10
$\begingroup$

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

$$\|A_n-B_n\|_2 \leq \text{constant} \cdot \|A_n^2-B_n^2\|_2$$

where the $2$-norm of a matrix $M$ is the maximum of the absolute value of its minimal eigenvalue and the absolute value of its maximal eigenvalue.

$\endgroup$
0

1 Answer 1

5
$\begingroup$

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).

$\endgroup$
2
  • 1
    $\begingroup$ "The norm you consider is usually called ..." or the spectral norm, or the norm induced by the $\ell^2$ norm, or... $\endgroup$ Dec 4, 2012 at 14:57
  • 1
    $\begingroup$ And of course strictly speaking the OP's definition of that norm is only correct for Hermitian matrices. $\endgroup$ Dec 4, 2012 at 14:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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