I am asked to show for which vectors the inequality $||Ax|| \le ||x||||A|| $ is an equality.
My intuition tells me that this happens when $x$ is in the direction of the right singular vector corresponding to the largest singular value of A. Because, then that happens, given the SVD of A as $A=U\Sigma V^*$, then we can say that $Av_1=\sigma_1u_1$, where the subscript 1 on $u$ and $v$, denotes that it is the vector corresponding to the first column of $U$ and $V$, respectively.
Because, we can take the norm of both sides and get the following:
$||Av_1||=||\sigma_1u_1||=\sigma_1||u_1||=||A||||u_1||$,
and, since $U$ and $V$ have orthonormal columns, we know that $||u_1||=||v_1||=1$, which means we can substitute and get :
$||Av_1||=||A||||v_1||$
And hence this is an equality when the vector $x$ is the right singular vector corresponding to the largest singular value.
Question: is the above logic correct? The reason I am unsure, is because if $||Av_1||=||A||||v_1||$, and $||v_1||$ has norm 1, then that would mean $||Av_1||=||A||$.
thanks.