Assume $V$ is a finite-dimensional vector space over $\mathbb{R}$, and $T: V \to V$ is a (linear) isomorphism. 



> 
> When is it possible to construct a norm on $V$ 
>  making $T$ an isometry?
> 
> 
> 


(Hopefully, I am looking for necessary & sufficient conditions $T$ should satisfy, i.e. a full characterization of the situation).


**What I have so far:**


A necessary condition: all the real eigenvalues of $T$ are of absolute value $1$. (Since $ T(v)=\lambda v \Rightarrow ||v||=||T(v)||=||\lambda v||=|\lambda| ||v||$ and an eigenvector $v$ must be nonzero).


**This condition is certainly not sufficient:**


For example look at $A$ = $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}: \mathbb{R}^2 \to \mathbb{R}^2$. It is an automorphism which has only one eigenvalue ($\lambda = 1$). However, $A\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+y \\ y \end{pmatrix}$, hence $A^n\begin{pmatrix} x \\ y \end{pmatrix}= \begin{pmatrix} x+ny \\ y \end{pmatrix}$. Now assume there exist a norm $||$ on $\mathbb{R}^2$ making $A$ an isometry. 


In particular $A$ must map the open unit ball $B$ to itself. By properties of norms $B$ must be a bounded open set containing the origin. (Note that since all the norms on a finite dimensional vector space are equivalent, the notions of boundedness and opennes are independent of the norm).


$B$ is open $\Rightarrow$ $\exists y>0$ such that $(0,y)\in B \Rightarrow A^n\begin{pmatrix} 0 \\ y \end{pmatrix}= \begin{pmatrix} ny \\ y \end{pmatrix} \in B$ for every $n \in \mathbb{N}$. This contradicts the boundedness of $B$ (w.r.t to the standard Euclidean norm for instance).


**Last remark:**


If we want to be more restrictive and require $T$ to be an isometry of some inner product, then the answer is [quite simple](https://math.stackexchange.com/a/1240323/104576).


$T$ preserves some inner product on $V$ if and only if $V$ admits a basis for which the matrix of $T$ is orthogonal (in other words the matrix of $T$ on an arbitrary basis is similar to an orthogonal matrix). The occurs if and only if the complexification of T is diagonalisable, and all its (complex) eigenvalues have absolute value 1.


**I am interested to know what additional potential isometries we can get when we allow more flexibility.** (That is we allow arbitrary norms, not just those that come from inner products).



Under what conditions a linear automorphism is an isometry of some norm?
For sake of completeness, I am writing a full answer based on the suggestion of
Pietro Majer.



> 
> **The following are equivalent:**
> 
> 
> 1) $A$ is an isometry w.r.to some norm.
> 
> 
> 2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1.
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> 
> 3) All orbits of $A$ are bounded ( $\sup\_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in \mathbb{R}^n$).
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> 
> 4) $A$ is an isometry w.r.to some inner product.
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> 
> 


**$(1)\iff (4)$ leads to an interesting point:** The union of all isometries of all norms equals the union of all isometries of all inner products. 


**Proof:**


$(1) \Rightarrow (2):$
Assume $||$ is a norm preserved by $A$. Then the operator norm of $A$ w.r.t to $||$ equals 1. Also $||A^n||\_{op}=1$. By the spectral radius formula: $\rho(A)=\lim\_{n\rightarrow\infty}||A^n||^{1/n}=1 $. The same argument for $A^{-1}$ implies $\rho(A^{-1})=1$. **This implies all the eigenvalues (including the complex ones) are of absolute value one.**


Also, it is easy to see that If $A$ is an isometry of the norm $| |\_1$ , $P∈GL(\mathbb{R^n})$, $P^{−1}AP$ is an isometry of the norm $||\_2$ where $||x||\_2=||Px||\_1$. Thus, the property that a given matrix admit such a norm is invariant under similarity. 


So it is enough to focus upon matrices of Jordan form. (which is available to us since we work over $\mathbb{C}$).


**Now assume $A$ is not diagonalizable**. By looking at Jordan form of a non-diagonalizable matrix, we can see there is a vector $v∈\mathbb{C}^n$ such that $||A^kv||\_{Euclidean}$ diverges. (Look at the example of $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}$ given in the question).


Since all the norms are equivalent This implies that $||A^kv||$ diverges. But since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $||A^kx||$ or $||A^ky||$ diverge. **This is impossible if $A$ is an isometry of $||$.**


$(2) \Rightarrow (4):
$ This is proved [here](https://math.stackexchange.com/a/1230213/104576) (The basic idea is to look at each Jordan block separately).


$(4) \Rightarrow (1):$ Obvious.


It remains to prove $(1) \iff (3)$:


$(3) \Rightarrow (1):$ If all orbits of $A$ are bounded, then we can renorm $\mathbb{R}^n$ by $\|x\|\_A:=\sup\_{k\in \mathbb{Z}} \|A^k x\|$, which is an $A$-invariant equivalent norm. 


$(1) \Rightarrow (3):$ This follows immediately by the fact all norms on a finite dimensional vector space are equivalent. The orbits of $A$ are all of constant norm ($\|x\|$) w.r.t to the norm $A$ preserves, hence are bounded. (w.r.t any other norm).


