# Under what conditions a linear automorphism is an isometry of some norm?

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.

$T$ 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).

• It seems to me that the phenomenon you describe with the matrix $A$ will repeat itself whenever you have a non-diagonalisable endomorphism (over $\mathbb C$). – Loïc Teyssier Jun 13 '15 at 9:21
• It seems the following. I shall follow similarly to example which you propose and I extend the comment by Loïc Teyssier. There exists a non-degenerated matrix complex matrix $T$ such that $T^{-1}AT=J$, where $J$ is a Jordan form of the matrix $A$. Then $J=\Lambda+\Delta$, where $\Lambda$ is a diagonal matrix and $D=\|d_{ij}\|$ is a $0-1$-matrix with $d_{ij}=0$ when $j\ne i+1$. Let $n$ be an arbitrary natural number. Then $$J^n=(\Lambda+\Delta)^n=\Lambda^n+ n\Lambda^{n-1}\Delta+\frac {n(n-1)}2 n\Lambda^{n-2}\Delta^2+\dots.$$ – Alex Ravsky Jun 13 '15 at 11:57
• So if the matrix $\Lambda$ is non-degenerated and $\Delta\ne 0$ then the matrix $n\Lambda^{n-1}\Delta$ (and so the matrix $J^n$ too) gets unbounded when $n$ goes to infinity. – Alex Ravsky Jun 13 '15 at 11:58
• Also: iff all orbits of $A$ are bounded (that is, $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$, for any $x\in\mathbb{R}^n$) . – Pietro Majer Jun 13 '15 at 16:14
• Use e.g. the spectral radius formula (for both $A$ and $A^{-1}$) to prove that the spectrum of $A$ is in the unit circle. If $A$ is not diagonalizable (over $\mathbb{C}$ ) there is a vector $v\in \mathbb{C}^n$ such that $A^kv$ diverges (use the Jordan form), 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 w.r.to some norm on $\mathbb{R}^n$. – Pietro Majer Jun 13 '15 at 19:08

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.

3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in \mathbb{R}^n$).

4) $A$ is an isometry w.r.to some inner product.

$(1)\iff (4)$ 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 (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).

• Also: 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. So the following are equivalent: (i) $A$ is an isometry w.r.to some norm (ii) all $A$-orbits are bounded (iii) $A$ is diagonalizable over $\mathbb{C}$ with all eigenvalues of modulus $1$. – Pietro Majer Jun 14 '15 at 6:29