $n$-th eigenvalue

  1. Is this correct? Do we need some hypotheses on $A$, $B$?

  2. Can somebody point me to a reference?

Thank you.

Improve this question
$\endgroup$
7
  • 4
    $\begingroup$ Seems a little tough if $B=-A$. $\endgroup$
    – Anthony Quas
    Commented Nov 17, 2020 at 16:20
  • $\begingroup$ I see. But in this case $\sigma_j(0)=0\geq \sigma_j(A)-\lambda_1(A)$. I think this is possible. $\endgroup$
    – Gabriele Oliva
    Commented Nov 17, 2020 at 16:28
  • 3
    $\begingroup$ No. Take $B$ the identity and $A=-B$. Your inequality reads $0 \geq 2$. $\endgroup$
    – Mikael de la Salle
    Commented Nov 17, 2020 at 17:16
  • 4
    $\begingroup$ I think the title of paper gives a hint that the matrices are expected to be positive semi definite. $\endgroup$
    – Anthony Quas
    Commented Nov 17, 2020 at 19:45
  • 1
    $\begingroup$ That's Weyl inequality when $A$ and $B$ are hermitian. $\endgroup$
    – Narutaka OZAWA
    Commented Nov 18, 2020 at 0:31

1 Answer 1

Reset to default
2
$\begingroup$

Jumping in many years later, but I believe there's a typo in the paper which originates from Zhang's 1999 matrix theory book.

The first clue that something is wrong is that $\lambda_n(B)$ for an arbitrary complex matrix $B$ may not be a real number, and so the inequality as a whole doesn't make any sense.

After going through the proof of Theorem 7.11 in Zhang's book (which is referenced in the paper you linked for this inequality), I believe that $\lambda_n(B)$ actually means $\lambda_n(\mathcal{H}(B))$, where $$\begin{align} \mathcal{H}(B) &= \begin{bmatrix} 0 & B\\ B^* & 0 \end{bmatrix} \end{align}$$ whence $\lambda_n(\mathcal{H}(B)) = -\sigma_1(B)$, and so the desired inequality actually reads $$ \sigma_j(A + B) \ge \sigma_j(A) - \sigma_1(B) $$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .