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Let us suppose that $A_{n}$ and $B_n$ are sequences of positive definite matrices satisfying

$c\leq \lambda_{\min}(A_n)\leq \lambda_{\max}(A_n)\leq C$

and

$c\leq \lambda_{\min}(B_n)\leq \lambda_{\max}(B_n)\leq C$

where $\lambda_{\min}$ and $\lambda_{\max}$ are minimal and maximal eigenvalues. Then, what's the relationship between $||A_n-B_n||_2$ and $||A_n^2-B_n^2||_2$? Is it true that $||A_n-B_n||_2 \leq \mathrm{constant}*||A_n^2-B_n^2||_2$, where the $\mathrm{L}_2$ norm of a matrix $M$ is the maximum of the absolute value of its minimal eigenvalue and the absolute value of its maximal eigenvalue.

edited Dec 3 at 21:22

J. H. S.
2,415●12●31

asked Dec 3 at 19:38

Jefflee
113●4

Denis Serre · Accepted Answer · 2012-12-04 07:23:20Z UTC

3

The norm you consider is usually called operator norm, or the norm subordinated to the $\ell^2$ norm over ${\mathbb C}^n$ (or ${\mathbb R}_n$). The correct inequality for positive Hermitian matrices $A$ an $B$ is $$\|A-B\|_2\le\sqrt{\|A^2-B^2\|_2}.$$ See Exercise 110 of my additional list accompagnying my book Matrices, GTM 216 (Springer-Verlag).

answered Dec 4 at 7:23

Denis Serre
18.7k●1●31●93

1

"The norm you consider is usually called ..." or the spectral norm, or the norm induced by the $\ell^2$ norm, or... – Mark Meckes Dec 4 at 14:57

1

And of course strictly speaking the OP's definition of that norm is only correct for Hermitian matrices. – Mark Meckes Dec 4 at 14:58

relationship between eigenvalues of (A-B) and eigenvalues of (A^2-B^2)

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