A singular value inequality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:35:53Z http://mathoverflow.net/feeds/question/77337 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/77337/a-singular-value-inequality A singular value inequality unknown (yahoo) 2011-10-06T07:34:56Z 2011-10-06T12:33:45Z <p>Let <code>$s_1,s_2: \mathbb{R}^{2\times 2} \mapsto \mathbb{R}_+$</code>, $s_{1}\left(\cdot\right)\ge s_{2}\left(\cdot\right)\ge 0$, 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 $N$?</p> <p>Thanks in advance for any helpful answers.</p> http://mathoverflow.net/questions/77337/a-singular-value-inequality/77340#77340 Answer by S. Sra for A singular value inequality S. Sra 2011-10-06T08:46:10Z 2011-10-06T12:33:45Z <p>Here is a more general result.</p> <blockquote> <p>Let $A$ and $B$ be arbitrary $n \times n$ complex matrices. Then, we have the <a href="http://en.wikipedia.org/wiki/Majorization" rel="nofollow">weak-majorization</a>:</p> </blockquote> <p>$$|s(A) - s(B)|\quad \prec_w\quad s(A-B)$$</p> <p>This result implies your alleged inequality as a special case. </p> <p>The above result follows from a famous theorem of <em>Lidskii</em>, which states that for Hermitian matrices $A$ and $B$,</p> <p>$$\lambda^\downarrow(A) - \lambda^\downarrow(B) \prec \lambda(A-B),$$ where $\lambda^\downarrow(A)$ lists eigenvalues of $A$ is decreasing order (notice that here the majorization is <em>strict</em>)</p> <p>For more details, see for example, Exercise IV.3.1 in <em>Matrix Analysis</em> by R. Bhatia.</p> <p>Alternatively, you can have a look (for the singular value majorization result) at Theorem 3.4.5 in <em>Topics in Matrix Analysis</em> by Horn and Johnson.</p>