Prove that $||A||_2 = \max_{||x||_2=1, ||y||_2 = 1} |y^TAx|$

inner-productslinear algebranormed-spaces

Prove that $||A||_2 = \max_{||x||_2=1, ||y||_2 = 1} |y^TAx|$.

Definitions:

$$ ||A||_2=\max_{||x||_2=1}||Ax||_2=\sqrt{\rho(A^T A)}.$$

My attempt:

  • $$||A||_2 = \max_{||x||_2=1}||Ax||_2 = \max_{||x||_2=1} \left( \sum_i \left(\sum_j a_{ij}x_j\right)^2\right)^{1/2},$$ and $\sum_j a_{ij} x_j$ can be seen as the inner product of the $i$th row of $A$ and $x$. The $i$th row of $A$ is $e_i^TA$, but from here on I have trouble writing the steps down. I want to find that $||A||_2 \le \max_{||x||_2 = 1} |e_i^T A x| $, as this would imply that $ ||A||_2\le \max_{||x||_2=1, ||y||_2 = 1} |y^TAx|$ (since the 2-norm of $e_i$ is 1, and the maximum is considered on a bigger set).

  • For the other inequality I tried a similar technique: $$ \max_{||x||_2=1, ||y||_2=1} |y^T A x| \ge \max_{||x||_2=1} |e_i^T A x|.$$ But I can't really find a lower bound.

How I can prove the equality?

Thanks.

Best Answer

  1. For any unit vectors $x$ and $y$ we have $$ |y^T Ax| \le \|y\|_2 \|Ax\|_2 \le \|A\|_2, $$ using Cauchy-Schwarz inequality and the definition of the operator norm.

  2. Notice that if $\rho(A^T A) = 0$ there is nothing to prove. Assume $\sigma = \sqrt{\rho(A^T A)} > 0$ from now on.

    Let $x$ be an unit eigenvector of $A^T A$ to the eigenvalue $\sigma^2$, that is $$ A^T A x = \sigma^2 x. $$ Further, let $y = \frac1\sigma A x$. Then, we have $\|y\|_2 = 1$ and $y^T A x = \sigma$.

Related Question