# 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).