I'm interested in calculating an operator norm of $T : \mathbb{R}^2 \to \mathbb{R}^2$, where $T(x,y)=(x+1, y)$, i.e. a translation by $1$ in direction $x$.
From wiki for bounded operator $$\left\lVert T \right\rVert_{op} = \inf \{ c \geq 0 : \lVert Tv \rVert \leq c \lVert v \rVert, \text{ for all } v \in V\}$$
But it could be equivalently defined as $$\lVert T \rVert_{op} = \sup\{\|Tv\| : v\in V \mbox{ with }\|v\| = 1\}$$
I'll further use a regular $\lVert \cdot \rVert_2$ on $\mathbb{R}^2$.
Lets define a sequence of points $a_n = (n, 0)$. $T$ is a shift-by-$1$ operator and we have $\lVert Ta_n \rVert_2 = n+1$, while $\lVert a_n \rVert_2 = n$. So it is obvious $$\lVert T \rVert_{op} \leq \inf\{\frac{n+1}{n} : n \in \mathbb{N}\} = 1$$
From the alternative definition take a point $(1, 0)$ it is shifted to $(2, 0)$, so $$\lVert T \rVert_{op} \geq \frac{\lVert T(1,0) \rVert_2}{\lVert (1,0) \rVert_2} = 2$$
Now, the (right) question, does this mean $T$ is not a bounded operator, despite the fact we use a matrix multiplication in homogeneous coordinates ? Or there is some other flaw in what am I doing?
I see, that $(0,0)$ is blown by $T$, so I'm confused.