Understanding operator norms $A=\left(\begin{array}{cc}
                     0 & 1 & 0\\
                     0 & 0 & 1\\
                     0 & 0 & 0\end{array}\right)$
Please help explain the fault in the following reasoning:
Let $x \in \mathbb{R}^3$, write $k = \frac{x}{\|x\|}$
Then $\|A\| = max_k\|Ak\|=\|e_1k_2 + e_2k_3\|=2$
The given answer is $1$, where is the mistake?
 A: Perhaps this will help: note that
$$
\|k_2e_1 + k_3 e_2\|^2 = 
k_2^2 + k_3^2 \leq
k_1^2 + k_2^2 + k_3^2 =
\|k\|^2
$$
A: The operator norm is defined (equivalently) as any of :
$\|A\| = \max_{\|x\|=1}\|Ax\|$
$\|A\| = \max_{\|x\|\leq1}\|Ax\|$
$\|A\| = \max_{\|x\|\neq0}\frac{\|Ax\|}{\|x\|}$
Above, you test against $k = (0,1,1)$ (I'm not sure what you mean by $e_1k_2+e_2k_3)$ - you're multiplying vectors here, which makes little sense in context, and none at all without specifying if you are using the cross or dot product), which has (Euclidian) norm $\sqrt{2}$, so if you wish to use it to test your operator norm, you need to use the third formula above, and $\frac{\|Ak\|_2}{\|k\|_2} = \frac{\|(1,1,0)\|_2}{\|(0,1,1)\|_2} = \frac{\sqrt{2}}{\sqrt{2}} = 1$, so you have a witness that $\|A\| \geq 1$. In order to prove that $\|A\| = 1$, you now need to show that $\|Ax\|_2 \leq \|x\|_2$ for all $x \in\mathbb{R}^3$ (or possibly faster, using the first formula, $\|Ax\|_2 \leq 1$ for all $x \in \mathbb{R}^3$ with $\|x\|_2 = 1$.
(NB: Throughout (except in the definition of the operator norm, which applies on any normed linear space) I've made the use of the Euclidian norm explicit by denoting it $\|\cdot\|_2$ - this is not actually important).
