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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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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? – Yemon Choi Feb 7 '12 at 5:45
Page 71 won't come up for me on Google Books, either. – Ryan Budney Feb 7 '12 at 5:55
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. – user21199 Feb 7 '12 at 5:56
[add to the comment above: also applies the gauge function to the right-hand side of the majorization.] – user21199 Feb 7 '12 at 6:07
Dear uj, it is best if you edit the question to contain all the information. – Mariano Suárez-Alvarez Feb 7 '12 at 6:13
up vote 1 down vote accepted

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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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|. – user21199 Feb 7 '12 at 18:58
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? – Suvrit Feb 7 '12 at 23:17
Also, note that $x \prec y$ notation sorts its arguments into descending order when applying inequalities that define majorization. – Suvrit Feb 7 '12 at 23:19

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