# Linear isometry and operator norm $=1$

For some reason I used to think that if $T$ is a linear operator on normed spaces $V \to W$ then saying $T$ is an isometry is the same as saying $\|T\|_{op} = 1$.

Well, I got stuck on a proof and subsequently looked up the definition and realised that the definition of isometry is that a linear operator $T$ is an isometry if $\|Tx\| = \|x\|$.

Now I've been wondering whether we have that $\|T\|_{op} = 1$ implies $T$ is an isometry?

The other direction holds: if $\|Tx\| = \|x\|$ for all $x$ then $\frac{\|Tx\|}{\|x\|} = 1$ for all $x \neq 0$ and hence $\sup_{\|x\|=1}\|Tx\| = \sup_{\|x\|=1} \frac{\|Tx\|}{\|x\|} = \|T\| = 1$.

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Already in two dimensions you'll find counter-examples. –  paul garrett Jul 31 '12 at 18:02
$\|T\|=1$ does not imply that $T$ is injective. –  Martin Sleziak Jul 31 '12 at 18:18
@MartinSleziak Ok, thanks for the hint. And is the proof in the other direction I gave, correct? –  Matt N. Jul 31 '12 at 18:22
Already for diagonal matrices you'll find counterexamples. –  Qiaochu Yuan Jul 31 '12 at 18:23
@Matt N.: Yes, the proof of the other direction you gave is correct. –  Jonas Meyer Jul 31 '12 at 18:25
For example, on a finite dimensional space, $\|T\|=1$ if and only if for all $x$, $\|Tx\|\leq \|x\|$, and there exists $x_0\neq 0$ such that $\|Tx_0\|=\|x_0\|$. On the other hand, for $T$ to be an isometry requires for all $x$, $\|Tx\|=\|x\|$. You can give many examples of the former that do not satisfy the latter if the dimension is greater than $1$, as paul garrett commented, even diagonal matrices as Qiaochu Yuan commented. The former condition even allows $T$ to have nontrivial kernel as Martin Sleziak commented.
As David Mitra commented, if $T\neq 0$ then $\dfrac{T}{\|T\|}$ has norm $1$, and note that if there exist nonzero vectors $x$ and $y$ such $\dfrac{\|Tx\|}{\|x\|}\neq \dfrac{\|Ty\|}{\|y\|}$, then $T$ is not a multiple of an isometry. For example, if $T$ is nonzero and noninjective, take $x\in\ker T$ and $y\not\in\ker T$.