# Norm-preserving map is linear

How can one show that a norm-preserving map $T: X \rightarrow X'$ where $X,X'$ are vector spaces and $T(0) = 0$ is linear? Thanks in advance.

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Could you clarify what you mean by norm-preserving? Note that if $T$ is not linear, then $|Tx| = |x|$ does not mean that $|Tx - Ty| = |x - y|$, which is what is usually meant (and as noted below, your claim is false if you only require $|Tx| = |x|$ and true in many cases if you actually meant distance-preserving). – Ben Millwood May 17 '12 at 11:30
– t.b. May 17 '12 at 12:03

The claim is false. For example, take $X=X'=\mathbb{R}$, and $T(x)=|x|$ for all $x$.

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I think, for nonlinear maps, "norm preserving" has to mean $\|T(x)-T(y)\| = \|x-y\|$. – GEdgar May 17 '12 at 12:46
@GEdgar, the OP assumes also $T(0)=0$ and so $\|T(x)-T(y)\| = \|x-y\|$ implies $\|T(x)\| = \|x\|$. – lhf May 17 '12 at 13:44
But not conversely. – GEdgar May 17 '12 at 14:30

This claim is true if your map $T$ is surjective and an isometry. In this case simply apply Mazur-Ulam theorem.

If we require that map to be surjective, but only norm-preserving, then we can construct counterexample $$T:\mathbb{C}\to\mathbb{C}:z\mapsto z e^{i |z|}$$ where $\mathbb{C}$ is consideres as vector spaces over $\mathbb{R}$.

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Hold on, the link you gave refers to isometries, but there are functions that are norm-preserving but not isometries (e.g. the norm function itself). Are you sure the theorem applies? – Ben Millwood May 17 '12 at 11:11
@benmachine So what is yours definition of norm-preserving? – Norbert May 17 '12 at 11:13
$|f(x)| = |x|$ for all $x$. Since $f(x)-f(y)$ need not be $f(x-y)$, this need not mean that $|f(x) - f(y)| = |x - y|$. – Ben Millwood May 17 '12 at 11:16
Oh, in this case you are right. Then even for surjective maps we can construct counterexamples. – Norbert May 17 '12 at 11:17
Yeah, on second thoughts, norm-preserving as I phrased it is a pretty silly condition. I can't imagine the OP doesn't mean isometry. – Ben Millwood May 17 '12 at 11:25