In the second paragraph on Page 71 of the book Matrix Analysis by Bhatia, 1997, it says ``as a consequence of (III.12) we have Theorem III 4.4''. How can one get the inequality in Theorem III 4.4 from (III.12) for $\Phi\left(x_{1},\cdots,x_{n}\right)=\left|x_{1}\right|+\cdots+\left|x_{n}\right|$?
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2$\begingroup$ The link you have provided does not seem to work for me. Why not tell us, in your own words, what III.12 and III.4.4 say? $\endgroup$– Yemon ChoiCommented Feb 7, 2012 at 5:45
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$\begingroup$ Page 71 won't come up for me on Google Books, either. $\endgroup$– Ryan BudneyCommented Feb 7, 2012 at 5:55
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1$\begingroup$ Basically, (III.12) says that the difference of the eigenvalues of two matrices is majorized by the eigenvalue of the difference of the two matrices, and III 4.4 says if one applies any gauge function to the difference of the eigenvalues of two matrices, the inequality for the majorization still holds. $\endgroup$– user21199Commented Feb 7, 2012 at 5:56
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$\begingroup$ [add to the comment above: also applies the gauge function to the right-hand side of the majorization.] $\endgroup$– user21199Commented Feb 7, 2012 at 6:07
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10$\begingroup$ Dear uj, it is best if you edit the question to contain all the information. $\endgroup$– Mariano Suárez-ÁlvarezCommented Feb 7, 2012 at 6:13
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Inequality III.12 is the famous Lidskii majorization for Hermitian matrices $A$ and $B$, which says that
$$\lambda^\downarrow(A) - \lambda^\downarrow(B) \prec \lambda(A-B) \prec \lambda^\downarrow(A) - \lambda^\uparrow(B).$$
Now, recall the following simple but crucial fact:
Fact. $x \prec y \implies$ $|x|\quad \prec_w\quad |y|$, where $\prec_w$ denotes weak majorization, and $|x|$ denotes the vector obtained from $x$ by taking elementwise absolute values.
Now Problem II.5.11-(vi) asks you to prove that $x\ \prec_w\ y\ $ iff $\Phi(x) \le \Phi(y)$ for every symmetric gauge function $\Phi$. Once you have proved / believed this, the inequality that you allude to follows after invoking the abovementioned "fact".
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$\begingroup$ I have actually tried to use II.5.11, but there is a condition for x and y that is, they need to be in $R_{+}^{n}$. For instance, if x=(−1,−1), y=(1,-1), then |x| is not weak majorized by |y|. $\endgroup$ Commented Feb 7, 2012 at 18:58
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$\begingroup$ note that $x \prec y$ requires $\sum_i x_i = \sum_i y_i$. Also, the in the example that you give, $|x|=|y|$, so the majorization holds trivially....maybe you are missing something else? $\endgroup$– SuvritCommented Feb 7, 2012 at 23:17
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$\begingroup$ Also, note that $x \prec y$ notation sorts its arguments into descending order when applying inequalities that define majorization. $\endgroup$– SuvritCommented Feb 7, 2012 at 23:19