# Characterisation of inner products preserved by an automorphism

Let $V$ be a finite dimensional vector space. Let us call an automorphism $T:V\rightarrow V$ admissible if there exists an inner product $\langle , \rangle$ on $V$ making $T$ an isometry. (You can see this question, where it is proved that $T$ preserves some inner product on $V$ if and only if $V$ admits a basis for which the representing matrix of $T$ is orthogonal)

My question: Assume $T$ is an admissible automorphism. "How many" inner products exists which are preserved by $T$?

I think I might have a solution,but I am not entirely sure. Also, there might be a shorter way of analyzing this problem, and I would like to see other perspectives. Here is what I have got so far: (A summary of the results shows below).

Some preliminary observations:

(a) It depends on the $T$. If $T=\pm Id$ then it preserves any inner product on $V$, but it's clearly not true for a rotation (see c) below).

(b) For any preserved inner product, any scalar multiple of it is also preserved.

(c) A (two-dimensional) proper rotation uniquely determines (up to scalar multiple) the preserved inner product. (a proof is provided at the end).

(d) The reflection w.r.t to $0$ ($T=-Id$) preserves any inner product.

A more detailed analysis:

By what I mentioned above, there exists a basis for $V$ w.r.t it $T$ has the form of an orthogonal matrix, and every such matrix has a canonical form. So. let $B=(v_1,...,v_n)$ be a suitable basis w.r.t it the matrix of $T$ has this canonical form.

We can obtain one preserved inner product by setting this basis to be orthonormal. Denote by $\langle, \rangle$ this inner product.

(1) Now note we can change freely the norm of every $v_i$ which corresponds to an eigenvalue $\pm 1$, and $T$ would still be an isometry.

(2) Also, if $v_1,v_2$ are eigenvectors which correspond to eigenvalues $\pm 1$ respectively, then: the requirement $\langle T(v_1),T(v_2) \rangle = \langle v_1,v_2 \rangle \Rightarrow \langle v_1,-v_2 \rangle = \langle v_1,v_2 \rangle \Rightarrow \langle v_1,v_2 \rangle = 0$.

Hence, $v_1,v_2$ must be orthogonal.

(3) Now let $v_1,v_2$ correspond to a non-real eigenvalue (i.e the restriction of $T$ to span{$v_1,v_2$} is a rotation $R_{\theta}$), and let $v_3$ correspond to eigenvalue $\pm 1$. Then it forces:

$\langle v_1,v_3 \rangle = \langle T(v_1),T(v_3) \rangle = \langle R_{\theta}(v_1),\pm v_3 \rangle \Rightarrow \langle v_1 \pm R_{\theta}(v_1),v_3 \rangle = 0$. Similarly, $\langle v_2 \pm R_{\theta}(v_2), v_3 \rangle = 0$.

Claim: $v_1 \pm R_{\theta}(v_1), v_2 \pm R_{\theta}(v_2)$ are linearly independent.

Proof: First note that since $-R_{\theta}=R_{\pi + \theta}$, then it is enough to prove this for the $+$ case. Assume $\alpha(v_1-R_{\theta}(v_1))+ \beta(v_2-R_{\theta}(v_2))=0$, s.t $(\alpha , \beta) \neq (0,0)$.

Since $\theta \neq 0, v_i \neq R_{\theta}(v_i)$, so $\alpha = 0 \iff \beta = 0$, hence both are nonzero, so we can assume $\alpha=1$. This implies: $(cos\theta v_1+sin\theta v_2 - v_1)+\beta (-sin \theta v_1 + cos \theta v_2 - v_2) = 0 \Rightarrow$

$cos\theta - 1 = \beta sin \theta , \beta (1-cos \theta ) = sin \theta$.

Putting the two equations together we get: $(cos \theta-1)^2=-sin^2 \theta \Rightarrow 2(1-cos \theta) = 0 \Rightarrow cos \theta = 0 \Rightarrow \theta = 0$. (But this is a degenerate case of rotation, contrary to our assumption about $v_1,v_2$).

Corollary: span{$v_1 \pm R_{\theta}(v_1), v_2 \pm R_{\theta}(v_2)$} $=$ span{$v_1,v_2$}. Hence, $v_3$ is in fact orthogonal to $v_1,v_2$.

Summary:

(2)&(3) together show that vectors from our chosen basis $B$ which correspond to different eigenvalues must remain orthogonal w.r.t every inner product which is preserved by $T$.

Statement $(c)$ then shows that the restriction of the inner product to any two-dimensional subspace where the restriction of $T$ is a rotation is determined uniquely.

Finally note that the restriction of $T$ to the eigenspaces of $1,-1$ are $Id,-Id$ hence according to observation (a) a preserved inner product does not have any constraints on these subspaces separately. (They still have to be mutually orthogonal).

Proof of statement (c): Take $T=\begin{pmatrix} cos\theta & -sin\theta \\\ sin\theta & cos\theta \end{pmatrix}: \mathbb{R}^2 \to \mathbb{R}^2$

If $\langle , \rangle$ is an inner product on $\mathbb{R}^2$ making $T$ an isometry, then: $T(e_1)= cos\theta e_1 + sin \theta e_2, T(e_2)=-sin\theta e_1 + cos \theta e_2$, hence:

$\lVert e_1 \rVert^2= \lVert T(e_1) \rVert^2= cos^2\theta \lVert e_1 \rVert^2 + sin^2\theta \lVert e_2 \rVert^2 + 2sin\theta cos\theta \langle e_1,e_2 \rangle \Rightarrow$

(*) $sin \theta \lVert e_1 \rVert^2 = sin\theta \lVert e_2 \rVert^2 + 2cos \theta \langle e_1,e_2 \rangle$

Similarly $\langle e_1,e_2 \rangle=\langle T(e_1),T(e_2) \rangle=sin\theta cos\theta (\lVert e_2 \rVert^2 - \lVert e_1 \rVert^2)+ \langle e_1,e_2 \rangle(cos^2\theta-sin^2\theta) \Rightarrow$

(**) $2sin\theta \langle e_1,e_2 \rangle = cos\theta (\lVert e_2 \rVert^2 - \lVert e_1 \rVert^2)$

Now using equations (*),(**) togetther we deduce:

$sin \theta \lVert e_1 \rVert^2 = sin\theta \lVert e_2 \rVert^2 + \frac{cos^2\theta}{sin\theta} (\lVert e_2 \rVert^2 - \lVert e_1 \rVert^2)$ so finally we obtain: $\lVert e_2 \rVert = \lVert e_1 \rVert$.

Now (**) implies $sin\theta \langle e_1,e_2 \rangle = 0$, so if our rotation is proper (not the identity or reflection w.r.t to the origin) then it must hold that $\langle e_1,e_2 \rangle = 0$.

So the inner product by a rotation in $\mathbb{R}^2$ must be a scalar multiple of the standard inner product.

In particular the isometry group of an inner product determines it completely. (Up to scalar multiple of course).