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

Let $s_1,s_2: \mathbb{R}^{2\times 2} \mapsto \mathbb{R}_+$, $s_{1}\left(\cdot\right)\ge s_{2}\left(\cdot\right)\ge 0$, be the singular values of a $2\times2$ matrix. Is it true that $$\left|s_{1}\left(M+N\right)-s_{1}\left(N\right)\right|+\left|s_{2}\left(M+N\right)-s_{2}\left(N\right)\right|\leq s_{1}\left(M\right)+s_{2}\left(M\right)$$ for any two $2\times2$ real matrices $M$ and $N$?

Thanks in advance for any helpful answers.

share|improve this question
    
The paper ams.org/notices/200102/fea-knutson.pdf is possibly relevant. –  Neil Strickland Oct 6 '11 at 8:43

1 Answer 1

Here is a more general result.

Let $A$ and $B$ be arbitrary $n \times n$ complex matrices. Then, we have the weak-majorization:

$$ |s(A) - s(B)|\quad \prec_w\quad s(A-B)$$

This result implies your alleged inequality as a special case.

The above result follows from a famous theorem of Lidskii, which states that for Hermitian matrices $A$ and $B$,

$$ \lambda^\downarrow(A) - \lambda^\downarrow(B) \prec \lambda(A-B),$$ where $\lambda^\downarrow(A)$ lists eigenvalues of $A$ is decreasing order (notice that here the majorization is strict)

For more details, see for example, Exercise IV.3.1 in Matrix Analysis by R. Bhatia.

Alternatively, you can have a look (for the singular value majorization result) at Theorem 3.4.5 in Topics in Matrix Analysis by Horn and Johnson.

share|improve this answer
    
Watch out, though, for the fact that Bhatia and Horn & Johnson use opposite conventions in the definiton of majorization. –  Mark Meckes Oct 6 '11 at 13:31
    
@Mark: Really? Seems like Def. 3.2.8 in HJ is the same as condition (II.2) in Bhatia's book (considering only weak majorization). –  Suvrit Oct 6 '11 at 13:57
    
Actually, I was thinking of Matrix Analysis by Horn and Johnson. It looks like they switched conventions between volumes. –  Mark Meckes Oct 11 '11 at 21:49

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.